Doc Brown's Chemistry

September 3, 2017 | Author: hirenpanchani | Category: Chemical Compounds, Molecules, Distillation, Chemical Bond, Chemical Elements
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chapter 1,2,3...

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FORMULAE, EQUATIONS AND AMOUNTS OF SUBSTANCE ELEMENTS, COMPOUNDS, MIXTURE separation, CHEMICAL REACTIONS & EQUATIONS Section 1.1 Introduction and Some keywords (see also pictures) 1.1a ATOM

(i) Even as far back as ancient Greece ~500BC philosophers had considered the concept of what would be formed on continuously dividing matter i.e. what was the smallest 'bit' left of any substance. In 1808 the English scientist-chemist Dalton proposed his 'atomic theory' - that all matter was made up of tiny individual units called atoms which could NOT be subdivided into simpler substances. What is more, he proposed the idea that there were different types of atoms which we now call 'elements' and combinations of them produce all the different substances which exist. The different types of atoms are called elements (examples below). An ATOM is the smallest particle of a substance which can have its own characteristic properties AND cannot be split into simpler substances. (ii) BUT, remember atoms are built up of even more fundamental sub-atomic particles - the electron, proton and neutron. The centre of an atom, called the nucleus, consists of proton and neutron particles and the electrons move around the nucleus in 'orbital' energy levels. For more details see the Atomic Structure Notes.

1.1b MOLECULES and their representation

A MOLECULE is a larger particle formed by the chemical combination of two or more atoms. The molecule may be an element e.g. hydrogen formula H2 (H-H, two atoms combined, all atoms the same) or a compound (more examples below) e.g. carbon dioxide formula CO2 (O=C=O, three atoms combined) and in each case the atoms are held together by chemical bonds. (detailed GCSE bonding notes and examples) You can represent molecule in various styles of diagram. For example, you can colour and size code the atoms of different elements, so in the molecule pictured on the upper left, you can tell there are five types of atom (elements) and six atoms in total in the molecule. The second molecule (lower left) shows the molecular structure of ethanol ('alcohol') which consists of two carbon atoms, six hydrogen atoms and one oxygen atoms. You will also come across shorthand versions of this diagrammatic style written like CH3CH2OH and even shorter C2H5OH but using these styles requires much more experience than is required when first learning the basic principles of chemistry. Why they are combined in this particular number and order depends primarily on an atoms combining power (its valency) an advanced concept dealt with in Part 3 equations, formula and valency. There are also styles to give a much greater '3D' impression of the shape of a molecule and they attempt to show the '3D' spatial arrangement of the atoms in a molecule and how the bonds connect them together.

On the left is the displayed formula of ethane, which shows all the atoms and individual bonds. Below left is stylised '3D' version of the displayed formula and you can see how it fits in with the 'ball and stick' (below it) and 'space-filling' model (below right) diagrams. Below are three '3D' representations of the hydrocarbon molecule called ethane.

The lower image is an example of a 'ball-and-stick' diagram and shows the chemical bonds which hold the atoms together. 1.1c ELEMENT and symbols

(i) Basic definition   

H I Th Er Ho W Ar U? Element Symbolname quizzes: easier-pictorial! or harder-no pictures!



 

CHEMICAL BOND

An ELEMENT is a pure substance made up of only one type of atom, 92 occur naturally and can be 'summarised' in the Periodic Table (detailed notes) i.e. from element 1 hydrogen H to 92. uranium U. Note that each element has symbol which is a single capital letter like H or U or a capital letter + small letter e.g. cobalt Co, calcium Ca or sodium Na. Each element has its own unique set of properties but the Periodic Table is a means of grouping similar elements together. They may exist as atoms like the Noble Gases e.g. helium He or as molecules e.g. hydrogen H2 or sulphur S8. (more examples applied to equations and see note about 'formula of elements')

(ii) Extended ideas

Metals and nonmetals

1.1d

is an example of a 'space-filling' diagram which gives a more accurate representation of the space the molecule actually occupies.



*At a higher level of thinking, all the atoms of the same element, have the same atomic or proton number. This number determines how many electrons the atom has, and so ultimately its chemistry. Any atom with 27 protons and electrons is cobalt! The diagram 60-Co-27 uses advanced notation - all explained on the atomic structure page. See also picture diagrams of elements/compounds and mixtures. Elements are broadly divided by physical and chemical character into metals and nonmetals. A few elements display characteristics of metallic and non-metallic elements and are referred to as semi-metals or metalloids. Elements can be more highly characterised and organised in the form of the Periodic Table. All of these points are discussed on the GCSE Periodic Table notes page.

The variety of chemical substances around you are all due to different combinations of atoms. Atoms combine or 'connect' together by means of chemical bonds of which there are various types, but all chemical bonds are based on the attraction of oppositely charged particles, i.e. the natural attraction of positive and negative particles - a fundamental law of physics!

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1.1e COMPOUNDS, FORMULA and MOLECULE

    

The result is that millions of different substances (molecules/compounds) can exist because of the huge variety of atom combinations possible. KS3 students do NOT need to know about chemical bonding except the idea that they exist - otherwise, what holds atoms together? Detailed GCSE bonding notes and examples for ionic, covalent and metallic bonds, o and includes a simplified introduction to chemical bonding

A COMPOUND is a chemical combination of two or more different elements. The chemical composition of a pure substance can be represented using the element symbols, and where necessary, subscripted numbers. A formula represents the relative numbers of atoms of each element in a substance. Note that the formula of a particular pure substance does NOT change just because it becomes part of a mixture. If it retains its own chemical identity in a mixture, i.e. it does not change chemically, then it must have the same formula. You need to be able to read a formula e.g. like the one below.

o o

o o

o o o o

1.1f



More on formulae and COMPOUNDS



methane CH4

(i) and (ii) The first 'picture' (i) is an example of a displayed formula, in which every individual atom is shown and how it is bonded ('connected') with other atoms in the molecule. All the dashes represent the covalent bonds between the atoms in the molecule. The dashes actually represent an electrical attractive force, but no need for any detail at all here. From this diagram you can tell there are four different elements in the molecules and the number of atoms of each element ... ... there are 4 carbon atoms (C), 8 hydrogen atoms (H), 1 bromine atom (Br) and 1 chlorine atom (Cl) and because there are at least 2 different elements chemically combined in the molecule or formula, this also tells you it is a compound (see below for more examples). A summary of all the atoms in the individual molecule is called the molecular formula, shown on the right. The number of atoms of each element is shown as a subscript number in the formula There are more examples in the next section which discusses the word compound further. Also, look at an example where they are used in chemical equations - burning methane. Note: A displayed formula is sometimes called a full structural formula or graphic formula.

As already stated, a COMPOUND is a pure substance formed by chemically combining at least two different elements by ionic or covalent bonding. Compounds can be represented by a FORMULA, which represents the whole number (integer) ratio of the atoms in a formula, and for molecules, a summary of all the atoms in one molecule. Examples: o sodium chloride NaCl, ionic compound, 2 elements, 1 atom of sodium to everyone 1 atom of chlorine). o methane CH4, covalent compound molecule, has 2 elements in it, 1 atoms of carbon and 4 atoms of hydrogen. The lines in the diagram of methane o ethane C2H6, two atoms of carbon combined with 6 atoms of hydrogen and note that the same two elements can form two different compounds because of different atom ratios i.e. compared with methane. o glucose C6H12O6 (covalent compound molecule, 3 elements in it, 6 atoms of carbon, 12 of hydrogen and 6 of oxygen). o carbon monoxide CO (1C + 1O atoms), carbon dioxide CO2 (1C + 2O atoms), note again that two elements can form two different compounds because of different atom ratios.  NOTE  There must be at least two different types of atom (elements) in a compound.



  ethane C2H6









  

1.1g



MIXTURE 

1.1h



PURE



SUBSTANCE



The number of atoms of each element is shown as a subscript number in the formula ...  ... except * the 1 is never written in the formula, no number means 1. Compounds have a fixed composition and therefore a fixed ratio of atoms represented by a fixed formula, however the compound is made or formed. In a compound the elements are not easily separated by physical means, and quite often not easily by chemical means either. The compound has properties quite different from the elements it is formed from. o For example the two elements soft silvery reactive sodium + reactive green gas chlorine ==> colourless, and not very reactive crystals of the compound sodium chloride. The FORMULA of a compound summarises the 'whole number' atomic ratio of what it is made up of e.g. o methane CH4 is composed of 1 carbon atom combined with 4 hydrogen atoms. Glucose has 6 carbon : 12 hydrogen : 6 oxygen atoms, sodium chloride is 1 sodium : 1 chlorine atom. Sometimes, a compound (usually ionic), is partly made up of two or more identical groups of atoms. To show this more accurately ( ) are used e.g. o Calcium hydroxide is Ca(OH)2 which makes more sense than CaO2H2 because the OH group of atoms is called hydroxide and exists in its own right in the compound.  Make sure you understand the use of (brackets). The subscripted 2 after the brackets doubles everything in the brackets BUT nothing else in the formula. o Similarly, aluminium sulphate has the formula  Al2(SO4)3 rather than Al2S3O12, because it consists of two aluminium ions 3+ 2Al and three sulphate ions SO4 . The sulphate ion is effectively a molecule that carries an overall surplus electrical charge - a 'molecular ion' if you like. The word formula or molecule can also apply to elements. e.g. hydrogen molecule H2, oxygen molecule O2, ozone molecule O3 (2nd unstable form of oxygen), phosphorus molecule P4, sulphur molecule S8, have 2, 2, 3, 4 and 8 atoms in their molecules. Elements like helium He are referred to as 'monatomic' because they exist as single uncombined atoms. o Incidentally, at GCSE level, and mainly at Advanced level too, phosphorus and sulfur are written as P and S respectively. However, in equations There are more examples and comments in equation section. Calculations involving empirical formula and molecular formula are dealt with in sections 5. and 8. on the calculations page. See also picture diagrams of elements/compounds and mixtures.

A MIXTURE is a material made up of at least two substances which may be elements or compounds. They are usually easily separated by physical means e.g. filtration, distillation, chromatography etc. Examples: air, soil, solutions. Separation methods are needed to separate useful materials and purify them. See also picture diagrams of elements/compounds and mixtures.

PURE means that only one substance is present in the material and can be a pure element or compound. A simple physical test for purity, and properties that can help identify a substance, is to measure the boiling point or melting point. Every pure substance melts and boils at a fixed temperature (though boiling point depends on the ambient air pressure). If a liquid is pure it should boil at a constant temperature called the boiling point e.g. o water boils at 100 C. Unfortunately, up on a very high mountain, at a lower air pressure, water boils at a constant, but lower temperature and it is difficult to make a good brew of tea! o An impure liquid will boil at a higher temperature if it contains a dissolved o solid impurity e.g. seawater, containing dissolved salts, boils at over 100 C. o An impure liquid can initially boil at a lower than the expected temperature, if it contains a lower boiling point liquid impurity. The boiling then takes place over a range of temperatures. For example, in the distillation of an alcohol-water mixture from a fermented yeast-sugar solution mixture, it boils away within a range starting o o at about 79 C (boiling point of alcohol) and the last drops distil over at 100 C (boiling point of pure water).



If a solid is pure, it melts sharply at its fixed melting point. o An impure solid melts below its expected melting point and the more impure, the wider the temperature melting range. o o e.g. a water and salt mixture melts below 0 C and butter, a mixture of fats, gradually melts more as the temperature rises on a hot summer's day.

1.1i



IMPURE



IMPURE usually means a mixture of mainly one substance plus one or more other substances physically mixed in. Some examples are mentioned above, in the discussion on the effect of impurities on the melting/boiling points of pure substances. The % purity of a compound is important, particularly in drug manufacture. Any impurities present are less cost-effective to the consumer and they may be harmful substances.



1.1j

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PURIFICATION  



1.1k



CHEMICAL EQUATION

    

PURIFICATION: Materials are purified by various separation techniques. The idea is to separate the desired material in as pure a form as possible from unwanted material or impurities, hence to produce the desired useful product or just a substance whose physical and chemical properties are to be investigated as part of a research project. Detailed examples of methods-examples of separating mixtures are described in later sections on this page. but they include: o Filtration to separate a solid from a liquid. You may want the solid or the liquid or both! o Simple distillation to separate a pure liquid from dissolved solid impurities which have a very high boiling point. o Fractional distillation to separate liquids with a range of different boiling points, especially if relatively close together. o Evaporation to remove a solvent to leave a solid behind. o Crystallisation to get a pure solid out of a solvent solution of it. o Chromatography can be used on a larger scale than spots' to separate out pure samples from a mixture. Methods of collecting gases are on a separate web page.

CHEMICAL EQUATION is an expression in words or symbols that describes precisely a particular chemical change eg the reaction in water between the alkali sodium hydroxide and the acid hydrochloric acid sodium hydroxide + hydrochloric acid ==> sodium chloride + water (word equation) NaOH + HCl ==> NaCl + H2O (symbol equation) NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) (symbol equation with state symbols) Detailed examples on How to write word & symbol equations

Section 1.2 Particle Picture examples of Elements, Compounds and Mixtures - useful visual images

Detailed notes on the physical 'States of Matter' - gases, liquids and solids - structure and properties

Section 1.3 PHYSICAL CHANGES - no new substance formed These are changes which do not lead to new substances being formed. Only the physical state of the material changes. The substance retains exactly the same chemical composition. Examples ... Melting, solid to liquid, easily reversed by cooling e.g. ice and liquid water are still the same H2O molecules. Dissolving, e.g. solid mixes completely with a liquid to form a solution, easily reversed by evaporating the liquid e.g. dissolving salt in water, on evaporation the original salt is regained. So freezing, evaporating, boiling, condensing are all physical changes and may be involved in separating a mixture. Separating a physical mixture e.g. chromatography, e.g. a coloured dye solution is easily separated on paper using a solvent, they can all be re-dissolved and mixed to form the original dye. So distillation, filtering are also physical changes. See also '3 States of Matter' - gases, liquids and solids for more examples and particle theory models

Section 1.4 CHEMICAL CHANGES - REACTIONS - reactants and products

Heating iron and sulphur is classic chemistry experiment to illustrate what is meant by CHEMICAL CHANGE and you can adapt the general conclusions described at the end of this section to any chemical reaction. A mixture of silvery grey iron filings and yellow sulphur powder is made. The iron can be plucked out with a magnet i.e. an easily achieved physical separation because the iron and sulphur are not chemically combined yet! They are still the same iron and sulphur. However, on heating the mixture, it eventually glows red on its own and a dark grey solid called iron sulphide is formed. Both observations indicate a chemical change is happening i.e. a new substance is being formed. We no longer have iron or sulphur BUT a new compound with different physical properties (e.g. colour) and chemical properties (unlike iron which forms hydrogen with acids, iron sulphide forms toxic nasty smelling hydrogen sulphide!). iron + sulphur (sulfur) ==> iron sulphide (iron sulfide) or in symbols: Fe + S ==> FeS AND it is no longer possible to separate the iron from the sulphur using a magnet! Further proof of a new substance formed: The original reactant iron, and the iron sulphide product, can be shown to be different substances by their reactions with dilute acid.

The products of a reaction are completely different from the reactants. This is illustrated by their different physical properties and different chemical reactions. Reactant: The silvery grey metallic iron forms a pale green solution of the salt iron sulphate with dilute sulphuric acid and evolves odourless hydrogen gas which gives a squeaky pop with a lit splint. The word and symbol equations are as follows ... iron + sulphuric acid ==> iron sulphate + hydrogen Fe + H2SO4 ==> FeSO4 + H2 Reactant: Yellow, solid non-metallic sulfur burns in air to form the choking acidic gas sulfur dioxide. sulfur + oxygen ==> sulfur dioxide S + O2 ==> SO

2

Iron sulphide also fizzes and dissolves in dilute sulphuric acid to form iron sulphate BUT produces the 'rotten eggs' smelly gas hydrogen sulphide which gives a black colour with lead ethanoate paper (old name lead acetate). iron sulphide + sulphuric acid ==> iron sulphate + hydrogen sulphide FeS + H2SO4 ==> FeSO4 + H2S This is NOT to be done by the student, hydrogen sulphide gas is highly poisonous. If hydrochloric acid is used, the same two colourless gases are produced but the salt formed would be iron chloride. So signs that a chemical reaction has happened include: change in appearance e.g. change in colour or texture. temperature changes because an energy change has taken place, change in mass e.g. some solids when burned in air gain mass in forming the oxide e.g. magnesium forms magnesium oxide. some solids lose mass when heated, e.g. carbonates lose carbon dioxide in thermal decomposition. and change in the chemical properties of the products compared to the original reactants. Therefore a chemical change is one in which a new substance is formed, by a process which is not easily reversed and usually accompanied by an energy (temperature) change. This is summarised as reactants ==> products as expressed in chemical equations in words or symbols. Apart from experiments and preparations in the laboratory, plenty of chemical changes occur in the home. For a start, you are an extremely complex chemical structure with lots of reactions going on in your body all the time, but others in the home include ... Cooking involves both physical and chemical changes, e.g. meat and potato change in both taste and texture and breakdown chemically to some extent, baking powder breaks down to release carbon dioxide gas which gives the 'rising action' in the production of cakes etc..

Acidic reagents dissolve limescale in the toilet. Candles burning at birthdays and Christmas and gas fire also involve combustion of hydrocarbons like methane. More advanced ideas [see GCSE notes on atomic structure and chemical bonding]: Atoms are held together in molecules or compounds by electrical forces of attraction between the positive nucleus and the outer negative electrons. Therefore, Atoms, ions or molecules react with each other to become electronically more stable. When chemical reactions occur chemical bonds are broken in the reactants and new bonds made in the formation of the products.

METHODS OF SEPARATING MIXTURES AND PURIFYING SUBSTANCES

2.1 Simple Distillation How do you separate two liquids that are miscible? Miscible means they completely mix and do not form two layers. Distillation involves 2 stages and both are physical state changes. (1) The liquid or solution mixture is boiled to vaporise the most volatile component in the mixture (liquid ==> gas). The ant-bumping granules give a smoother boiling action. (2) The vapour passes up from the flask and down into the condenser, where it is cooled by cold water and condenses (gas ==> liquid) back to a liquid (the distillate) which is collected in the flask. This can be used to purify water because the dissolved solids have a much higher boiling point and will not evaporate with the steam, BUT it is too simple a method to separate a mixture of liquids especially if the boiling points of the components are relatively close e.g. as in crude oil and a water-ethanol mixture obtained on fermenting sugar to alcohol.

2.2 Fractional Distillation How can you separate a complex mixture of liquids by a method of distillation? Simple distillation isn't good enough to do an efficient job of separating liquids with boiling points that may be relatively close together. Fractional distillation involves 2 main stages and both are physical state changes. It can only work with liquids with different boiling points. However, this method only works if all the liquids in the mixture are miscible (e.g. alcohol/water, crude oil etc.) and do NOT separate out into layers like oil/water. (1) The liquid or solution mixture is boiled to vaporise the most volatile component in the mixture (liquid ==> gas). The ant-bumping granules give a smoother boiling action. (2) The vapour passes up through a fractionating column, where the separation takes place (theory at the end). This column is not used in the simple distillation described above. (3) The vapour is cooled by cold water in the condenser to condense (gas ==> liquid) it back to a liquid (the distillate) which is collected. This can be used to separate alcohol from a fermented sugar solution. It is used on a large scale to separate the components of crude oil, because the different hydrocarbons have different boiling and condensation points (see oil). FRACTIONAL DISTILLATION THEORY: Imagine green liquid is a mixture of a blue liquid (boiling point o o 80 C) and a yellow liquid (boiling point 100 C), so we have a coloured diagram simulation of a colourless alcohol and water mixture! As the vapour from the boiling mixture enters the fractionating column it begins to cool and condense. The highest boiling or least volatile liquid tends to condense more i.e. the yellow liquid (water). The lower boiling more volatile blue liquid gets further up the column. Gradually up the column the blue and yellow separate from each other so that yellow condenses back into the flask and pure blue distils over to be collected. The 1st liquid, the lowest boiling point, is called the 1st fraction and each liquid distils over when the top of the column reaches its particular boiling point to give the 2nd, 3rd fraction etc. To increase the separation efficiency of the tall fractionating column, it is usually packed with glass beads, short glass tubes or glass rings etc. which greatly increase the surface area for evaporation and condensation. In the distillation of crude oil the different fractions are condensed out at different points in a huge fractionating column. At the top are the very low boiling fuel gases like butane and at the bottom are the high boiling big molecules of waxes and tar. More details on the Fractional distillation of crude oil & uses of fractions

2.3 Paper or Thin Layer Chromatography This method of separation is used to see what coloured materials make up e.g. a food dye analysis. The different food colourings in confectionary products e.g. in the icing top of a cake, the sugar coating on smarties etc. can all be separated and identified using paper chromatography. The coloured material mixture to be separated e.g. a food dye (6) is dissolved in a solvent like ethanol ('alcohol') and carefully spotted onto chromatography paper or a thin layer of a white mineral material on a glass sheet. Alongside it are spotted known colours on the reference 'start line' (1-5), which is drawn in pencil so it doesn't 'run or smudge'. The paper is carefully dipped into the solvent and suspended so the start line is above the liquid solvent. The solvent is absorbed into the paper and rises up it as it soaks into the paper. The solvent may be water (aqueous solvent) or an organic liquid non-aqueous solvent like an alcohol (e.g. ethanol, butanol) or a hydrocarbon (e.g. hexane). For accurate work the distance moved by the solvent is marked on carefully with a pencil and the distances moved by each 'centre' of the coloured spots is also measured. These can be compared with known substances BUT if so, the identical paper and solvent must be used (See R f values below). Due to different solubilities and different molecular 'adhesion' some colours move more than others up the paper, so effecting the separation of the different coloured molecules. The final result is the vertical separation of the spots up the paper which is now referred to as the chromatogram. Any colour which horizontally matches another is likely to be the same molecule i.e. red (1 and 6), brown (3 and 6) and blue (4 and 6) match, showing these three are all in the food dye (6). In the diagram, think of dye (colouring) spots 1 to 5 as the known food dye colours, therefore food colouring dye 6 must be a mixture of three dyes, that is food colourings 1, 3 and 4 because these are the spots that line up horizontally with known standard samples. The distance a substance moves, compared to the distance the solvent front moves (top of grey area on 2nd diagram) is called the reference or Rf value and has a value of 0.0 (not moved - no good), to 1.0 (too soluble - no good either), but Rf ratio values between 0.1 and 0.9 can be useful for analysis and identification. Rf = distance moved by dissolved substance (solute) / distance moved by solvent. Some technical terms: The substances (solutes) to be analysed must dissolve in the solvent, which is called the mobile phase because it moves. The paper or thin layer of material on which the separation takes place is called the stationary or immobile phase because it doesn't move. It is possible to analyse colourless mixture if the components can be made coloured e.g. protein can be broken down into amino acids and coloured purple by a chemical reagent called Ninhydrin and many colourless organic molecules fluoresce when ultra-violet light is shone on them. These are called locating agents. Thin layer chromatography (t.l.c) is where a layer of paste is thinly and evenly spread on e.g. a glass plate. The paste consists of the solid immobile phase like aluminium oxide dispersed in a liquid such as water. The plate is allowed to dry and then used in the same way as paper chromatography. Gas-liquid chromatography is described below

2.6 GAS CHROMATOGRAPHY 

Gas-liquid chromatography (gc/glc/g.l.c.) can be used to analyse liquid mixtures which can be

vapourised (e.g. petrol, blood for alcohol content). The instrument is called a gas chromatograph. o o A sample of the substance under investigation is injected and vapourised into a tube containing a carrier gas (called the mobile phase, it moves). The gas carries the vaporised substance through a long 'separating' tube or column wound around inside a thermostated oven. o The substances in the mixture are partially absorbed by an absorbent material held in the or column (called the immobile phase or stationary phase, which doesn't move), but only temporarily. However different substances are held back, or 'retained', for different times so that the mixture separates out in the carrier gas stream.  There is a dynamic equilibrium between the stationary and mobile phases and the separation of the components of a mixture by chromatography depends on the distribution of the components in the sample between the mobile and stationary phases.  The column is filled with a porous solid so gas can get through but passes over a large surface area OR it is coated in a very high boiling organic liquid which can also provide a large absorbing surface but still allows gas flow. o The gases emerge from the oven into a detector system which electronically records the different signal as each substance comes through. A printout or computer display of the results from the gas chromatograph, called the gas chromatogram, shows a series of peaks in the graph line imposed on a steady baseline when only the carrier gas is passing through the detector. o The time it takes for a substance to come through is called the retention time and is unique for each substance for a particular set of conditions (flow rate, length of separating column, nature of separating column material, temperature etc.). Generally speaking, the greater the molecular mass of the mixture molecule, the longer the retention time. This is because the component molecule - immobile phase intermolecular force of attraction increases with the size of the component molecule, so it is absorbed/retained temporarily a bit more strongly (see right of diagram). o The height of the peak, or more strictly speaking, the area under the peak, is proportional to the amount of that particular substance in the mixture. o Therefore it is possible to identify components in a mixture and calculate their relative proportions in the mixture. o The chromatogram shown above (right of diagram) illustrates the separation of some alkane hydrocarbons in petrol (in reality it is far more complicated with dozens of hydrocarbon molecule peaks on the chromatogram). The different peak heights give the relative proportions i.e. hexane >pentane > heptane. o The retention time order follows the trend of increasing molecular mass gives increasing retention time i.e. in time heptane C7H16 > C6H14 > C5H12 o The gas chromatographic instrument can be calibrated with known amounts of known substances.  So, the timing position of the peak identifies the component X in the gas and the height of the peak tells you much of X is in the mixture.

2.4 Three techniques used in a particular and separation and purification procedure e.g. How can we separate a mixture of sand and salt? or, how do we separate a salt from a salt preparation? 2.4a FILTRATION

2.4b EVAPORATION

2.4c CRYSTALLISATION

Filtration use a filter paper or fine porous ceramic to separate a solid from a liquid. It works because the tiny dissolved particles are too small to be filtered BUT any insoluble 'non-dissolved' solid particles are too big to go through! Evaporation means a liquid changing to a gas or vapour. In separation, its removing the liquid from a solution, usually to leave a solid. It can be done quickly with gentle heating or left out to 'dry up' slowly. The solid will almost certainly be less volatile than the solvent and will remain as a crystalline residue. Crystallisation can mean a liquid substance changing to its solid form. However, the term usually means what happens when the liquid from a solution has evaporated to a point beyond the solubility limit. Then solid crystals will 'grow' out of the solution because the solution is too concentrated for all the solid to remain dissolved at that temperature. Crystallisation is often done from a hot concentrated solution, because most substance are more soluble the hotter the liquid. Consequently on cooling a hot concentrated solution, crystals form as the solubility gets less and less. These separation methods are involved in e.g. (1) separation of a sand and salt mixture or (2) salt preparations (1) The sand/salt mixture is stirred with water to dissolve the salt. The sand is filtered off and washed with pure water to remove remaining traces of salt solution. The salt solution (filtrate) is carefully heated in a dish to evaporate the water and eventually the salt crystals form. Here the solvent is water, but other mixtures can be separated using the same sequence of procedures using a different solvent. e.g. copper and sulphur can be separated using an organic solvent like tetrachloromethane which will dissolve the sulphur (hazardous chemical solvent). (2a) When the water insoluble base (e.g. a metal oxide) is dissolved in an acid, the excess solid base is filtered off and the filtrate solution heated to evaporate the water to produce the salt crystals. (2b) Two solutions of soluble substances are mixed and react to form an insoluble salt. The insoluble salt is filtered off to separate it from the solution, washed with pure water to remove any residual salt solution. The solid is then removed from the filter paper and dried to give the pure dry insoluble salt. Some important words-phrases to do with the above procedures. A solvent is a liquid that dissolves things. The solute is the solid that dissolves in a solvent. A solution is a mixture of a liquid with something dissolved in it. The technique of solvent extraction involves using a liquid to dissolve a solid to separate it from a mixture (e.g. in purifying salt in the experiment described above. A saturated solution is one in which no more substance will dissolve in the liquid. Soluble means the substance (gas, liquid or solid) dissolves in a liquid to form a solution.

Insoluble means a substance won't dissolve in a particular liquid. Remember, a solid may dissolve in one liquid (soluble) but not in another (insoluble). See how these methods are used in making salts

2.4d DECANTATION Decanting is the simplest possible way of separating a liquid (pure or a solution) from an insoluble 3 solid which has a density greater than water (i.e. > 1.0 g/cm ). The solid-liquid mixture is allowed to stand e.g. in a beaker, until all the solid settles out to the bottom of the container. Then the liquid is carefully poured off to leave the insoluble solid behind. However it is inefficient e.g. a small amount of liquid is always left in the solid residue and very fine solid particles take some time to settle out and any disturbance of the liquid can mix them in with the liquid being poured off. Wine may be served in a decanter to leave the undesirable solids behind - no good for bits of cork though, they float!

2.5 Miscellaneous Separation Methods and other apparatus uses

How can we separate two liquids that do not mix? Distillation, described above is used to separate miscible liquids that dissolve in each other. If two liquids do NOT mix, they form two separate layers and are known as immiscible liquids (e.g. oil/water). This is illustrated in the diagram on the left, where the lower grey liquid will be more dense than the upper layer of the yellow liquid and shows how you can separate these two liquids using a separating funnel. (particle picture on gas-liquid-solid page) 1. The mixture is put in the separating funnel with the stopper on and the tap closed and the layers left to settle out. Separating funnel

2. The stopper is removed, and the tap is opened so that you can carefully run the lower grey layer off first into a beaker. 3. This leaves behind the upper yellow layer liquid, so separating the two immiscible liquids.

MAGNET

How can we separate pieces of iron from a mixture of solids? e.g. in scrap iron/steel metal from non-magnetic metals like copper or aluminium. You can retrieve scrap iron or steel form domestic waste.

A magnet can be used to separate iron filings from a mixture with sulfur powder. It is used in recycling to recover iron and steel from domestic waster i.e. the 'rubbish' is on a conveyer belt that passes a powerful magnet which pluck's out magnetic materials. GASES

Methods of collecting gases are on a separate web page. Includes the preparation of ammonia, carbon dioxide, sulphur dioxide, hydrogen and a cracking experiment.

Use of U tube to collect things in e.g. condensing out water in a combustion investigation

In its simplest form these techniques involve using a liquid to dissolve a solid to separate it from a mixture. The extraction of pure salt from a sand-salt mixture is a simple example of the technique. Solvent Extraction

Solvent extraction may dissolving out a desired product where the mixture involves two immiscible liquids or solution. For more complex examples see the advanced level chemistry page. Advanced level chemistry - solute distribution between two immiscible liquids, partition coefficient , calculations and uses How can we separate fine particles of an insoluble solid from a liquid? Centrifuges are devices or apparatus that can be used to separate insoluble materials (usually a solid) from a liquid, where normal filtration does not work well e.g. a suspension of very fine (tiny) solid particles. The centrifuge consists of carriage or glass tube holder, mounted on an electrically motor driven vertical axle.

Centrifuges and centrifuging

The carriage holds the balanced glass tubes of equal amounts of the solid-liquid mixture in each tube, all tubes initially in a horizontal position before the motor is switched on. The tube carriage is rotated at high speed safely in a fully enclosed container. Unbalanced tubes can break with the extra vibration and this situation has a 'knock on' effect, quite literally, as other tubes are likely to shatter with the erratic high speed unbalanced motion. High velocity glass fragments are not good for you! On rapid rotation of the carriage the tubes whirl round horizontally and the centrifugal force causes the more dense insoluble material

particles to move outwards, separating from the liquid. When rotation ceases the solid particles end up at the 'bottom' of the glass tubes with the liquid above. After the centrifuging operation the liquid can be decanted off and the solid is left at the bottom of the glass tube. You might be interested in the solid, liquid or both products depending on the context. Centrifuges come in all sizes and centrifuge technology has many applications in the separation of mixtures and the purification of materials. If [ ] represents the glass tubes, the horizontal rotation situation is shown below .. [solid/liquid] [liquid\solid] Uses-applications: In biology cells can be separated from fluids. A waste 'sludge' can be treated e.g. removing toxic solids from contaminated water from an industrial process. Milk can be separated from whey. Edible oils, wines and spirits can be cleaned or 'clarified' of solid impurities. Expensive oils and other fluids used as lubricants in machining metal parts in industry become contaminated with tiny metal fragments. The larger pieces of metal are easily removed by filtration or sedimentation (allowing to settle out) but the very fine metal particles can only be removed by using a centrifuge. This is likely to be a cheaper option than buying more machine fluid AND reducing pollution since the fluid is recycled leaving less waste to dispose of.

HOW TO WRITE EQUATIONS, WORK OUT FORMULA AND NAME COMPOUNDS

3.1 THE CONSTRUCTION OF CHEMICAL EQUATIONS How do we write equations? "How to write and understand chemical equations" (3rd draft) Seven equations are presented, but approached in the following way The individual symbols and formulae are explained The word equation is presented to summarise the change of reactants to products. A balanced 'picture' equation which helps you understand reading formulae and atom counting to balance the equation. The fully written out symbol equation with state symbols (often optional for starter students). 3.1a Chemical Symbols and Formula

For any reaction, what you start with are called the reactants, and what you form are called the products. So any chemical equation shows in some way the overall chemical change of ... REACTANTS ==> PRODUCTS, which can be written in words or symbols/formulae. It is most important you read about formula in an earlier section of this page. Empirical formula and molecular formula are dealt with on another page. In the equations outlined below several things have been deliberately simplified. This is to allow the 'starter' chemistry student to concentrate on understanding formulae and balancing chemical equations. Some teachers may disagree with this approach BUT my simplifications are: The word 'molecule' is sometimes loosely used to mean a 'formula'. The real 3D shape of the 'molecule' and the 'relative size' of the different element atoms is ignored. If the compound is ionic, the ion structure and charge is ignored, its just treated as a 'formula'.

3.1b Chemical word equations ==> means the direction of change from reactants == to ==> products No symbols or numbers are used in word equations. Always try to fit all the words neatly lined up from left to right, especially if its a long word equation.

3.1c Balancing Symbol equations Writing the correct symbol or formula for each equation component. Numbers in a formula are written as subscripts after the number of atoms of the element concerned e.g. H2SO4 means 2 H's, 1 S and 4 O's or the subscript number can double, treble etc. a part of the formula e.g. Ca(OH)2 means 1 Ca and 2 OH's (or 2 O's and 2 H's in total) Numbers before a formula double or treble it etc. e.g. 2NaCl means 2 Na's and 2 Cl's in total or 2H2SO4 means 2 x H2SO4 = 4 H's, 2 S's and 8 O's in total NOTE: If the number is 1 itself, by convention, no number is shown in a formula or before a formula.

Using numbers if necessary to balance the equation. If all is correct, then the sum of atoms for each element should be the same on both side of the equation arrow ..... in other words: atoms of products = atoms of reactants This is a chemical conservation law of atoms and later it may be described as the 'law of conservation of mass. the 7 equations are first presented in 'picture' style and then written out fully with state symbols The individual formulas involved and the word equations will be been presented in the examples below. NEVER alter a formula to balance an equation! BUT use the CORRECT FORMULA and only put NUMBERS BEFORE THE FORMULA if needed to balance the number of atoms to balance the equation. PRACTICE QUESTIONS - on words and symbol equations (on other web pages) Multiple choice quiz on balancing numbers Balancing number/formula-fill exercises Reactions of acids with metals, oxides, hydroxides, carbonates and ammonia.

3.1d EXAMPLES of CONSTRUCTING WORD or SYMBOL EQUATIONS Remember from the 'Law of Conservation of Mass' the mass of products = mass of original reactants, which means that the number of atoms of each element in the reactants must be equal to those in the products and that is the basis of writing a correctly balanced symbol equation, BUT don't forget, you must write the correct formula for each species in the equation, otherwise you may write a correctly balanced equation which is totally wrong! so beware!

3.1d(1) A single symbol means an uncombined single atom of the element, iron, etc.)

or Fe 1 atom of

or S 1 atom of sulphur (2Fe would mean two atoms, 5S would mean five sulphur atoms

or the formula FeS means one atom of iron is chemically combined with 1 atom of sulphur to form the compound called iron sulphide iron + sulphur ==> iron sulphide

on average one atom of iron chemically combines with one atom of iron forming one molecule of iron sulphide two elements chemically combining to form a new compound Fe + S ==> FeS

Fe(s) + S(s) ==> FeS(s) (with state symbols) Atom balancing, sum left = sum right: 1Fe + 1S = (1Fe combined with 1S) For a balanced equation on both sides of the equation you should have 1 iron atom and 1 sulfur atom combined in their particular way in the reactants or products All the reactants (what you start with) and all the products (what is formed) are all solids in this case. When first learning symbol equations you probably won't use state symbols like (s) at first (see end note).

3.1d(2) or the formula NaOH means 1 atom of sodium is combined with 1 atom of oxygen and 1 atom of hydrogen to form the compound called sodium hydroxide

or the formula HCl means 1 atom of hydrogen is combined with 1 atom of chlorine to form 1 molecule of the compound called hydrochloric acid

or the formula NaCl means 1 atom of sodium are combined with 1 atom chlorine to form the compound called sodium chloride

or the formula H2O means 2 atoms of hydrogen are chemically combined with 1 atom of oxygen to form the compound called water. sodium hydroxide + hydrochloric acid ==> sodium chloride + water

the reactants are one molecule of sodium hydroxide and one molecule of hydrochloric acid the products are one molecule of sodium chloride and one molecule of water all chemicals involved are compounds NaOH + HCl ==> NaCl + H2O NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) (with state symbols) atom balancing, sum left = right: (1Na + 1O + 1H) + (1H + 1Cl) = (1Na + 1Cl) + (2H + 1O) For a balanced equation on both sides of the equation you should have 1 sodium atom, 1 oxygen atom, 1 chlorine atom and 2 hydrogen atoms combined in their particular way in the reactants or products

3.1d(3)

or the symbol Mg means 1 atom of the element called magnesium

or 2HCl means two separate molecules of the compound called

hydrochloric acid (see example 2)

or the formula MgCl2 means 1 formula of the compound called magnesium chloride, made of one atom of magnesium and two atoms of chlorine.

or the formula H2 means 1 molecule of the element called hydrogen made up of two joined hydrogen atoms magnesium + hydrochloric acid ==> magnesium chloride + hydrogen

one atom of magnesium reacts with two molecules of hydrochloric acid the products are one molecule of magnesium chloride and one molecule of hydrogen Mg and H-H are elements, H-Cl and Cl-Mg-Cl are compounds Mg + 2HCl ==> MgCl2 + H2 Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) (with state symbols) atom balancing, sum left = right: (1Mg) + 2 x (1H + 1Cl) = (1Mg + 2Cl) + (2H) For a balanced equation on both sides of the equation you should have 1 magnesium atom, 2 hydrogen atoms and 2 chlorine atoms combined in their particular way in the reactants or products

3.1d(4) or the formula CuCO3 means one formula of the compound called copper carbonate, made up of one atom of copper is combined with one atom of carbon and three atoms of oxygen to form the compound copper carbonate

or the formula H2SO4 means one formula of the compound called sulphuric acid, which is made up of two atoms of hydrogen, one atom of sulphur and four atoms of oxygen

or the formula CuSO4 means one formula of the compound called copper sulphate which is made up of one atom of copper, one atom of sulphur and four atoms of oxygen H2O (example 2)

or the formula CO2 means one molecule of the compound called carbon dioxide which is a chemical combination of one atom of carbon and two atoms of oxygen.

copper carbonate + sulphuric acid ==> copper sulphate + water + carbon dioxide

the reactants are one formula of copper carbonate and one molecule of sulphuric acid the products are one formula of copper sulphate, one molecule of water and one molecule of carbon dioxide all molecules are compounds in this reaction CuCO3 + H2SO4 ==> CuSO4 + H2O + CO2 CuCO3(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l) + CO2(g) (with state symbols) balancing sum left = sum right: (1Cu + 1C + 3O) + (2H + 1S + 4O) = (1Cu + 1S + 4O) + (2H + 1O) + (1C + 2O) For a balanced equation on both sides of the equation you should have 1 copper atom, 1 carbon atom, 7 oxygen atoms, 2 hydrogen atoms, 1 sulphur atom combined in their particular way in the reactants or products

3.1d(5) or the formula CH4 means one molecule of the compound called methane which is made of one atom of carbon combined with four atoms of hydrogen

or 2O2 means two separate molecules of the element called oxygen, and each oxygen molecule consists of two atoms of oxygen CO2 (see also example 4)

or 2H2O means two separate molecules of the compound called water (see also example 2) methane + oxygen ==> carbon dioxide + water

Using displayed formula the equation would look like this ...

... in which every individual atom is shown and how it is bonded ('connected') with other atoms in the molecule. All the dashes represent the covalent bonds between the atoms in the molecules.

one molecule of methane is completely burned by two molecules of oxygen to form one molecule of carbon dioxide and two molecules of water CH4 + 2O2 ==> CO2 + 2H2O CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l) (with state symbols) atom balancing, sum left = sum right: (1C + 4H) + 2 x (2O) = (1C + 2O) + 2 x (2H + 1O) For a balanced equation on both sides of the equation you should have 1 carbon atom, 4 hydrogen atoms, 4 oxygen atoms combined in their particular way in the reactants or products

3.1d(6) or the formula Mg(OH)2 is the compound magnesium hydroxide made up of one magnesium, two oxygen and two hydrogen atoms BUT the OH is a particular combination called hydroxide within a compound, so it is best to think of this compound as a combination of an Mg and two OH's, hence the use of the ( ). The subscripted 2 doubles everything in the brackets.

or 2HNO3 means two separate molecules of the compound nitric acid, each molecule is made up of one hydrogen atom, one nitrogen atom and three oxygen atoms.

or the formula Mg(NO3)2 is the compound magnesium nitrate, it consists of a magnesium (ion) and two 'nitrates' (ions), each nitrate consists of one nitrogen and three oxygen atoms, again the nitrate is a particular combination of atoms within a compound and hence the use of ( ) again.

or 2H2O meaning two molecules of the compound water (see also examples 2 and 5) magnesium hydroxide + nitric acid ==> magnesium nitrate + water

one formula of magnesium hydroxide reacts with two molecules of nitric acid to form one formula of magnesium nitrate and two molecules of water (all compounds) Mg(OH)2 + 2HNO3 ==> Mg(NO3)2 + 2H2O Mg(OH)2(aq) + 2HNO3(aq) ==> Mg(NO3)2(aq) + 2H2O(l) (with state symbols) atom balancing, sum left = sum right: (1Mg + 2O + 2H) + 2 x (1H + 1N + 3O) = (1Mg + 2N + 6O) + 2 x (2H + 1O) For a balanced equation on both sides of the equation you should have 1 magnesium atom, 8 oxygen atoms, 4 hydrogen atoms, 2 nitrogen atoms combined in their particular way in the reactants or products

3.1d(7) or the formula Al2O3 means one formula of the compound called aluminium oxide, made up of two atoms of aluminium Al and three atoms of oxygen O

or 3H2SO4 meaning three molecules of the compound called sulphuric acid (see also example 4)

or the formula Al2(SO4)3 means one formula of the compound called aluminium sulphate, it consists of two aluminium, three sulphur and twelve oxygen atoms BUT the SO4 is a particular grouping called sulphate, so it is best to think of the compound as a combination of two Al's and three SO4's

or 3H2O means three separate molecules of the compound called water (see also examples 2 and 5) aluminium oxide + sulphuric acid ==> aluminium sulphate + water

one formula of aluminium oxide reacts with three molecules of sulphuric acid to form one formula of aluminium sulphate and three molecules of water note the first use of numbers (3) for the sulphuric acid and water! so picture three of them in your head, otherwise the picture gets a bit big! Al2O3 + 3H2SO4 ==> Al2(SO4)3 + 3H2O Al2O3(s) + 3H2SO4(aq) ==> Al2(SO4)3(aq) + 3H2O(l) (with state symbols) atom balancing, sum left = sum right: (2Al + 3O) + 3 x (2H + 1S + 4O) = (2Al + 3S + 12O) + 3 x (2H + 1O) For a balanced equation on both sides of the equation you should have 2 aluminium atoms, 15 oxygen atoms, 6 hydrogen atoms, 3 sulfur atoms combined in their particular way in the reactants or products 3+

2-

GCSE-AS-A2-IB note: Aluminium sulfate is actually an ionic compound (Al )2(SO4 )3 NOTE 1: means a reversible reaction, it can be made to go the 'other way' if the conditions are changed. Example: nitrogen + hydrogen N2(g) + 3H2(g)

ammonia

2NH3(g) (with state symbols)

balancing: 2 nitrogen's and 6 hydrogen's on both sides of equation

Note 2 on the state symbols X(?) of reactants or products in equations (g) means gas, (l) means liquid, (s) means solid and (aq) means aqueous solution or dissolved in water e.g. carbon dioxide gas CO2(g), liquid water H2O(l), solid sodium chloride 'salt' NaCl(s) and copper sulphate solution CuSO4(aq) 3.1e IONIC EQUATIONS (for higher GCSE and AS students) What is an 'ionic equation'? How do we construct and write ionic equations? In many reactions only certain ions change their 'chemical state' but other ions remain in exactly the same original physical and chemical state. The ions that do not change physically or chemically are called 'spectator ions'. The ionic equation represents the 'actual' chemical change and omits the spectator ions. Five types of examples of ionic equations are presented below including neutralisation, salt precipitation and redox equations. (1) Acid-base reactions: Acids can be defined as proton donors. A base can be defined as a proton acceptor. e.g. any acid-alkali neutralisation involves the hydroxide ion is (base) and this accepts a proton from an acid. HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l) which can be re-written ionically as +

-

+

-

+

-

H Cl (aq) + Na OH (aq) ==> Na Cl (aq) + H2O(l) +

-

+

-

+

-

or: H + Cl (aq) + Na + OH (aq) ==> Na + Cl (aq) + H2O(l) H

+

-

(aq)

+ OH (aq) ==> H2O(l) which is the ionic equation for neutralisation -

+

the spectator ions are chloride Cl and sodium Na

(2) Insoluble salt formation: An insoluble salt is made by mixing two solutions of soluble compounds to form the insoluble compound in a process called 'precipitation'. A precipitation reaction is generally defined as 'the formation of an insoluble solid on mixing two solutions or a bubbling a gas into a solution'. (a) Silver chloride is made by mixing solutions of solutions of silver nitrate and sodium chloride. silver nitrate + sodium chloride ==> silver chloride + sodium nitrate AgNO3(aq) + NaCl(aq) ==> AgCl(s) + NaNO3(aq) in terms of ions it could be written as

+

-

+

-

+

-

Ag NO3 (aq) + Na Cl (aq) ==> AgCl(s) + Na NO3 (aq) +

-

+

-

+

-

or: Ag + NO3 (aq) + Na + Cl (aq) ==> AgCl(s) + Na + NO3 (aq) -

+

but the spectator ions are nitrate NO3 and sodium Na which do not change at all, so the ionic equation is simply: Ag

+

-

(aq)

+ Cl (aq) ==> AgCl(s)

Note that ionic equations omit ions that do not change there chemical or physical state. -

In this case the nitrate (NO3 (aq)) and sodium (Na are called spectator ions,

+

(aq))

ions do not change physically or chemically and

+

-

BUT the aqueous silver ion, Ag (aq), combines with the aqueous chloride ion, Cl (aq), to form the insoluble salt silver chloride, AgCl(s), thereby changing their states both chemically and physically. (b) Lead(II) iodide, a yellow precipitate (insoluble in water!) can be made by mixing lead(II) nitrate solution with e.g. potassium iodide solution. lead(II) nitrate + potassium iodide ==> lead(II) iodide + potassium nitrate Pb(NO3)2(aq) + 2KI(aq) ==> PbI2(s) + 2KNO3(aq) which can be written as Pb

2+

-

(aq)

+ 2NO3 (aq) + 2K

+

-

(aq)

the ionic equation is: Pb

+ 2I (aq) ==> PbI2(s) + 2K

2+

+

-

(aq)

+ 2NO3 (aq)

-

(aq)

+ 2I (aq) ==> PbI2(s) -

+

because the spectator ions are nitrate NO3 and potassium K . (c) Calcium carbonate, a white precipitate, forms on e.g. mixing calcium chloride and sodium carbonate solutions ... calcium chloride + sodium carbonate ==> calcium carbonate + sodium chloride CaCl2(aq) + Na2CO3(aq) ==> CaCO3(s) + 2NaCl(aq) 2+

Ca

-

(aq)

+

+ 2Cl (aq) + 2Na 2+

ionically: Ca

2(aq) + CO3 (aq)

2-

(aq)

+ CO3

(aq)

+

==> CaCO3(s) + 2Na

-

(aq)

+ 2Cl (aq)

==> CaCO3(s) -

+

because the spectator ions are chloride Cl and sodium Na . (d) Barium sulphate, a white precipitate, forms on mixing e.g. barium chloride and dilute sulphuric acid ... barium chloride + sulphuric acid ==> barium sulphate + hydrochloric acid BaCl2(aq) + H2SO4(aq) ==> BaSO4(s) + 2HCl(aq) 2+

Ba

-

(aq)

+ 2Cl (aq) + 2H

+

2+

ionic equation: Ba

2-

(aq) +

SO4

(aq)

2-

(aq)

+ SO4

(aq)

==> BaSO4(s) + 2H

+

-

(aq)

+ 2Cl (aq)

==> BaSO4(s) -

+

because the spectator ions are chloride Cl and hydrogen H . (3) Redox reaction analysis: (a) magnesium + iron(II) sulphate ==> magnesium sulphate + iron

Mg(s) + FeSO4(aq) ==> MgSO4(aq) + Fe(s) this is the 'ordinary molecular' equation for a typical metal displacement reaction, but this does not really show what happens in terms of atoms, ions and electrons, so we use ionic equations like the one shown below. 2+

2-

Mg(s) + Fe SO4

2+

(aq)

2-

==> Mg SO4

(aq)

+ Fe(s)

2-

The sulphate ion SO4 (aq) is the spectator ion, because it doesn't change in the reaction and can be omitted from the ionic equation. No electrons show up in the full equations because electrons lost by Mg must equal the electrons gained by Fe. so the ionic-redox equation is Mg(s) + Fe

2+

==> Mg

(aq)

2+ (aq)

+ Fe(s)

Mg oxidised by electron loss, Fe

2+

reduced by electron gain

(b) zinc + hydrochloric acid ==> zinc chloride + hydrogen Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g) Zn(s) + 2H

+

-

(aq)

+ 2Cl (aq) ==> Zn

2+

-

(aq)

+ 2Cl (aq) + H2(g)

-

the chloride ion Cl is the spectator ion Zn(s) + 2H

+ (aq)

==> Zn

2+ (aq)

+ H2(g) +

Zinc atoms, Zn, oxidised by electron loss and hydrogen ions, H , are reduced by electron gain (c) copper + silver nitrate ==> silver + copper(II) nitrate Cu(s) + 2AgNO3(aq) ==> 2Ag + Cu(NO3)2(aq) -

the nitrate ion NO3 is the spectator ion Cu(s) + 2Ag

+ (aq)

==> 2Ag(s) + Cu

2+ (aq) +

Cu oxidised by electron loss, Ag reduced by electron gain (d) halogen (more reactive) + halide salt (of less reactive halogen) ==> halide salt (of more reactive halogen) + halogen (less reactive) +

+

X2(aq) + 2K Y(aq) ==> 2K X(aq) + Y2(aq) -

-

X2(aq) + 2Y (aq) ==> 2X (aq) + Y2(aq) +

the potassium ion K is the spectator ion halogen X is more reactive than halogen Y, F > Cl > Br > I) X is the oxidising agent (electron acceptor, so is reduced) -

KY or Y is the reducing agent (electron donor, so is oxidised) (4) Ion Exchange Resins: Ion exchange polymer resin columns hold hydrogen ions or sodium ions. These can be replaced by calcium and magnesium ions when hard water passes down the column. The calcium or magnesium ions are held on the negatively charged resin. The freed hydrogen or sodium ions do not form a scum with soap. -

e.g. 2[resin] H

+

2+

(s)

+ Ca

-

(aq)

2+

-

+

==> [resin] Ca [resin] (s) + 2H

(aq)

-

+

or 2[resin] Na

(s)

+ Mg

2+

-

(aq)

2+

-

+

==> [resin] Mg [resin] (s) + 2Na

(aq)

etc.

(5) Scum formation with hard water: On mixing hard water with soaps made from the sodium salts of fatty acids, insoluble calcium or magnesium salts of the soap are formed as a grey precipitate ... CaSO4(aq) + 2C17H35COONa(aq) ==> (C17H35COO)2Ca(s for scum!) + Na2SO4(aq) or more simply ionically: Ca

2+

-

(aq) 2-

-

2+

+ 2C17H35COO (aq) ==> (C17H35COO )2Ca

the spectator ions are SO4 and Na

(s)

+

3.2 VALENCY - COMBINING POWER - FORMULA DEDUCTION

3.2a Introduction What is valency? How do you use valency to work out the formula of a compound? The valency of an atom or group of atoms is its numerical combining power with other atoms or groups of atoms. i.e. its numerical capacity to combine with other atoms. The theory behind this, is all about stable electron structures! The combining power or valency is related to the number of outer electrons. You need to consult the page on "Bonding" to get the electronic background. A group of atoms, which is part of a formula, with a definite composition, is sometimes referred to as a radical. In the case of ions, the charge on the ion is its valency or combining power (list below). To work out a formula by combining 'A' with 'B' the rule is: number of atom 'A' x valency of 'A' = number of atom 'B' x valency of 'B', However it is easier perhaps? to grasp with ionic compound formulae. In the electrically balanced stable formula, the total positive ionic charge must equal the total negative ionic charge. number of ion 'A' x charge of ion 'A' = number of ion 'B' x charge of ion 'B' (you ignore charge sign) Example: As difficult an example as any you will have to work out! Aluminium oxide consists of aluminium ions Al 3+

number of Al

x charge on Al

3+

3+

2-

and oxide ions O

= number of O x charge on O

the simplest numbers are 2 of Al

3+

2-

2-

2-

x 3 = 3 of O x 2 (total 6+ balances total 6-)

so the simplest whole number formula for aluminium oxide is Al2O3

3.2b Examples of COVALENT and IONIC COMPOUND FORMULAE Selected combining power of ions (table left) valency = numerical ion charge value and examples of covalent combining power of atoms ie valencies (selection below).       

Hydrogen H (1) Chlorine Cl and other halogens (1) Oxygen O and sulphur S (2) Boron B and aluminium Al (3) Nitrogen (3, 4, 5) Carbon C and silicon Si (4) Phosphorus (P 3,5)

3.2c Examples of working out covalent formulae 'A' (valency)

'B' (valency)

deduced formula of A + B

1 of carbon C (4)

balances 4 of hydrogen H (1)

1 x 4 = 4 x 1 = CH4

1 of nitrogen (3)

balances 3 of chlorine Cl (1)

1 x 3 = 3 x 1 = NCl3

1 of carbon C (4)

balances 2 of oxygen O (2)

1 x 4 = 2 x 2 = CO2

The diagram on the left illustrates the three covalent examples above for methane CH4 nitrogen trichloride NCl3 carbon dioxide CO2

6 more examples of working out an ionic formula numerically charge = valency of A or B to deduce the formula

valency or ionic charge = the combining power of the ion 'molecular' or ionic style of formula and compound name +

-

+

-

1 of K balances 1 of Br because 1 x 1 = 1 x 1 gives KBr or K Br potassium bromide +

2-

+

2 of Na balances 1 of O because 2 x 1 = 1 x 2 gives Na2O or (Na )2O 1 of Mg

2+

-

3+

-

-

3+

-

balances 3 of F because 1 x 3 = 3 x 1 gives FeF3 or Fe (F )3 iron(III) fluoride -

1 of Ca

sodium oxide

balances 2 of Cl because 1 x 2 = 2 x 1 gives MgCl2 or Mg (Cl )2 magnesium chloride

1 of Fe 2+

2+

2-

2+

-

balances 2 of NO3 because 1 x 2 = 2 x 1 gives Ca(NO3)2 or Ca (NO3 )2 calcium nitrate 3+

2 of Fe

2-

3+

2-

balances 3 of SO4 because 2 x 3 = 3 x 2 gives Fe2(SO4)3 or (Fe )2(SO4 )3 iron(III) sulphate

3.3 KS3-GCSE note on naming compounds When combined with other elements in simple compounds the name of the non-metallic element changes slightly from ...??? to ...ide. 2-

2-

Sulphur forms a sulphide (ion S ), oxygen forms an oxide (ion O ), fluorine forms a fluoride (ion F ), chlorine forms a chloride (ion Cl ), bromine a bromide (ion Br ) and iodine an iodide (ion I ). The other element at the start of the compound name e.g. hydrogen or a metal like sodium, potassium, magnesium, calcium, etc. usually remains unchanged in simple compounds at KS3GCSE level. So typical compound names are, sodium sulphide, hydrogen sulphide, magnesium oxide, potassium fluoride, hydrogen chloride, sodium chloride, calcium bromide, magnesium iodide etc. However, even at GCSE level the complications will arise e.g. (i) Where an element can form two different compounds with different formulae with the same element there needs to be a way of expressing it in the name as well as in the formula e.g. iron(II) chloride, FeCl2 and iron(III) chloride, FeCl3 copper(I) oxide, Cu2O and copper(II) oxide, CuO Hear chlorine has a combining power of 1 (valency 1) and oxygen 2 in both compounds. However, iron can have a valency of 2 or 3 and copper 1 or 2 and these also correspond 2+ 3+ + 2+ numerically to the charge on the metal ions in such compounds e.g. Fe and Fe , Cu and Cu . Therefore the 'Roman numerals' number in () gives the valency of the element in that particular compound. At a higher academic level this is known as the oxidation state. (ii) When the non-metal is combined with oxygen to form a negative ion (anion) ion which combines with a positive ion (cation) from hydrogen or a metal, then the end of the 2nd part of the name ends in ...ate or ...ite e.g.

-

NO3 in a compound formula is nitrate e.g. KNO3, potassium nitrate. SO3 in a formula is sulphite, e.g. Na2SO3, sodium sulphite, SO4 is sulphate, e.g. MgSO4, magnesium sulphate, PO4 is phosphate, e.g. Na2HPO4, disodium hydrogen phosphate

TYPES OF CHEMICAL REACTIONS OR CHEMICAL PROCESSES Electrolysis , anodising and electroplating 







 

When substances which are made of ions are dissolved in water, or melted, they can be broken down (decomposed) into simpler substances by passing an electric current through them. This process is called electrolysis. The electrical conducting solution or melt of ions is called the electrolyte. When an ionic substance is melted or dissolved in water the ions are free to move about and move to the electrical contacts called electrodes. The electrodes are usually inert e.g. carbon or platinum. The electron rich or negative electrode is called the cathode. The electron deficient or positive electrode is called the anode. During electrolysis ions move to their oppositely charged electrode. o Positively charged ions, usually hydrogen or metal ions move to the negative electrode. 2+ Depending on the voltage, the positive ions may be reduced (e.g. Cu ) by electron gain to deposit a + metal (eg. Cu) or release hydrogen gas (H2) from hydrogen ions (H ). o At the same time negatively charged ions move to the positive electrode. The negative ions may be oxidised by electron loss. This usually results in the release of a non-metallic gas e.g. oxygen (O2) from hydroxide ions (OH ) or chlorine from chloride ions (Cl )etc. Electroplating is the process of coating a conducting material with a layer of metal using the process of electrolysis. The object to be coated is made the negative cathode and dipped into a salt solution of the metal ions of the metal to form the coating. On passing a low d.c. voltage the metal is deposited on the conducting negative cathode. For more details see 4th link below. Anodising a metal, like aluminium, is done by making it the positive anode in an electrolysis system. When the electrolyte is sulphuric acid solution, the oxygen formed at the anode oxidises the metal surface to make a thicker metal oxide layer. For more detailed examples see .. o Electrolysis of sodium chloride solution o Extraction of Aluminium o Purification of copper o Electroplating o A more detailed introduction to Electrochemistry o See also below for oxidation-reduction and electrode equations

Exothermic and Endothermic Reactions or Changes



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EXOTHERMIC CHANGES o Heat is released or given out to the surroundings by the materials involved, so the temperature rises. o chemical change examples involve a new substance being formed and lots of examples (i) to (vi) below (but they are not always exothermic - see endothermic below). o physical change examples e.g. condensation, freezing etc. all require the removal of energy from the material e.g. water, to the surroundings to produce the change in state (its the same as releasing heat, but it doesn't seem like it!).  Note: Dissolving substances in water can release heat giving a warm/hot solution e.g. diluting concentrated sulphuric acid.  At KS3-GCSE level this exothermic 'dissolving' is considered a physical change, but at ASA2 level they may be considered a chemical change too! At a higher level of thinking for exothermic chemical changes: The net energy change when the energy needed to break bonds in the reactants is less than the energy released when new bonds are formed in the products. o See also KS4 Science GCSE/IGCSE Chemistry Notes on Energy Changes for more details A burning or combustion reaction usually means a very fast exothermic reaction where a flame is observed. It involves a highly energetic oxidation of 'fuels' where the temperature generated is so high the atoms give off light from the luminous flame zone e.g. o (i) bunsen flame as methane gas fuel burns ...  methane + oxygen ==> carbon dioxide + water  CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l)  This is complete combustion with a pale blue flame and the products cannot react any further with oxygen.  If the oxygen supply is limited the flame is more yellow and can be 'smokey' due to soot formation (C) and dangerous since carbon monoxide (CO) can be formed.  These are examples of incomplete combustion.  methane + oxygen ==> carbon monoxide + water  2CH4(g) + 3O2(g) ==> 2CO(g) + 4H2O(l) (carbon monoxide formation)  Most people who die in house fires are poisoned by carbon monoxide (and other toxic gases) in the thick smoke rather than from burns.  or  methane + oxygen ==> carbon (soot) + water  CH4(g) + O2(g) ==> C(s) + 2H2O(l) (soot formation)  The sooty carbon particles e.g. in a candle flame, are heated to such a high temperature they become incandescent and give out yellow light, but as far as I know virtually no carbon monoxide is formed!  See also KS4 Science GCSE/IGCSE Chemistry Notes on fossil fuel combustion o (ii) passing chlorine over hot aluminium metal to make aluminium chloride, the aluminium burns to form the chloride ...  aluminium + chlorine ==> aluminium chloride  2Al(s) + 3Cl2(g) ==> 2AlCl3(s) o (iii) burning magnesium ribbon with a bright white flame ...  magnesium + oxygen ==> magnesium oxide  2Mg(s) + O2(g) ==> 2MgO(s) o (i) to (iii) are all oxidation reactions, as are all 'fuel' burning reactions. Continuous combustion requires the 'fire triangle' of heat + fuel + oxidant (oxidants like oxygen, air or other reactive gases like chlorine or fluorine and in rockets liquids like hydrogen peroxide) o Very fast or explosive combustion:  A roaring bunsen flame (of methane burning) is an example of fast combustion and when the air (oxygen) - methane (natural gas) mixture is first ignited it is a small explosion! (equation above). It seems contradictory, but a source of ignition is needed because the C-H and O=O bonds are very strong giving a high activation energy. However, once ignited, the heat from the flame keeps the burning going.  Another explosive example is the 'squeaky pop test for hydrogen'. When a lit splint is applied there is a faint blue flame for a fraction of a second as the two gases explode to form water + heat, light and sound energy!  hydrogen + oxygen ==> water  2H2(g) + O2(g) ==> 2H2O(l)  In all these cases the high temperature reaction zone is seen as flame and an initial high energy source for ignition is needed to initiate the reaction e.g. a match or an electrical discharge. o Slow or smouldering combustion:  In these cases no flame is seen, but a high temperature heat source is still required to start the reaction and the reaction zone is still at a high temperature e.g. the red hot slow burning of charcoal (mainly carbon), but the main combustion product is still carbon dioxide. You can only get this slow/smouldering combustion with solid combustible reactants.  Gases will tend to explode unless controlled in a burner and liquids will vaporise in the heat from the flame and so will also burn very fast with a flame.  carbon + oxygen ==> carbon dioxide

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 

 C(s) + O2(g) ==> CO2(g)  This is an example of complete combustion.  BUT quite often, with limited air/oxygen supply, carbon monoxide is readily formed,  carbon + oxygen ==> carbon monoxide  2C(s) + O2(g) ==> 2CO(g)  This is an example of incomplete combustion. o Spontaneous combustion:  This is when combustion occurs without any application of a high energy ignition source, sometimes described as self-ignition, though in some cases heat is generated in some way which triggers the reaction.  For example, it is possible to prepare a very finely divided black powder form of iron(II) oxide. When the powder is dropped through air lots of tiny flashes of light are seen as it burns to form another iron oxide (probably Fe3O4). The reaction is triggered by heat from friction. The powder has such a large surface area that the friction caused by just falling in air produces enough heat to initiate the reaction. Powdered coal dust or very fine flour can behave in the same way and both have been responsible for serious accidents in industry.  Other substances can spontaneously ignite in air because the activation energies required are so low and the kinetic energy of the particles is sufficient for the reaction to happen without help! e.g. the highly reactive Group 1 Alkali Metal caesium and the silicon-hydrogen compounds called silanes (SiH4, Si2H6 etc. which are like organic alkanes with the C's replaced by Si and far less stable).  Potassium and all the alkali metals below it ignite in water (Rb and Cs explosively) because the reaction is so exothermic and ignites metal vapour and the hydrogen gas produced. See other web page for more detailed notes on energy changes and calculations. BUT many exothermic reactions are not as dramatic as burning with a flame! e.g. o (iv) Respiration: the relatively slow 'burning' of carbohydrates in animals/plants, but it releases plenty of o energy at 37 C!  glucose + oxygen ==> carbon dioxide + water + energy  C6H12O6(aq) + 6O2(g) ==> 6CO2(g) + 6H2O(l) + energy o (v) Neutralisation: acid + alkali ==> salt + water  e.g. hydrochloric acid + sodium hydroxide ==> sodium chloride + water o  which is one of the fastest reactions in water, but the mixture only warms up by 5 to 10 C! A bunsen o flame reaches 1200 C in the main combustion zone!  More details further down. o (vi) Rusting in which iron slowly reacts with water and oxygen (from air) to form the orange-brown hydrated iron oxide we call rust.

ENDOTHERMIC CHANGES o Heat is absorbed or taken in by the materials involved from the surroundings, the system cools or has to be heated to effect the change. o Chemical change examples of endothermic reactions e.g. thermal decomposition of limestone, cracking oil fractions, decomposition by electrolysis etc.  (i) Photosynthesis: input of energy from sunlight needed  carbon dioxide + water ==> glucose + oxygen  6CO2 + 6H2O + sunlight energy ==> C6H12O6 + 6O2 o  (ii) Making lime by heating limestone to over 900 C where a net input/absorption of energy is needed to bring about this thermal decomposition ...  limestone ==> quicklime + carbon dioxide  calcium carbonate ==> calcium oxide + carbon dioxide  CaCO3(s) ==> CaO(s) + CO2(g)  (iii) Cracking hydrocarbon molecules from oil to make smaller molecules, also requires this absorption of heat by the reactant molecules to break em' up', or to put it 'poshly', another example thermal decomposition e.g.  hexane => ethene + butane  C6H14 ==> C2H4 + C4H10 o Physical change examples e.g. melting, boiling, evaporation etc. all require the input of energy to effect the change of state of the material.  Note: Dissolving substances in water can absorb heat giving a cool solution e.g. dissolving ammonium nitrate salt in water.  At KS3-GCSE level this endothermic 'dissolving' is considered a physical change, but at AS-A2 level they may be considered a chemical change too! At a higher level of thinking for endothermic chemical changes. The net energy change when the energy needed to break bonds in the reactants is more than the energy released when new bonds are formed in the products. See also KS4 Science GCSE/IGCSE Chemistry Notes on fossil fuel combustion

Decomposition and Thermal Decomposition 

Decomposition in general means to break down into small species e.g. natural organic matter decomposes with enzymes into carbon dioxide, water and nitrogen etc. o Fermentation is form of biological degradation, catalysed by enzymes, to break down glucose sugar into the smaller molecules of ethanol ('alcohol') and carbon dioxide ...  C6H12O6(aq) ==> 2C2H5OH(aq) + 2CO2(g) Light can cause decomposition e.g. in photography is a sort of photo-decomposition. o silver chloride + light ==> silver + chlorine o 2AgCl ==> 2Ag + Cl2 Thermal decomposition means to break down substances into two or more substances by heat (usually endothermic reactions at temperatures well above room temperature) e.g. o The decomposition of calcium carbonate (limestone) into calcium oxide (lime) and carbon dioxide in a high temperature lime kiln. o calcium carbonate ==> calcium oxide + carbon dioxide o CaCO3(s) ==> CaO(s) + CO2(g)  For more details see the Extra Industrial Chemistry notes. o The breaking down of hydrocarbons into smaller ones using a catalyst as well as a high temperature. This reaction is also known as cracking.  For more details see the Oil and its useful Products notes. o e.g. octane ==> hexane + ethene  C8H18 ==> C6H14 + C2H4 o Other thermal decompositions which are examples of reversible reactions.

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OXIDATION and REDUCTION - REDOX REACTIONS OXIDATION - definition and examples

REDUCTION - definition and examples

The gain or addition of oxygen by an atom, molecule or ion e.g. ...

The loss or removal of oxygen from a compound etc. e.g. ...

(1) S + O2 ==> SO2 [burning sulphur - oxidised to sulphur dioxide]

(1) CuO + H2 ==> Cu + H2O [loss of oxygen from copper(II) oxide shows it to be reduced to copper atoms]

(2) CH4 + O2 ==> CO2 + 2H2O [burning methane to water and (2) Fe2O3 + 3CO ==> Fe + 3CO2 [iron(III) oxide ore is carbon dioxide, methane oxidised as the C and H atoms gain reduced to iron metal by oxygen loss in the blast furnace] O] (3) 2CO + 2NO ==> CO2 + N2 [nitrogen monoxide reduced to (3) 2NO + O2 ==> 2NO2 [nitrogen monoxide is oxidised to nitrogen by losing oxygen] nitrogen dioxide by gaining oxygen] (4) CuO + Mg ==> Cu + MgO [loss of oxygen from copper(II) 22(4) SO3 + [O] ==> SO4 [oxidising the sulphite ion to oxide shows it to be reduced to copper atoms] the sulphate ion] The loss or removal of electrons from an atom, ion or molecule e.g. 2+

-

(1) Fe ==> Fe iron(II) ion] 2+

+ 2e [iron atom loses 2 electrons to form the

3+

(2) Fe ==> Fe the iron(III) ion] -

The gain or addition of electrons by an atom, ion or molecule e.g. ... 2+

-

(1) Cu + 2e ==> Cu [the copper(II) ion gains 2 electrons to form neutral copper atoms e.g. in electrolysis or metal displacement reactions)

-

+ e [the iron(II) ion loses 1 electron to form

3+

-

2+

(2) Fe + e ==> Fe [the iron(III) ion gains an electron and is reduced to the iron(II) ion] -

(3) 2Cl ==> Cl2 + 2e [the loss of electrons by chloride ions to + (3) 2H + 2e ==> H2 [hydrogen ions gain electrons to form form chlorine molecules in electrolysis of chlorides or

halogen displace]

neutral hydrogen molecules] -

-

(4) Cl2 + 2e ==> 2Cl [chlorine molecules gain electrons to form chloride ions An oxidising agent is the species that gives the oxygen or removes the electrons

A reducing agent is the species that removes the oxygen or acts as the electron donor

REDOX REACTIONS - in a reaction overall, reduction and oxidation must go together Redox reaction analysis based on the oxygen definitions

1. copper(II) oxide + hydrogen ==> copper + water o o

2.

3. 4.

5.

CuO(s) + H2(g) ==> Cu(s) + H2O(g) copper oxide reduced to copper, hydrogen is oxidised to water, hydrogen is the reducing agent (removes O from CuO) and copper oxide is the oxidising agent (donates O to hydrogen) iron(III) oxide + carbon monoxide ==> iron + carbon dioxide o Fe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g) o the iron(III) oxide is reduced to iron, the carbon monoxide is oxidised to carbon dioxide, CO is the reducing agent (O remover from Fe2O3) and the Fe2O3 is the oxidising agent (O donator to CO) 2NO(g) + 2CO(g) ==> N2(g) + 2CO2(g) o nitrogen monoxide is reduced to nitrogen, carbon monoxide is oxidised to carbon dioxide, CO is the reducing agent, NO is the oxidising agent iron(III) oxide + aluminium ==> aluminium oxide + iron o Fe2O3(s) + 2Al(s) ==> Al2O3(s) + 2Fe(s) o iron(III) oxide is reduced and is the oxidising agent, aluminium is oxidised and is the reducing agent - incidentally, this is the 'thermit' reaction! For more details of some of these and similar reactions see ... o Metal Reactivity notes and Metal Extraction notes. Redox reaction analysis based on the electron definitions

1. magnesium + iron(II) sulphate ==> magnesium sulphate + iron o o

2.

3.

4.

5.

Mg(s) + FeSO4(aq) ==> MgSO4(aq) + Fe(s) [this is the 'ordinary' equation but this does not really show what happens in terms of atoms, ions and 2electrons, so we use ionic equations like the one shown below. The sulphate ion SO4 (aq) is called a spectator ion, because it doesn't change in the reaction and can be omitted from the ionic equation. No electrons show up in the full equations because electrons lost by x = electrons gained by y!!] 2+ 2+ o Mg(s) + Fe (aq) ==> Mg (aq) + Fe(s) o the magnesium atom loses 2 electrons (oxidation) to form the magnesium ion, the iron(II) ion gains 2 2+ electrons (reduced) to form iron atoms. Mg is the reducing agent and the Fe is the oxidising agent](2)(i) zinc + hydrochloric acid ==> zinc chloride + hydrogen zinc + hydrochloric acid ==> zinc chloride + hydrogen o Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g) o the chloride ion Cl is the spectator ion] + 2+ o Zn(s) + 2H (aq) ==> Zn (aq) + H2(g) o Zinc atoms are oxidised to zinc ions by electron loss and zinc is the reducing agent, hydrogen ions are the oxidising agent (gaining the electrons) and are reduced to form hydrogen molecules] copper + silver nitrate ==> silver + copper(II) nitrate o Cu(s) + 2AgNO3(aq) ==> 2Ag(s) + Cu(NO3)2(aq) o the nitrate ion NO3 is the spectator ion + 2+ o Cu(s) + 2Ag (aq) ==> 2Ag(s) + Cu (aq) o copper atoms are oxidised by the silver ion, electrons are transferred from the copper atoms to the silver ions, which are reduced. Silver ions are the oxidising agent and copper atoms are the reducing agent chlorine + potassium iodide ==> potassium chloride + iodine o Cl2(aq) + 2KI(aq) ==> 2KCl(aq) + I2(aq) o or Cl2(aq) + 2I (aq) ==> 2Cl (aq) + I2(aq) o a halogen displacement reaction, the more reactive chlorine displaces the less reactive iodine. o Chlorine is reduced by electron gain and the iodide ions are oxidised by electron loss. For more details of similar reactions see the Metal Reactivity, Metal Extraction and Group 7 The Halogens notes.

POLYMERISATION means joining many small molecules called monomers into a long molecules of many units called a polymer and there are two principal types of polymerisation process (1) Addition polymers are formed by (e.g. alkene) monomers adding together and forming no other products except the polymer e.g. two examples of addition polymerisation are ethene ==> poly(ethene) phenylethene ==> poly(phenylethene), old name polystyrene (2) Condensation polymers are formed by one or more monomers add together, forming the polymer BUT in forming the polymer small molecules are eliminated 'between' the monomers e.g. two examples of condensation polymerisation are ... dicarboxylic acid + diol ==> polyester + water diamine + dicarboxylic acid ==> nylon + water (1) Example equations showing addition polymerisation

(Ex. 1a) formation of poly(ethene) or 'polythene' from polymerising ethene to form an addition polymer. No other molecule is formed - just simple addition polymerisation.

(Ex. 1b) formation of poly(chloroethene) or 'PVC' from polymerizing chloroethene to form an addition polymer. No other molecule is formed - just simple addition polymerization.

For more examples and details of addition polymers see Useful Oil Products Part 7 (2) Example equation illustrating condensation polymerisation

+ small molecules eliminated In the case of Nylon, for each 'red' monomer - 'blue' monomer, a link is formed at each end of each monomer molecule by eliminating a water molecule e.g. where [R] = 'rest of molecule' a single link formation reaction can be shown as

(Example of 2) representation of a Nylon made from two different monomers (shown as red and green + linking atoms) joining by eliminating a small molecule between the two monomers, therefore Nylon is a condensation polymer.

[R]-COOH + HO-[R] ==> [R]-CO-O-[R] + H2O For more examples and details of condensation polymers see Useful Oil Products Part 11

NEUTRALISATION   





NEUTRALISATION usually involves mixing an acid (pH 7 if soluble) which react to form a neutral salt solution of around pH7 Two situations are common: (1) Water soluble bases, called alkalis and often hydroxides, are mixed with a soluble acid such as hydrochloric, citric, sulphuric or nitric acid. o acid + base/alkali ==> salt + water o e.g. sodium hydroxide + hydrochloric acid ==> sodium chloride + water  NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) o certain carbonates like sodium carbonate, are also soluble to form alkaline solutions, and they will be similarly neutralised with 'fizzing' as carbon dioxide is formed as a 3rd product o e.g. sodium carbonate + hydrochloric acid ==> sodium chloride + water + carbon dioxide  Na2CO3(aq) + 2HCl(aq) ==> 2NaCl(aq) + H2O(l) + CO2(g) (2) Dissolving a water insoluble base (often an oxide) in an acid o e.g. copper oxide + sulphuric acid ==> copper sulphate + water  CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l) o the acid can also be neutralised with a metal or a carbonate to give a salt solution o metal + acid ==> salt + hydrogen (this is also a redox reaction) o e.g. zinc + hydrochloric acid ==> zinc chloride + hydrogen  Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g) o insoluble carbonate + acid ==> (often soluble) salt + water + carbon dioxide o e.g. magnesium carbonate + sulphuric acid ==> magnesium sulphate + water + carbon dioxide  MgCO3(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) + CO2(g) More details of these types of reactions involving acids and bases-alkalis

PRECIPITATION REACTIONS  





A precipitation reaction is generally defined as 'the formation of an insoluble solid either by mixing two solutions or bubbling a gas into a solution. The silver halide salts are used in photography and can be made by precipitation on mixing solutions of two soluble salts e.g. o silver nitrate + potassium chloride ==> silver chloride + potassium nitrate o AgNO3(aq) + KCl(aq) ==> AgCl(s) + KNO3(aq) + o Ag (aq) + Cl (aq) ==> AgCl(s) +  is the ionic equation (NO3 and K are spectator ions - they don't take part in the reaction) Insoluble barium sulphate can be made in a similar manner by mixing dilute sulfuric acid with barium chloride solution o barium chloride + sulphuric acid ==> barium sulphate + hydrochloric acid o BaCl2(aq) + H2SO4(aq) ==> BaSO4(s) + 2HCl(aq) 2+ 2o Ba (aq) + SO4 (aq) ==> BaSO4(s) +  is the ionic equation (Cl and H are spectator ions - they don't take part in the reaction) Calcium carbonate is precipitated when carbon dioxide gas is bubbled into calcium hydroxide solution o calcium hydroxide + carbon dioxide ==> calcium carbonate + water o Ca(OH)2(aq) + CO2(g) ==> CaCO3(s) + H2O(l)

Reversible Reactions 

Generally speaking most chemical reactions are irreversible, that mean they go 100% one way to the products and that's that! e.g. magnesium fizzing away in hydrochloric acid to form magnesium chloride and hydrogen. There

  

is no feasible way of reacting hydrogen and magnesium chloride to re-form magnesium metal and hydrochloric acid! However a reversible reaction is a chemical reaction in which the products can be converted back to reactants under suitable conditions and there are quite a few examples of them. A reversible reaction is shown by the sign i.e. a half-arrow to the right (forward reaction) and a half-arrow to the left (backward reaction). As pointed out above, most reactions are not reversible and have the usual complete arrow right i.e. in the direction the reaction goes 100%.

only pointing to the

Two examples of reversible reactions are given below: (a) The thermal decomposition of ammonium chloride On heating strongly, the white solid ammonium chloride, decomposes into a mixture of two colourless gases - ammonia and hydrogen chloride. On cooling the reaction is reversed and solid ammonium chloride reforms. Ammonium chloride + heat NH4Cl(s)

ammonia + hydrogen chloride NH3(g) + HCl(g)

(b) The thermal decomposition of hydrated copper(II) sulphate  

On heating the blue solid, hydrated copper(II) sulphate, steam is given off and the white solid of anhydrous copper(II) sulphate is formed (left to right reaction). When the white solid is cooled and water added, blue hydrated copper(II) sulphate is reformed (right to left reaction). blue hydrated copper(II) sulphate + heat

white anhydrous copper(II) sulphate + water

CuSO4.5H2O(s)

CuSO4(s) + 5H2O(g)

For more details of reversible reactions for IGCSE/GCSE science-chemistry Reversible reactions and Chemical Equilibrium (more details for GCSE)      

When a reversible reaction occurs in a closed system an equilibrium is formed, in which the original reactants and products formed coexist. In an equilibrium there is a state of balance between the concentrations of the reactants and products. At equilibrium the rate at which the reactants change into products is exactly equal to the rate at which the products change back to the original reactants. The result is that that the concentrations of the reactants and products remain the same BUT the reactions don't stop! However the relative amounts of the reactants and products depend on the reaction conditions e.g. the temperature and pressure. You can change the conditions to favour a particular reaction direction e.g. in a limekiln the carbon dioxide is vented out so all the limestone changes to lime avoiding an. (a) The formation of calcium oxide (lime) from calcium carbonate (limestone) calcium carbonate (limestone) CaCO3(s)

calcium oxide (lime) + carbon dioxide CaO(s) + CO2(g)

(b) The formation of ammonia nitrogen + hydrogen N2(g) + 3H2(g)

ammonia 2NH3(g)

Hydration and dehydration (often reversible under the right conditions)

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Hydration means the addition of water or combining with water. Dehydration means the losing or removal of water. Example 1: On heating the blue solid, hydrated copper(II) sulphate, steam is given off and the white solid of anhydrous copper(II) sulphate is formed (left to right reaction is a dehydration, water lost). When the white solid is cooled and water added, blue hydrated copper(II) sulphate is reformed (right to left reaction is a hydration, water gained). o blue hydrated copper(II) sulphate + heat white anhydrous copper(II) sulphate + water  CuSO4.5H2O(s) CuSO4(s) + 5H2O(g)  The water in the blue crystals (the 5H2O) is called water of crystallisation and becomes part of the crystal structure when a concentrated solution of copper sulphate is slowly evaporated to allow crystallisation e.g. in a salt preparation.  The left to right dehydration reaction also happens if blue hydrated copper sulphate is treated with cold/warm concentrated sulphuric acid, which is a powerful dehydrating agent (see sugar reaction further down in this section).  Summing up for this example:  From left to right is dehydration and from right to left is hydration and this connection will always be so for this type of reversible reaction. Example 2: If the alcohol ethanol is heated with concentrated sulphuric acid it is dehydrated to form the alkene gas ethene plus water. The same reaction happens if ethanol vapour is passed over very hot aluminium oxide which also catalyses the dehydration of the ethanol. o ethanol ==> ethene + water o a dehydration reaction - also a particular type of elimination reaction

o ==> + H2O Example 3: If ethene gas is dissolved in concentrated sulphuric acid and diluted with water, on gentle boiling the alcohol ethanol is formed (the reverse of example 2). The same reaction occurs if ethene gas and water vapour are o passed over a silica gel-phosphoric acid catalyst at 300 C. In both cases the ethene is hydrated to form ethanol by water addition. o ethene + water ==> ethanol o a hydration reaction - also a particular type of an addition reaction

o + H2O ==> Example 4: If white crystalline sugar is heated with concentrated sulphuric acid a spongy mass of carbon is formed. o The acid dehydrates the sugar, i.e. it removes the equivalent of 'H2O' and leaves the 'C' atoms!  e.g. glucose ==> carbon + water  C6H12O6 ==> 6C + 6H2O  a dehydration reaction - the removal or elimination of water

DISPLACEMENT REACTIONS (usually with a 'reactivity series')     

Displacement essentially means one part of a compound is replaced by another. The reactants and products are similar BUT one 'bit' has been swapped with another, a sort of 'swap around'. Displacement reactions occur when a more reactive metal displaces a less reactive metal from one of its compounds e.g. from the oxide, sulphate, chloride or nitrate. Displacement reactions also occur when a more reactive non-metal displaces a less reactive non-metal from one of its compounds e.g. from the metal salt. These experiments can be used to establish a so called 'Reactivity Series'.

Non-metal displacements

Metal displacements 

     

e.g. the spectacular very exothermic 'thermit reaction' which shows aluminium is more reactive than iron. When a mixture of aluminium and iron(III) oxide is ignited with a magnesium fuse (high activation energy), the mixture burns furiously with a shower of sparks to leave a red hot blob of iron and a white as of aluminium oxide ... iron(III) oxide + aluminium ==> aluminium oxide + iron Fe2O3(s) + 2Al(s) ==> Al2O3(s) + 2Fe(s) or more reactive copper 'gently' displaces silver from silver nitrate solution to make silvery plated copper copper + silver nitrate ==> copper(II) nitrate + silver Cu(s) + 2AgNO3(aq) ==> 2Ag + Cu(NO3)2(aq) more details in Metal Reactivity Notes

      

e.g. halogen displacement, more reactive chlorine displaces bromine from potassium bromide chlorine + potassium bromide ==> potassium chloride + bromine Cl2(aq) + 2KBr(aq) ==> 2KCl(aq) + Br2(aq) or more reactive bromine displaces iodine from sodium iodide bromine + sodium iodide ==> sodium bromide + iodine Br2(aq) + 2NaI(aq) ==> 2NaCl(aq) + I2(aq) more details in KS4 Science GCSE/IGCSE Chemistry Notes on Group 7 Halogen Notes

Other processes and reactions Galvanising Haber Process

Contact Process

Double decomposition

This means to coat iron or steel with a layer of zinc to stop it rusting (more details on Metal Reactivity page) The synthesis of ammonia by combining nitrogen and hydrogen using high temperature, pressure and an iron catalyst. (all the details)     

Double decomposition is chemical reaction that takes place between two compounds, in which the first part of one compound combines with the second part of another compound. The bits left over combine to form the second compound. One of the compounds is usually insoluble.    

Catalytic Conversion (car exhaust)



 Esterification

Part of the manufacture of sulphuric acid from: (Full details of Contact Process) sulphur ==> sulphur dioxide ==V catalyst==> sulphur trioxide* ==> sulphuric acid. S(s) + O2 (g) ==> SO2 (g), * 2SO2 (g) + O2 (g) ==> 2SO3 (g), followed by, indirectly, SO3 + H2O ==> H2SO4 where the sulfur trioxide combines with water to form sulfuric acid..



e.g. if you mix solutions of sodium chromate with lead nitrate you get a yellow precipitate of lead chromate and sodium nitrate is left in solution. sodium chromate + lead nitrate ==> lead chromate + sodium nitrate Na2CrO4(aq) + Pb(NO3)2(aq) ==> PbCrO4(s) + 2NaNO3(aq) This is the yellow pigment Van Gogh used in his paintings and you see it as the road markings you don't park on!

One way of reducing pollutants from a car exhaust is to use transition metal catalysts (Pt+Rh set on a heat resistant base = the catalytic converter). A platinum/rhodium catalyst converts toxic carbon monoxide and nitrogen monoxide gases into harmless carbon dioxide and nitrogen gases. 2CO (g) + 2NO (g) == Pt/Rh ==> 2CO2 (g) + N2 (g) Combining an organic carboxylic acid with an alcohol, produces an pleasant smelling ester (which are used in perfumes and flavourings) and water. o e.g. ethanoic acid + ethanol ethyl ethanoate + water

o o

Rusting

    

Substitution

   

Addition

  

+ + H2O 2 Its an equilibrium, /3 rds conversion, and the reaction is catalysed by a few drops of concentrated sulphuric acid.

Iron (or steel) corrodes more quickly than most other transition metals and readily does so in the presence of both oxygen (in air) and water to form an iron oxide. This means rusting is an oxidation reaction. iron + oxygen + water ==> hydrated iron(III) oxide 4Fe(s) + 3O2(g) + xH2O(l) ==> 2Fe2O3.xH2O(s) i.e. rust is an orange-brown solid of hydrated iron(III) oxide formed from the reaction with oxygen and water (the equation is not meant to be balanced and the amount of water x is variable, from dry to soggy!). For more details of the chemistry of rusting and its prevention go to the corrosion section on the Metal Reactivity Series page.

A substitution reaction is where one part of a molecule is replaced by something else. e.g. when chlorine reacts with methane, a hydrogen atom in methane is replaced by a chlorine atom from the chlorine molecule. methane + chlorine ==> chloromethane + hydrogen chloride CH4 + Cl2 ==> CH3Cl + HCl An addition reaction is when one molecule adds to another molecule resulting in a single product e.g. ethene + bromine ==> 1,2-dibromoethane C2H4 + Br2 ==> C2H4Br2



Chemical Synthesis     

The word synthesis in chemistry, means to build up a larger molecules from simpler molecules or atoms ... e.g. iron sulphide is synthesised by heating together iron and sulphur ... o Fe(s) + S(s) ==> FeS(s) or The Haber Synthesis of ammonia from its constituent elements ... o nitrogen + hydrogen ==> ammonia o N2(g) + 3H2(g) ==> 2NH3(g) In organic chemistry, usually at more advanced level than GCSE, complex molecules are made by doing a whole sequence of steps, modifying one molecule to another. This would be described as a multi-stage synthesis.

Deliquescent and Hygroscopic Two overlapping terms involving physical changes rather than chemical changes Deliquescent

   

Hygroscopic

  

A substance is described as showing 'deliquescence' if it absorbs so much water from the air it forms a solution i.e. dissolves in the absorbed water. Examples: the salts calcium chloride and potassium fluoride. It is the extreme case of being hygroscopic (below). At a higher level, technically it has a lower water vapour pressure than the surrounding air, so there is a net gain in water.

A substance is described as hygroscopic if it absorbs water vapour from the air to form a damp or moist solid or even a solution - see deliquescence above. These substances can be used as drying agents (dessicants) for air or liquids. e.g. the anhydrous salts calcium chloride and sodium sulphate both readily absorb water to form the hydrated salt.

SECTION 1. RELATIVE ATOMIC MASS (AR) AND ISOTOPIC MASS

1. Explaining and how to calculate the relative atomic mass RAM or Ar of an element How to calculate relative atomic mass Introduction



Every atom has its own unique relative atomic mass (RAM) based on a standard comparison or relative scale e.g. it has been based on hydrogen H = 1 amu and oxygen O = 16 amu in the past (amu = relative atomic mass unit).



The relative atomic mass scale is now based on an isotope of carbon, carbon-12, , which is given the value of 12.0000 amu. o In this standard nuclide notation, the top left number is the mass number (12) and the bottom left number is the atomic/proton number (6). In other words the relative atomic mass of an element is now based on the arbitrary value of the carbon-12 isotope being assigned a mass of 12.0000 by international agreement! o Examples are shown in the Periodic Table diagram above. o Note  (i) Because of the presence of neutrons in the nucleus, the relative atomic mass is usually at least double the atomic/proton number because there are usually more neutrons than protons in the nucleus (mass proton = 1, neutron = 1). Just scan the periodic table above and examine the pairs of numbers.  You should also notice that generally speaking the numerical difference between the atomic/proton number and the relative atomic mass tends to increase with increasing atomic number. This has consequences for nuclear stability.  (ii) For many calculation purposes, relative atomic masses are usually quoted and used at this academic level to zero or one decimal place eg.  e.g. hydrogen H = 1.0 or ~1, calcium Ca= 40.0 or ~40, chlorine Cl = 35.5, copper Cu = 63.6 or ~64, silver Ag 107.9 or ~108.  Sometimes at A level, values of relative atomic masses may be quoted to two decimal places.  Many atomic masses are known to an accuracy of four decimal places, but for some elements, isotopic composition varies depending on the mineralogical source, so four decimal places isn't necessarily more accurate! In using the symbol Ar for RAM, you should bear in mind that the letter A on its own usually means the mass number of a particular isotope and amu is the acronym shorthand for atomic mass units. However there are complications due to isotopes and so very accurate atomic masses are never whole integer numbers. Isotopes are atoms of the same element with different masses due to different numbers of neutrons. The very 12 accurate relative atomic mass scale is based on a specific isotope of carbon, carbon-12, C = 12.0000 units exactly, for most purposes C = 12 is used for simplicity.



  





For example hydrogen-1, hydrogen-2, and hydrogen-3, are the nuclide notation for the three isotopes of hydrogen, though the vast majority of hydrogen atoms have a mass of 1. When their accurate isotopic masses, and their % abundance are taken into account the average accurate relative mass for hydrogen = 1.008, but for most purposes H = 1 is good enough! o See also GCSE/IGCSE/AS Atomic Structure Notes The strict definition of relative atomic mass (Ar) is that it equals the average mass of all the isotopic atoms 1 present in the element compared to /12th the mass of a carbon-12 atom (relative isotopic mass of 12.0000). o So, in calculating relative atomic mass you must take into account the different isotopic masses of the same elements, but also their % abundance in the element. o Therefore you need to know the percentage (%) of each isotope of an element in order to accurately calculate the element's relative atomic mass. o For approximate calculations of relative atomic mass you can just use the mass numbers of the isotopes, which are obviously all integers ('whole numbers'!) e.g. in the two calculations below. o To the nearest whole number, isotopic mass = mass number for a specific isotope.

Examples of relative atomic mass calculations for GCSE/IGCSE/AS level students How do I calculate relative atomic mass?



Example 1.1 Calculating the relative atomic mass of bromine and 79 81 o bromine consists of two isotopes, 50% Br and 50% Br, calculate the Ar of bromine from the mass numbers (top left numbers). o Ar = [ (50 x 79) + (50 x 81) ] /100 = 80 o So the relative atomic mass of bromine is 80 or RAM or Ar(Br) = 80 o Note the full working shown. Yes, ok, you can do it in your head BUT many students ignore the %'s and just average all the isotopic masses (mass numbers) given, in this case bromine-79 and bromine-81. o This is the only case I know where averaging the isotopic masses



Example 1.2 Calculating the relative atomic mass of chlorine and o chlorine consists of two isotopes, 75% chlorine-35 and 25% chlorine-37, so using these two mass numbers ... o ... think of the data based on 100 atoms, so 75 have a mass of 35 and 25 atoms have a mass of 37. o The average mass = [ (75 x 35) + (25 x 37) ] / 100 = 35.5 o So the relative atomic mass of chlorine is 35.5 or RAM or Ar(Cl) = 35.5 35 37 o Note: Cl and Cl are the most common isotopes of chlorine, but, there are tiny percentages of other chlorine isotopes which are usually ignored at GCSE/IGCSE and Advanced GCE AS/A2 A level.



Example 1.3:

Examples for Advanced Level Chemistry students only How to calculate relative atomic mass with accurate relative isotopic masses Using data from modern very accurate mass spectrometers

(a) Accurate calculation of relative atomic mass (need to know and define what relative isotopic mass is) 1

Relative isotopic mass is defined as the accurate mass of a single isotope of an element compared to /12th the mass of a carbon-12 atom e.g. the accurate relative isotopic mass of the cobalt-5

is 58.9332

This definition of relative isotopic mass is a completely different from the definition of relative atomic mass, except both are 12 based on the same international standard of atomic mass i.e. 1 unit = 1/12th the mass of a carbon-12 isotope ( C). If we were to redo the calculation of the relative atomic mass of chlorine (example 1.1 above), which is quite adequate for GCSE purposes (and maybe A level too), but more accurately at A level, we might do .... chlorine is 75.77%

35

Cl of isotopic mass 34.9689 and 24.23%

37

Cl of isotopic mass 36.9658

so Ar(Cl) = [(75.77 x 34.9689) + (24.23 x 36.9658)] / 100 = 35.4527 (but 35.5 is usually ok in calculations pre-university!)

See also Mass Spectrometer and isotope analysis on the GCSE-AS(basic) Atomic Structure Notes, with further RAM calculations. (b) Calculations of % composition of isotopes It is possible to do the reverse of a relative atomic mass calculation if you know the A r and which isotopes are present. It involves a little bit of arithmetical algebra. The Ar of boron is 10.81 and consists of only two isotopes, boron-10 and boron-11 The relative atomic mass of boron was obtained accurately in the past from chemical analysis of reacting masses but now mass spectrometers can sort out all of the isotopes present and their relative abundance. If you let X = % of boron 10, then 100-X is equal to % of boron-11 Therefore Ar(B) = (X x 10) + [(100-X) x 11)] / 100 = 10.81 so, 10X -11X +1100 =100 x 10.81 -X + 1100 = 1081, 1100 - 1081 = X (change sides change sign!)

therefore X = 19 10

11

so naturally occurring boron consists of 19% B and 81% B (the data books actually quote 18.7 and 81.3, but we didn't use the very accurate relative isotopic masses)

SECTION 2. RELATIVE FORMULA MASS AND RELATIVE MOLECULAR MASS (M R)

2. How to calculate relative formula mass or relative molecular mass RFM/RMM or M r How do I calculate relative molecular mass? RMM How to calculate relative formula mass? RFM Is there any difference between RMM and RFM? Does it matter whether the compound is ionic or covalent? If all the individual atomic masses of all the atoms in a formula are added together you have calculated the relative formula mass (for ionic compounds e.g. NaCl = 58.5) or molecular mass (for covalent elements e.g. N2 = 28 or compounds e.g. C6H12O6 = 180). To be honest, the term relative formula mass can be used with any compound whether it be ionic or covalent - it just seems NOT correct to talk about the molecular mass of an ionic compound when it doesn't consist of molecules, but is that one for the purists! The shorthand Mr can be used for the formula of any element or compound and to repeat, 'it doesn't matter whether a compound is ionic or covalent'. Mr = Relative formula mass = relative molecular mass = the sum of all the atomic masses for all the atoms in a given formula NOTE: You cannot successfully calculate formula/molecular masses if you cannot read a formula correctly! Whereas relative atomic mass (section 1. Relative Atomic Mass) only applies to a single atom, anything with at least two atoms in the formula requires the term relative formula mass or relative molecular mass to be used. The most common error is to use atomic/proton numbers instead of atomic masses, unfortunately, except for hydrogen, they are different! Examples of relative formula/molecular mass calculations: How to calculate relative molecular mass = How to calculate relative formula mass Recap: Molecular/formula mass = total of all the atomic masses of all the atoms in the molecule/compound. 

 

Molecular/formula mass calculation Example 2.1 o The diatomic molecules of the elements hydrogen H2 and chlorine Cl2 o relative atomic masses, Ar: H = 1, Cl = 35.5 o Formula masses, RMM or Mr  are for hydrogen H2 = 2 x 1 = 2  and for chlorine Cl2 = 2 x 35.5 = 71 respectively. Molecular/formula mass calculation Example 2.2 o The element phosphorus consists of P4 molecules. o RMM or Mr of phosphorus = 4 x its atomic mass = 4 x 31 = 124 Molecular/formula mass calculation Example 2.3: The compound water H2O o relative atomic masses are H=1 and O=16











o RMM or Mr = (1x2) + 16 = 18 (molecular mass of water) Molecular/formula mass calculation Example 2.4 o The compound sulphuric acid H2SO4 o relative atomic masses are H=1, S=32 and O=16 o RMM or Mr = (1x2) + 32 + (4x16) = 98 (molecular mass of sulphuric acid) Molecular/formula mass calculation Example 2.5 o The compound calcium hydroxide Ca(OH)2 (ionic) o relative atomic masses are Ca=40, H=1 and O=16 o RMM or Mr = 40 + 2 x (16+1) = 74 Molecular/formula mass calculation Example 2.6 3+ 2o The ionic compound aluminium oxide (Al )2(O )3 or just the plain formula Al2O3, o but it makes no difference to the calculation of relative formula mass or relative molecular mass. o relative atomic masses are Al = 27 and O = 16 o so the formula mass RFM or Mr = (2 x 27) + (2 x 16) = 102 Molecular/formula mass calculation Example 2.7 o Calcium phosphate is also ionic but a more tricky formula to work out! 2+ 3o (Ca )3(PO4 )2 or Ca3(PO4)3, but it makes no difference to the calculation of relative formula mass or relative molecular mass. o atomic masses: Ca = 40, P = 31, O =16 o RFM or Mr = (3 x 40) + 3 x {31 + (4 x 16)} = (120) + (3 x 95) = 405 Molecular/formula mass calculation Example 2.7 o Glucose C6H12O6 o atomic masses: C = 12, O= 16, H = 1 o Molecular mass of glucose Mr(C6H12O6) = (6 x 12) + (12 x 1) + (6 x 16) = 180

SECTION 3. LAW OF CONSERVATION OF MASS AND SIMPLE REACTING MASS CALCULATIONS

3. Law of Conservation of mass calculations     

What is the Law of Conservation of Mass? When elements and compounds react to form new products, mass cannot be lost or gained. "The Law of Conservation of Mass" definition states that mass cannot be created or destroyed, but changed into different forms. So, in a chemical change, the total mass of reactants must equal the total mass of products. By using this law, together with atomic and formula masses, you can calculate the quantities of reactants and products involved in a reaction and the simplest formula of a compound o See also Section 5. which shows how to use this law to get to a compound's formula

NOTE: (1) the symbol equation must be correctly balanced to get the right answer! (2) There are good reasons why, when doing a real chemical preparation-reaction to make a substance you will not get 100% of what you theoretically calculate. 





Law of conservation of mass calculation Example 3.1 o Magnesium + Oxygen ==> Magnesium oxide o 2Mg + O2 ==> 2MgO (atomic masses required: Mg=24 and O=16) o think of the ==> as an = sign, so the mass changes in the reaction are: o (2 x 24) + (2 x 16) = 2 x (24 + 16) o 48 + 32 = 2 x 40 and so 80 mass units of reactants = or produces 80 mass units of products (you can work with any mass units such as g, kg or tonne (1 tonne = 1000 kg) Law of conservation of mass calculation Example 3.2 o iron + sulphur ==> iron sulphide (see the diagram at the top of the page!) o Fe + S ==> FeS (atomic masses: Fe = 56, S = 32) o If 59g of iron is heated with 32g of sulphur to form iron sulphide, how much iron is left unreacted? (assuming all the sulphur reacted) o From the atomic masses, 56g of Fe combines with 32g of S to give 88g FeS. o This means 59 - 56 = 3g Fe unreacted. Law of conservation of mass calculation Example 3.3 o When limestone (calcium carbonate) is strongly heated, it undergoes thermal decomposition to form lime (calcium oxide) and carbon dioxide gas. o CaCO3 ==> CaO + CO2 (relative atomic masses: Ca = 40, C = 12 and O = 16) o Calculate the mass of calcium oxide and the mass of carbon dioxide formed by decomposing 50 tonnes of calcium carbonate.

o o o o o o

(40 + 12 + 3x16) ==> (40 + 16) + (12 + 2x16) 100 ==> 56 + 44 scaling down by a factor of two gives 50 ==> 28 + 22 so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of carbon dioxide gas.

SECTION 4. % COMPOSITION BY MASS OF A COMPOUND

4. Method of calculating the % percentage by mass of the elements in a compounds The 'percent' % by mass composition of a compound in terms of its constituent elements is calculated in three easy steps Chemistry calculations 4. How do you calculate the percent (%) by mass of an element in a compound formula? How do you calculate the percent (%) by mass of water or an ion in a compound formula? (i) Calculate the formula or molecular mass of the compound (ii) Calculate the mass of the specified element (for its %) in the compound, taking into account the number of atoms of the element in the compound formula (iii) Calculate (ii) as a percentage of (i) Percentage of an element Z in a compound = 100 x atomic mass of Z x number of atoms of Z in the compound formula / formula mass of the compound containing atoms of Z % by mass of Z = 100 x Ar(Z) x atoms of Z / M r(compound) 





Calculation of % composition Example 4.1 o Calculate the % of copper in copper sulphate, CuSO4 o Relative atomic masses: Cu = 64, S = 32 and O = 16 o relative formula mass = 64 + 32 + (4x16) = 160 o only one copper atom of relative atomic mass 64 o % Cu = 100 x 64 / 160 o = 40% copper by mass in the compound  Note that similarly, you can calculate the % of the other elements in the compound e.g.  % sulfur = (32/160) x 100 = 20% S  % oxygen = (64/160) x 100 = 40% O  Also note that if you haven't made any errors, they should add up to 100% ! Calculation of % composition Example 4.2 o Calculate the % of oxygen in aluminium sulphate, Al 2(SO4)3 o Relative atomic masses: Al = 27, S = 32 and O = 16 o relative formula mass = 2x27 + 3x(32 + 4x16) = 342 o there are 4 x 3 = 12 oxygen atoms, each of relative atomic mass 16, o giving a total mass of oxygen in the formula of 12 x 16 = 192 o % O = 100 x 192 / 342 = 56.1% oxygen by mass in aluminium sulphate Calculation of % composition Example 4.3 o The next two examples extend the idea of % element composition to include % composition of part of a compound, in these cases water in a hydrated salt and the sulfate ion in a potassium salt. o Calculate the % of water in hydrated magnesium sulphate MgSO 4.7H2O o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1 o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246 o 7 x 18 = 126 is the mass of water o so % water = 100 x 126 / 246 = 51.2 % H2O

o



Note: The determination and calculation of the formula of a hydrated salt like MgSO4.7H2O is covered in Calculations section 14.4. Calculation of % composition Example 4.4 o Calculate the percentage by mass, of sulfate ion in sodium sulfate o formula of sodium sulfate Na2SO4, atomic masses: Na = 23, S = 32, O = 16 o Formula mass Na2SO4 = (2 x 23) + 32 + (4 x 16) = 142 2o Formula mass of sulfate ion SO4 (or just SO4 will do for the calculation) = 32 + (4 x 16) = 96 o Therefore % sulfate ion in sodium sulfate = (96/142) x 100 = 67.6% SO4

SECTION 5. EMPIRICAL FORMULA FROM REACTING MASSES

5. Empirical formula and formula mass from reacting masses (easy start, no moles!) The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound. The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound. (see section 3. for some simpler examples). Here the word 'empirical' means from experimental data. Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule. The molecular formula and empirical formula can be different or the same. They are the same if the molecular formula cannot be simplified on a whole number basis. Examples where molecular formula = empirical formula e.g. sodium sulfate Na2SO4 and propane C3H8 You can simplify the atomic ratios 2 : 1 : 4 or 3 : 8 to smaller whole number (integer) ratios Examples of where molecular formula and empirical formula are different e.g. butane molecular formula C4H10, empirical formula C2H5 numerically, the empirical formula of butane is 'half' of its molecular formula 4 : 10 ==> 2 : 5 glucose molecular formula C6H12O6, empirical formula CH2O 1

numerically, the empirical formula of glucose is ' /6th' of the molecular formula 6 : 12 : 6 ==> 1 : 2 : 1 Where the empirical formula and molecular formula are different, you need extra information to deduce the molecular formula from the empirical formula (see link below). For more advanced students see Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition) The following examples illustrate the ideas using numbers more easily appreciated than in real experiments. In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and a more 'tricky' mole calculation method is required than shown here (dealt with later for higher students in section 8).

However the examples below show in principal how formulae are worked out from experiments. Any calculation method must take into account the different relative atomic masses of the elements in order to get to the actual ratio of the atoms in the formula. For example, just because 10g of X combines with 20g of Y, it does not mean that the formula of the compound is XY 2 ! 

Empirical formula calculation Example 5.1 The compound formed between lead and sulfur o It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide. o From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32) o In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass) o so the formula is simply PbS o You can set out the calculation in a simple table format, in this case the numbers are very easy to deal with! o RATIOS ...

lead (Ar = 207)

sulphur S (Ar = 32)

Comments and tips

Reacting mass

207g

32g

not the real atom ratio

atom ratio from mass / atomic mass values

207/207 = 1

32/32 = 1

simplest whole number atom ratio by trial & error

1

work out the simplest whole number ratio 1

therefore the integer simplest ratio of 1 : 1 gives the empirical formula for lead sulphide as PbS



o Empirical formula calculation Example 5.2 The empirical formula of a lead oxide o It is found that 207g of lead combined with oxygen to form 239g of a lead oxide. o From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16) o In this case, you first have to work out the amount of oxygen combined with the lead. o By simple logic from the law of conservation of mass, this is 239 - 207 = 32g o In atomic ratio terms, the 207 is equivalent to 1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16), o so the formula is simply PbO2 o Note: The mass of oxygen combined with the lead is deduced by subtracting the original mass of lead from final total mass of lead oxide. o RATIOS ...

lead (Ar = 207)

oxygen O (Ar = 16)

Comments and tips

Reacting mass

207g

239-207 = 32g

not the real atom ratio

atom ratio from mass / atomic mass values

207/207 = 1

32/16 = 2

simplest whole number atom ratio by trial & error

1

work out the simplest whole number ratio 2

therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for this lead oxide as PbO2 Its actually called lead(IV) oxide



o Empirical formula calculation Example 5.3 The empirical formula of aluminium sulfide o It is found that 54g of aluminium forms 150g of aluminium sulphide. o Work out the formula of aluminium sulphide. (Relative atomic masses: Al = 27 and S = 32).

o o o o o

Amount of sulphur combined with the aluminium = 150 - 54 = 96g By atomic ratio, the 54 of aluminium is equivalent to 2 atoms of aluminium and the 96 of sulphur is equivalent to 3 atoms of sulphur. Therefore the atomic ratio is 2 to 3, so the formula of aluminium sulphide is Al2S3

RATIOS ...

aluminium (Ar = 27)

sulfur S (Ar = 32)

Comments and tips

Reacting mass

54g

150-54 = 96g

not the real atom ratio

atom ratio from mass / atomic mass values

54/27

96/32

simplest whole number atom ratio by trial & error

2

work out the simplest whole number ratio 3

therefore the simplest integer ratio of 2 : 3 gives the empirical formula for aluminium sulphide as Al2S3





o Empirical formula calculation Example 5.4: This is a much more elaborate reacting mass calculation involving solution concentrations to arrive at a formula mass. o I've used HCl and MOH as shorthand in the question and answers. o The idea is to use the analysis data to work out the atomic mass of metal M, that forms a hydroxide of formula MOH. 3 o A solution of hydrochloric contained 3.65 g/dm . 3 o A solution of a metal hydroxide of formula MOH was prepared by dissolving 5.0g of MOH in 1 dm of water (M is an unknown metal). 3 3 o 25 cm of the MOH solution required 22.3 cm of the HCl acid solution to neutralise it in a titration procedure using a pipette (MOH) and burette (HCl).  The equation for the neutralisation reaction is: MOH + HCl ==> MCl + H2O  Atomic masses: H = 1, Cl = 35.5, O = 16, M = ?  (a) Calculate the mass of MOH neutralised in each titration. 3 3  5 x 25 / 1000 = 0.125g MOH (remember 1 dm = 1000 cm )  (b) Calculate the mass of HCl reacting in each titration.  3.65 x 22.3 / 1000 = 0.0814 g HCl  (d) Calculate the formula mass of HCl  Formula mass of HCl = 1 + 35.5 = 36.5  (c) Calculate the mass of MOH that reacts with 36.5 g HCl and hence the formula mass of MOH.  If 0.125 g MOH reacts with 0.0814 g HCl  z g MOH reacts with 36.5 g HCl  solving the ratio, z = 36.5 x 0.125 / 0.0814 = 56.1 g MOH  Therefore the experimental formula mass of MOH is 56.1 (~56) because from the equation 1 HCl reacts with 1 MOH.  (d) If the metal is in the Group 1 of Alkali Metals, what is the atomic mass of M and what metal is M?  If the formula mass of MOH is 56, atomic mass of M = 56 - 1 -16 = 39  The atomic mass of potassium is 39, so M is potassium.  (So 'fictitious' MOH is really KOH, potassium hydroxide. You are likely to very familiar with another in the same group, sodium hydroxide NaOH) o Empirical formula calculation Example 5.5 From now on, questions just using the table method to work out empirical formula from more awkward numbers! In this case a compound formed between copper and chlorine. o A compound of copper contained 47.4% copper and 52.6% chlorine. o The atomic masses are: Cu = 64 and Cl = 35.5 o Think of the percentages as masses in grams to solve the empirical formula problem. o RATIOS ...

Cu

Cl

Comments and tips

Reacting mass

47.4

52.6

not the real atom ratio

atom ratio from mass /

47.4/64 = 0.74

52.6/35.5 = 1.48

work out the simplest

atomic mass values simplest whole number atom ratio by trial & error

whole number ratio

0.74/0.74 = 1.0

1.48/0.74 = 2.0

therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for copper chloride as CuCl2 Its actually called copper(II) chloride



o Empirical formula calculation Example 5.6 The empirical formula of a compound of carbon and chlorine o It was found that 0.39 g of carbon was combined with 4.61g of chlorine. o Atomic masses: C = 12 and Cl = 35.5 o RATIOS ...

C

Cl

Comments and tips

Reacting mass

0.39

4.61

not the real atom ratio

atom ratio from mass / atomic mass values

0.39/12 = 0.0325

4.61/35.5 = 0.130

simplest whole number atom ratio by trial & error

0.0325/0.0325 = 1.0

work out the simplest whole number ratio 0.130/0.0325 = 4.0

therefore the simplest ratio of 1 : 4 gives the empirical formula is CCl4 Its actually called tetrachloromethane



o Empirical formula calculation Example 5.7 The formula of a hydrocarbon. o It was that that 0.75g of carbon was combined with 0.25g of hydrogen. o Atomic masses: C = 12 and H = 1 o Calculate the empirical formula of the hydrocarbon o RATIOS ...

C

H

Comments and tips

Reacting mass

0.75g

0.25g

not the real atom ratio

atom ratio from mass / atomic mass values

0.75/12 = 0.0625

0.25/1 = 0.25

simplest whole number atom ratio by trial & error

0.0625/0.0625 = 1.0

work out the simplest whole number ratio 0.25/0.0625 = 4.0

therefore the simplest ratio gives the empirical formula for the hydrocarbon = 1 : 4, so formula is CH4 This is the simplest hydrocarbon molecule called methane (main constituent in natural gas)



o Empirical formula calculation Example 5.8 The analysis of sodium sulfate, calculating its empirical formula from the % composition by mass. o On analysis, the salt sodium sulfate was found to contain 32.4% sodium, 22.5% sulfur and 45.1% oxygen. o Atomic masses: Na = 23, S = 32 and O = 16 o Calculate the empirical formula of sodium sulfate o

RATIOS ...

Na

S

O

Comments and tips

Reacting mass

32.4

22.5

45.1

not the real atom ratio

atom ratio from mass / atomic mass values

32.4/23 = 1.41

22.5/32 = 0.70

simplest whole number atom ratio by trial & error

1.41/7 = 2.01 ~2.0

0.70/0.70 = 1.0

work out the simplest whole number ratio, in this case you have to make a reasonable judgement as to the 2.82/0.70 = 4.03 ~4.0 values of the integers 45.1/16 = 2.82

the simplest ratio gives the empirical formula for sodium sulfate = 2 : 1 : 4, formula is Na2SO4

SECTION 6. REACTING MASSES OF REACTANTS AND PRODUCTS AND ACTUAL PERCENT % YIELD, THEORETICAL YIELD AND ATOM ECONOMY CALCULATING MASSES IN REACTIONS How do we calculate mass of products formed? How do we calculate mass of reactants needed? 

Reacting mass calculation Example 6a.1 o Burning magnesium to form magnesium oxide o What mass of magnesium oxide is formed on burning 2.00g of magnesium?  e.g. by heating magnesium metal ribbon in a crucible.  2Mg + O2 ==> 2MgO o (atomic masses Mg =24, O = 16) o converting the equation into reacting masses gives ...  (2 x 24) + (2 x 16) ==> 2 x (24 + 16) o and this gives a basic reacting mass ratio of ...  48g Mg + 32g O2 ==> 80g MgO o The ratio can be used, no matter what the units, to calculate and predict quite a lot! and you don't necessarily have to work out and use all the numbers in the ratio.  BUT all units must be the same e.g. all masses in grams, all masses in kg or all masses in tonnes. o What you must be able to do is solve a ratio! o e.g. 24g Mg will make 40g MgO, why?, 24 is half of 48, so half of 80 is 40. o You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ... Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

comments

2Mg

==>

2MgO

only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

2 x 24 = 48g

==>

2 x 40 = 80g

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

1g

==>

80/48 = 1.667g

divide by 48 to scale down to 1g of Mg reactant

2 x 1 = 2g

==>

2 x 1.667 = 3.334g

then scale up to 2g of Mg reactant, x 2 factor

Therefore on burning 2g of magnesium you make 3.33g of magnesium oxide



o Reacting mass calculation Example 6a.2 o The neutralisation of sulfuric acid with sodium hydroxide. o 2NaOH + H2SO4 ==> Na2SO4 + 2H2O o (atomic masses Na = 23, O = 16, H = 1, S = 32) o mass ratio is: (2 x 40) + (98) ==> (142) + (2 x 18) = (80) + (98) ==> (142) + (36),  but pick the ratio needed to solve the particular problem  e.g. reacting mass ratio of 2NaOH : Na2SO4 is 80 (2 x 40) : 142 o (a) calculate how much sodium hydroxide is needed to make 5.00g of sodium sulphate. o from the reacting mass equation: 142g Na2SO4 is formed from 80g of NaOH o 5g Na2SO4 is formed from 5g x 80 / 142 = 2.82 g of NaOH by scaling down from 142 => 5 o You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ... Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

comments

2NaOH

==>

Na2SO4

only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

2 x 40 = 80 g

==>

142 g

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

80/142 = 0.563 g

==>

142/142 = 1.0 g

divide by 142 to scale down to 1g of Na2SO4 product

5.0 x 0.563 = 2.82g

==>

5.0 x 1.0 = 5.00g

then scale up to the 5g of Na2SO4 product by multiplying by 5

Therefore you need 2.82g of sodium hydroxide to make 5.00g of sodium sulfate

o o o

(b) calculate how much water is formed when 10g of sulphuric acid reacts with sodium hydroxide. from the reacting mass equation: 98g of H2SO4 forms 36g of H2O 10g of H2SO4 forms 10g x 36 / 98 = 3.67g of H2O by scaling down from 98 => 10 You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ... Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

H2SO4

==>

2H2O

comments

only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

98g

==>

2 x 18 = 36g

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

1g

==>

36/98 = 0.367g

divide by 98 to scale down to 1g of H2SO4 reactant

10 x 1 = 10g

==>

10 x 0.367 = 3.67g

then scale up to the 10g of H2SO4 reactant, factor x 10

Therefore 3.67g of water is formed when 10g of sulfuric acid reacts with sulfuric acid



o Reacting mass calculation Example 6a.3 o The reduction of copper(II) oxide by heating with carbon o 2CuO(s) + C(s) ==> 2Cu(s) + CO2(g) o (atomic masses Cu=64, O=16, C=12) o Formula Mass ratio is 2 x (64+16) + (12) ==> 2 x (64) + (12 + 2x16) o = Reacting mass ratio 160 + 12 ==> 128 + 44  (in the calculation, impurities are ignored) o (a) In a copper smelter, how many tonne of carbon (charcoal, coke) is needed to make 16.00 tonne of copper? o from the reacting mass equation: 12 of C makes 128 of Cu o scaling down numerically: mass of carbon needed = 12 x 16 / 128 = 1.5 tonne of C You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ... Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

comments

C

==>

2Cu

only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

12 tonne

==>

2 x 64 = 128 tonne

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

12/128 = 0.09375 tonne

==>

128/128 = 1.0 tonne

divide by 128 to scale down to 1 tonne of copper product

16 x 0.09375 = 1.50 tonne

==>

16 x 1.0 = 16 tonne

then scale up to the 16 tonne of Cu product, the factor is x 16

Therefore 1.50 tonne of carbon is needed to reduce 16 tonne of copper oxide to copper

o o o

(b) How many tonne of copper can be made from 800 tonne of copper oxide ore? from the reacting mass equation: 160 of CuO makes 128 of Cu (or direct from formula 80 CuO ==> 64 Cu) scaling up numerically: mass copper formed = 800 x 128 / 160 = 640 tonne Cu You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ... Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant

comments

and product

2CuO

==>

2Cu

only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

2 x 80 = 160 tonne

==>

2 x 64 = 128 tonne

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

1 tonne

==>

128/160 = 0.80 tonne

divide by 160 to scale down to 1 tonne of CuO reactant

800 x 1 = 800 tonne

==>

800 x 0.80 = 640 tonne

then scale by a factor of 800 for the copper oxide reactant

Therefore 640 tonne of copper can be extracted from 800 tonne of copper(II) oxide



o Reacting mass calculation Example 6a.4 o The reduction of iron oxide ore in a furnace by heating with carbon o What mass of carbon is required to reduce 20.0 tonne of iron(III) oxide ore if carbon monoxide is formed in the process as well as iron? o (atomic masses: Fe = 56, O = 16) o reaction equation: Fe2O3 + 3C ==> 2Fe + 3CO o formula mass Fe2O3 = (2x56) + (3x16) = 160 o 160 mass units of iron oxide reacts with 3 x 12 = 36 mass units of carbon o So the reacting mass ratio is 160 : 36 o So the ratio to solve is 20 : x, scaling down, x = 36 x 20/160 = 4.5 tonne carbon needed. o Note: Fe2O3 + 3CO ==> 2Fe + 3CO2 is the other most likely reaction that reduces the iron ore to iron. You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ... Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

comments

Fe2O3

+

3C

only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

160 tonne

+

3 x 12 = 36 tonne

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

160/160 = 1.0 tonne

+

36/160 = 0.225 tonne

divide by 160 to scale down to 1 tonne of iron oxide reactant

20 x 1.0 = 20.0 tonne

+

20 x 0.225 = 4.5 tonne

then scale up by factor of 20

Therefore 4.5 tonne of carbon is needed to reduce 20 tonne of the iron oxide



o Reacting mass calculation Example 6a.5 o The production of copper from a copper ore

o o o o o o

(a) Theoretically how much copper can be obtained from 2000 tonne of pure chalcopyrite ore, formula CuFeS2 ? Chalcopyrite is a copper-iron sulphide compound and one of the most important and common ores containing copper. Atomic masses: Cu = 64, Fe = 56 and S = 32 For every one CuFeS2 ==> one Cu can be extracted, formula mass of ore = 64 + 56 + (2x32) = 184 Therefore the reacting mass ratio is: 184 ==> 64 so, solving the ratio ...  2000 CuFeS2 ==> 2000 x 64 / 184 Cu = 695.7 tonne copper  This is the maximum amount of copper that can be theoretically extracted from the 'pure' ore.  In reality there are impurities in the ore (e.g. other minerals) and in the extracted molten copper. You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ... Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

CuFeS2

==>

comments

Cu

only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

184 tonne

==>

64 tonne

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

184/184 = 1.0 tonne

==>

64/184 = 0.3478 tonne

divide by 184 to scale down to 1 tonne of copper ore reactant

2000 x 1.0 = 2000 tonne

==>

2000 x 0.0.3478 = 695.6 tonne

then scale up by factor of 2000 for the initial 2000 tonne of ore

Therefore 695.6 tonne of copper can be extracted from 2000 tonne of chalcopyrite copper ore

o



(b) If only 670.2 tonne of pure copper is finally obtained after further purification of the extracted copper by electrolysis, what is the % yield of the overall process? o % yield = actual yield x 100/theoretical yield o % yield = 100 x 670.2 / 695.7 = 96.3% o More on % yield and atom economy in calculations section 14. o Reacting mass calculation Example 6a.6 o Calculating the theoretical yield of iron from an impure iron oxide ore. o A sample of magnetite iron ore contains 76% of the iron oxide compound Fe 3O4 and 24% of waste silicate minerals. o (a) What is the maximum theoretical mass of iron that can be extracted from each tonne (1000 kg) of magnetite ore by carbon reduction? [ Atomic masses: Fe = 56, C = 12 and O = 16 ] o The reduction equation is: Fe3O4 + 2C ==> 3Fe + 2CO2 o Before doing the reacting mass calculation, you need to do simple calculation to take into account the lack of purity of the ore. o 76% of 1 tonne is 0.76 tonne (760 kg). o For the reacting mass ratio: 1 Fe3O4 ==> 3 Fe (you can ignore rest of equation) o Therefore in reacting mass units: (3 x 56) + (4 x 16) ==> 3 x 56 o so, from the reacting mass equation: 232 Fe3O4 ==> 168 Fe o 0.76 Fe3O4 tonne ==> x tonne Fe o solving the ratio, x = 0.76 x 168/232 = 0.55 o = 0.55 tonne Fe (550 kg)/tonne (1000 kg) of magnetite ore o o (b) What is the atom economy of the carbon reduction reaction? o You can use some of the data from part (a). o % atom economy = 00 x total mass of useful product / total mass of products

o o o o o o o





Because of the law on conservation of mass, total mass reactants = total mass of reactants % atom economy = 100 x total mass of useful product / total mass of products It doesn't matter which version you use for the atom economy calculation, you should get the same answer! = 100 x 168 / (232 + 2x12) = 100 x 168/256 = 65.6% More on % yield and atom economy in calculations section 14. (c) Will the atom economy be smaller, the same, or greater, if the reduction involves carbon monoxide (CO) rather than carbon (C)? explain? o The atom economy will be smaller because CO is a bigger molecular/reactant mass than C and 4 molecules would be needed per 'molecule' of Fe3O4, so the mass of reactants is greater for the same product mass of iron (i.e. bottom line numerically bigger, so % smaller). This is bound to be so because the carbon in CO is already chemically bound to some oxygen and can't remove as much oxygen as carbon itself. o Fe3O4 + 4CO ==> 3Fe + 4CO2 o so the atom economy = 100 x 168 / (232 + 4x28) = 48.8 % o Note reactants mass (232 + 4x28) = (3x56 + 4x44) products mass o More on % yield and atom economy in calculations section 14. o Reacting mass calculation Example 6a.7 o Deposition from hard water samples o On analysis, a sample of hard water was found to contain 0.056 mg of calcium 3 hydrogen carbonate per cm (0.056 mg/ml). o If the water is boiled, calcium hydrogencarbonate Ca(HCO 3)2, decomposes to give a precipitate of calcium carbonate CaCO3, water and carbon dioxide. o (a) Give the symbol equation of the decomposition complete with state symbols.  Ca(HCO3)2(aq) ==> CaCO3(s) + H2O(l) + CO2(g) 3 3 o (b) Calculate the mass of calcium carbonate in grammes deposited if 2 litres (2 dm , 2000 cm or ml) is boiled in a kettle.  [ atomic masses: Ca = 40, H = 1, C = 12, O = 16 ]  the relevant ratio is based on: Ca(HCO3)2 ==> CaCO3  The formula masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40 + 12 + 16x3) respectively  there the reacting mass ratio is 162 units of Ca(HCO3)2 ==> 100 units of CaCO3 3  the mass of Ca(HCO3)2 in 2000 cm (ml) = 2000 x 0.056 = 112 mg  therefore solving the ratio 162 : 100 and 112 : z mg CaCO 3  where z = unknown mass of calcium carbonate  z = 112 x 100/162 = 69.1 mg CaCO3  since 1g = 1000 mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate o (c) Comment on the result, its consequences and why is it often referred to as 'limescale'?  This precipitate of calcium carbonate will cause a white/grey deposit to be formed on the side of the kettle, especially on the heating element.  Although 0.0691 g doesn't seem much, it will build up appreciably after many cups of tea!  The precipitate is calcium carbonate, which occurs naturally as the rock limestone, which dissolved in rain water containing carbon dioxide, to give the calcium hydrogen carbonate in the first place.  Since the deposit of 'limestone' builds up in layers it is called 'limescale'. o Reacting mass calculation Example 6a.8 o This is a much more elaborate reacting mass calculation involving solution concentrations and extended ideas from the results. o In this exemplar Q I've used the formulae a lot for short-hand. 3 o A solution of hydrochloric contained 7.3 g HCl/dm . 3 o A solution of a metal hydroxide of formula MOH was prepared by dissolving 4.0g of MOH in 250 cm of water. + 3 o M is an unknown metal but it is known that the ionic formula of the hydroxide is M OH . 25cm samples of the MOH solution were pipetted into a conical flask and titrated with the hydrochloric solution using a burette and a few drops of phenolphthalein indicator. o All the MOH is neutralised as soon as the pink indicator colour disappears (i.e. the indicator becomes colourless). 3 3 o On average 19.7 cm of the HCl acid solution was required to completely neutralise 25.0 cm of the MOH solution. o [Atomic masses: H = 1, Cl = 35.5, O = 16, M = ?]  (a) Give the equation for the reaction between the metal hydroxide and the hydrochloric acid.  MOH(aq) + HCl(aq) ==> MCl(aq) + H2O(l)  You may or may not be required to give the state symbols in (), or you may be just asked to complete the equation given part of it.  (b) Calculate the mass of HCl used in each titration. 3 3 3  1 dm = 1000 cm , so in 19.7 cm of the HCl solution there will be  19.7 x 7.3 / 1000 = 0.1438 g HCl  (c) Calculate the mass of MOH that reacts with the mass of HCl calculated in (b). 3 3  25cm of the 250 cm MOH solution was used, so the mass of MOH titrated is

 

 





 25 x 4 / 250 = 0.40 g MOH (d) Calculate the formula mass of HCl.  formula mass = 1 + 35.5 = 36.5 (e) Calculate the mass in g of MOH that reacts with 36.5g of HCl and hence the formula mass of MOH.  0.1438 g HCl reacts with 0.40 g MOH  therefore 36.5g HCl reacts with z g of MOH  solving the ratio for z, z = 36.5 x 0.40 / 0.1438 = 101.5 g MOH  Since the formula mass of HCl is 36.5 and from the equation, 1 MOH reacts with 1 HCl the experimental formula mass of MOH is found to be 101.5 (f) What is the atomic mass of the metal?  Since O = 16 and H = 1, total of 17, atomic mass of M = 101.5 - 17 = 84.5 (g) From the formula information on the metal hydroxide deduce the following giving reasons:  What group of the periodic table is M likely to be from?  Group 1 Alkali Metals because they form singly charged positive ions (they readily lose their outer electron to do so).  Which metal is M likely to be?  Rubidium. The atomic mass for rubidium is quoted as 85.5 and the experimental value of 84.5 is pretty close - don't forget the experimental procedures will not be perfect, especially at GCSE level!

o Reacting mass calculation 6a.9, including calculating atom economy and percentage yield o The displacement reaction between iron and copper(II) sulfate solution o (a) Write out the equation for this reaction  You can write the equation out in several different ways, and any can be used to do a reacting mass calculation  Fe + CuSO4 ==> Cu + FeSO4 (simple 'molecular' equation)  Fe(s) + CuSO4(aq) ==> Cu(s) + FeSO4(aq) (with state symbols) 2+ 2+  Fe + Cu ==> Cu + Fe (ionic equation) 2+ 2+  Fe(s) + Cu (aq) ==> Cu(s) + Fe (aq) (ionic equation with state symbols)  o (b) Calculate the maximum amount of copper that can be displaced by 2.8g of iron?  Atomic masses: Fe = 56, Cu = 64, S = 32, O =16  From the equation 1 atom of iron displaces 1 atom of copper,  therefore 56g of iron will displace 64g of copper  therefore 2.8g of iron will displace x g of copper  by scaling down  x = 2.8 x 64/56 = 3.2 g Cu can be theoretically displaced  o (c) What is the atom economy of the reaction based on the ionic equation?  Based on atomic mass units  total mass of products = 64 + 56 = 120  mass of useful product = 64  atom economy = 100 x 64/120 = 53.3 %  More on % yield and atom economy in calculations section 14.  o (d) If only 2.7 g of pure copper was recovered from the experiment, what was the % yield?  % yield = 100 x actual yield/maximum theoretical yield  % yield = 100 x 2.7/3.2 = 84.4%  Reacting mass calculation 6a.10, including calculating atom economy and percentage yield o The displacement reaction between copper and silver nitrate solution. o The equation for this reaction is  Cu + 2AgNO3 ==> Cu(NO3)2 + 2Ag  atomic masses: Cu = 64, Ag = 108, N = 14, O = 16  o (a) What is the atom economy of the reaction?  You can add up the total mass of reactants or total mass of products, its all the same (because of the law of conservation of mass!)  Based on atomic mass units  mass of reactants = 64 + 2 x [108 + 14 + (3 x 16)] = 64 + (2 x 170) = 404  mass of useful product = 2 x 108 = 216  atom economy = 100 x 216/ = 100 x 216/404 = 53.5 %  More on % yield and atom economy in calculations section 14.  o (b) 500g of copper was used to displace silver from a silver nitrate solution.  (i) What is the maximum amount of silver that could be obtained from the process?  From the equation 1 atom Cu ==> 2 atoms silver  therefore 64g Cu ==> 2 x 108 g Ag



 64g Cu ==> 216g Ag  500 g Cu ==> x g Ag  x = 500 x 216/64 = 1687.5 g Ag   (ii) If 1.5 kg g of pure silver was extracted, what was the percentage yield of silver?  1.5 kg = 1500g  % yield = 100 x actual yield/theoretical maximum yield  % yield = 100 x 1500/1687.5 = 88.9 % yield of silver  More on % yield and atom economy in calculations section 14.  Reacting mass calculation Appendix 1 Solving Ratios For ... Ex 6a1. 2Mg + O2 ==> MgO Ex 6a2a./2b. 2NaOH + H2SO4 ==> Na2SO4 + 2H2O

SECTION 7. INTRODUCING MOLES - THE CONNECTION OF MOLES, MASS AND FORMULA MASS (a) WHAT IS THE MOLE CONCEPT? and WHAT IS ONE MOLE OF A SUBSTANCE? The mole concept is an invaluable way of solving many quantitative problems in chemistry! Its a very important way of doing chemical calculations! The theoretical basis is explained in section (b). The mole is most simply expressed as the relative 'formula mass in g' or the 'molecular mass in g' of the defined chemical 'species', and that is how it is used in most chemical calculations. The formula mass in grams = one mole of the defined substance. One mole of a substance equals the molecular mass in grams. moles of species = (actual mass of species in g) / (atomic/formula mass of species) therefore (using triangle on right if necessary) mass of species in g = moles species x atomic/formula mass of species atomic/formula mass of species = mass of species in g / moles of species Note these equations are for either an element or a compound, but, whatever, you must clearly define the chemical species you mean for any mole calculation e.g. +

-

Al metal element atom, H2O covalent molecule, an element O2 molecule, Na Cl ionic compound or just any compound formula like CuSO4 etc. etc. Mr is 'shorthand' for relative formula mass or molecular mass in amu (atomic mass units). The term relative molecular mass (sum of the atomic masses of the atoms in a single molecule of the substance) is usually applied to definite molecular species e.g. molecular mass 18 for the water molecule H2O, 17 for the ammonia molecule NH3 16 for the methane molecule CH4 and 180 for the glucose sugar molecule C6H12O6 (atomic masses for these examples H = 1, O = 16, N = 14, C = 12, S = 32, Na = 23, Cl = 35.5) The term relative formula mass (sum of the atomic masses of the atoms in a specified formula) can be used for ANY specified formula of ANY chemical substance, though it is most often applied to ionic substances. +

-

e.g. mass of 1 mole of ionic sodium chloride NaCl or Na Cl is 58.5g (from 23 + 35.5) +

2-

mass of 1 mole of ammonium sulfate (ionic salt) (NH4)2SO4 or (NH4 )2(SO4 ) = 130g or 18g (1 + 1 + 16) for the mass of 1 mole of water H2O, and 17g (14 + 1 + 1 + 1) is the mass of 1 mole of the ammonia molecule NH3

(b) THE AVOGADRO CONSTANT If you don't need to know about the Avogadro Constant, you can skip this section. Every mole of any substance contains the same number of the defined species.

The actual particle number is known and is called the Avogadro Constant, denoted NA). It is equal to 6.023 x 10

23

'defined species' per mole i.e. 6.023 x 10

23

-1

mol

This means there are that many atoms in 12g of carbon (C = 12) or that many molecules in 18g water (H2O = 1+1+16 = 18, H = 1; O = 16) *. 3

3

* This is about 18cm , so picture this number of molecules in a nearly full 20cm measuring cylinder or a 100ml 1 beaker less than /5th full! The Avogadro number is 6.023 x 10

23

= 602 300 000 000 000 000 000 000 atoms or molecules per mole!

= six hundred and two thousand and three hundred million million million 'particles' per mole ! 3

3

A thimble full of water is about 1cm , 1 mole of water = 18g and ~ 18cm because the density of water is ~1.0 g/cm 23

Therefore in a thimble full of water there are ~6.023 x 10 /18 = ~3.3 x 10

22

3

= 33 000 000 000 000 000 000 000 molecules!

= thirty three thousand million million million molecules of water! So, just think how many molecules of water are in your body! AND just think how useful the 'mole' is, to make life 'simple' in calculations! (well sort of!)

The real importance of the mole is that it allows you to compare ratios of the relative amounts of reactants and products, or the element composition of a compound, at the atomic and molecular level. 1 mole of any defined chemical species has an identical number of that species to any other defined chemical species. If you have a mole ratio for A : B of 1 : 3, it means 1 particle of A to 3 particles of B irrespective of the atomic or formula masses of A and B. It also means that you can read equations in terms of a mole ratio e.g. 2NaOH + H2SO4 ==> Na2SO4 + 2H2O can be read as 2 moles of sodium hydroxide neutralises 1 mole of sulfuric acid to form 1 mole of the salt sodium sulfate and 2 moles of water, BUT the equation can be read in terms of any molar quantities, as long as you keep the ratios the same! 1

e.g. by only taking /20th of a mole of sodium hydroxide you can deduce (yes predict!) 0.05 moles of NaOH reacts with 0.025 moles of H2SO4 to form 0.025 moles of Na2SO4 and 0.05 moles of H2O What more, since you can convert moles to mass, you can do deduce the mass of product formed or the mass of reactants need. Also, since you can go from mass to moles, you can deduce equations from measuring reacting masses. (see also section 6. for reacting masses not using moles) (c) EXAMPLES OF MOLE CALCULATIONS to illustrate the introduction section (a) 

Mole calculation Example 7.1.1 o Consider the formation of 1 mole of ammonia, NH 3, o which consists of 1 mole of nitrogen atoms combined with 3 moles of hydrogen atoms. o Or you could say 2 moles of ammonia is formed from 1 mole of nitrogen molecules (N 2) combining with 3 moles of hydrogen molecules (H2).









The latter is a better way to look at ammonia formation because nitrogen and hydrogen exist as diatomic molecules and NOT individual atoms. o N2(g) + 3H2(g) ==> 2NH3(g) o You can then think in any ratio you want e.g. 0.05 mol nitrogen combines with 0.15 mol hydrogen to form 0.10 mol of ammonia. o So, you can calculate using any mole ratio on the basis of the 1 : 3 : 2 ratio (or 1 : 3 ==> 2) of the reactants and products in the balanced symbol equation. Mole calculation Example 7.1.2 o Consider the formation of 1 or 2 moles of aluminium oxide, o Al2O3, consists of 2 moles of aluminium atoms combined with 3 moles of oxygen atoms (or 1.5 moles of O2 molecules) to form 1 mole of aluminium oxide. 3  2Al(s) + /2O2(g) ==> Al2O3(s) o To avoid awkward fractions in equations you can say 4 moles of aluminium atoms combine with exactly 3 moles of oxygen molecules to form 2 moles of aluminium.  4Al(s) + 3O2(g) ==> 2Al2O3(s)  So the simplest whole number reacting mole ratio is 4 : 3 : 2 (or 4 : 3 ==> 2) Mole calculation Example 7.1.3 o How do you go from a reacting mole ratio to reacting mass ratio? o You read the equation in relative numbers of moles and convert the moles into mass.  mass moles x formula mass - see triangle on right  e.g. the formation of copper(II) chloride from copper(II) oxide and hydrochloric acid.  CuO(s) + 2HCl(aq) ==> CuCl2(aq) + H2O(l)  1 mole + 2 moles ==> 1 mole + 1 mole  Atomic masses: Cu = 64, O = 16, H = 1, Cl =35.5  Therefore ...  (64 + 16)g CuO + [2 x (1 + 35.5)]g HCl ==> [64 + (2 x 35.5)]g CuCl2 + [(2 x 1) + 16)]g H2O  80g CuO + 73g HCl ==> 135g CuCl2 + 18g H2O  So, from a mole ratio of 1 : 2 ==> 1 : 1  you get a mass ratio of 80 : 73 ==> 135 : 18 Mole calculation Example 7.1.4 o This can be useful for calculating the quantities of chemicals you need for e.g. the chemical preparation of a compound. o Using the concept of mole ratio and the exemplar reactions above ... o (a) Calculate how many grams of copper(II) oxide you need to dissolve in hydrochloric acid to make 0.25 moles of copper(II) chloride?  From the equation, 1 mole of copper oxide makes 1 mole of copper chloride,  therefore you need 0.25 moles of CuO  since mass = mass of 1 mole x formula mass  you need 0.25 x 80 = 20g of CuO o (b) What mass of aluminium metal do you need to make 0.1 moles of aluminium oxide?  4Al(s) + 3O2(g) ==> 2Al2O3(s) and the atomic mass of aluminium is 27  4 moles of aluminium makes 2 moles of aluminium oxide, (ratio 4:2 or 2:1)  therefore 0.2 moles of aluminium metal makes 0.1 moles of aluminium oxide (keeping the ratio of 2:1)  mass of aluminium metal needed = 0.2 x 27 = 5.4g of Al o Note that you can pick out the ratio you need to solve a problem - you DON'T need all the numbers of the full molar ratio, all you do is pick out the relevant ratio! For calculation purposes learn the following formula for 'Z' and use a triangle if necessary.

REMINDER For a substance 'Z' i.e. a specifically defined chemical species (1) mole of Z = g of Z / atomic or formula mass of Z, (2) or g of Z = mole of Z x atomic or formula mass of Z (3) or atomic or formula mass of Z = g of Z / mole of Z where Z represents atoms, molecules or formula of the particular element or compound defined in the question and all masses quoted in grams (g). 

Mole calculation Example 7.2.1 o How many moles of potassium ions and bromide ions are there in 0.25 moles of potassium bromide? +  1 mole of KBr contains 1 mole of potassium ions (K ) and 1 mole of bromide ions (Br ).  So there will be 0.25 moles of each ion.









 







Mole calculation Example 7.2.2 o How many moles of calcium ions and chloride ions are there in 2.5 moles of calcium chloride? 2+  1 mole of CaCl2 consists of 1 mole of calcium ions (Ca ) and 2 moles of chloride ion (Cl ).  So there will be 2.5 x 1 = 2.5 moles of calcium ions and 2.5 x 2 = 5 moles chloride ions. Mole calculation Example 7.2.3 o How many moles of lead and oxygen atoms are needed to make 5 moles of lead dioxide?  1 mole of PbO2 contains 1 mole of lead combined with 2 moles of oxygen atoms (or 1 mole of oxygen molecules O2).  So 1 x 5 = 5 mol of lead atoms and 2 x 5 = 10 mol of oxygen atoms (or 5 mol oxygen molecules) are needed. Mole calculation Example 7.2.4 o How many moles of aluminium ions and sulphate ions are in 2 moles of aluminium sulphate? 3+ 2 1 mole of Al2(SO4)3 contains 2 moles of aluminium ions (Al ) and 3 moles of sulphate ion (SO4 ).  So there will be 2 x 2 = 4 mol aluminium ions and 2 x 3 = 6 mol of sulphate ion. Mole calculation Example 7.2.5 o How many moles of chlorine gas are there in 6.5g of the gas? Ar(Cl) = 35.5)  chlorine consists of Cl2 molecules, so Mr = 2 x 35.5 = 71  moles chlorine = mass / Mr = 6.5 / 71 = 0.0944 mol Mole calculation Example 7.2.6 o How many moles of iron in 20g of the metal? (Fe = 56)  iron consists of Fe atoms, so moles iron = mass/Ar = 20/56 = 0.357 mol Fe Mole calculation Example 7.2.7 o How many grams of propane C3H8 are there in 0.21 moles of the gas? (C = 12, H = 1)  Mr of propane = (3 x 12) + (1 x 8) = 44  so g propane = moles x Mr = 0.21 x 44 = 9.24g Mole calculation Example 7.2.8 o 0.25 moles of molecule X was found to have a mass of 28g.  Calculate its molecular mass.  Mr = mass X / moles of X = 28 / 0.25 = 112 Mole calculation Example 7.2.9 o What mass and moles of magnesium chloride are formed when 5g of magnesium oxide is dissolved in excess hydrochloric acid?  reaction equation: MgO + 2HCl ==> MgCl2 + H2O  means 1 mole magnesium oxide forms 1 mole of magnesium chloride (1 : 1 molar ratio)  formula mass MgCl2 = 24+(2x35.5) = 95  MgO = 24+16 = 40  1 mole MgO = 40g  so 5g MgO = 5/40 = 0.125 mol  which means 0.125 mol MgO forms 0.125 mol MgCl2,  Mass = moles x formula mass = 0.125 x 95 = 11.9g MgCl2 Mole calculation Example 7.2.10 o What mass and moles of sodium chloride is formed when 21.2g of sodium carbonate is reacted with excess dilute hydrochloric acid?  reaction equation: Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2  means 1 mole sodium carbonate gives 2 moles of sodium chloride (1:2 ratio in equation)  Formula mass of Na2CO3 = (2x23) + 12 + (3x16) = 106  Formula mass of NaCl = 23 + 35.5 = 58.5  moles Na2CO3 = 21.2/106 = 0.2 mole  therefore 2 x 0.2 = 0.4 mol of NaCl formed.  mass of NaCl formed = moles x formula mass = 0.4 x 58.5 = 23.4g NaCl

(d) More advanced calculations using the Avogadro Constant to illustrate the introduction section (b) where you can actually calculate the number of particles in known quantity of material ! 



Avogadro constant Mole calculation Example 7.3.1 o How many water molecules are there in 1g of water, H2O ?  formula mass of water = (2 x 1) + 16 = 18 23  every mole of a substance contains 6 x 10 particles of 'it' (the Avogadro Constant).  moles water = 1 / 18 = 0.0556 23 22  molecules of water = 0.0556 x 6 x 10 = 3.34 x 10 3 3  Since water has a density of 1g/cm , it means in every cm or ml there are  33 400 000 000 000 000 000 000 individual H2O molecules or particles! Avogadro number calculation Example 7.3.2 o How many atoms of iron (Fe = 56) are there in an iron filing of mass 0.001g ?



 0.001g of iron = 0.001 / 56 = 0.00001786 mol 19  atoms of iron in the nail = 0.00001786 x 6 x 1023 = 1.07 x 10 actual Fe atoms  (10.7 million million million atoms!) Avogadro constant calculation Example 7.3.2 o (a) How many particles of 'Al2O3' in 51g of aluminium oxide?  Atomic masses: Al =27, O = 16, f. mass Al2O3 = (2x27) + (3x16) = 102  moles 'Al2O3' = 51/102 = 0.5 mol 23 23  Number of 'Al2O3' particles = 0.5 x 6 x 10 = 3 x 10 3+ o (b) Aluminium oxide is an ionic compound. Calculate the number of individual aluminium ions (Al ) 2and oxide ions (O ) in the same 51g of the substance. 3+ 2 For every Al2O3 there are two Al and three O ions.  So in 51g of Al2O3 there are ... 23 23 3+  0.5 x 2 x 6 x 10 = 6 x 10 Al ions, and 23 23 2 0.5 x 3 x 6 x 10 = 9 x 10 O ions.

(e) More advanced use of the mole and Avogadro Number concepts (for advanced level students only)    

You can have a mole of whatever you want in terms of chemical species e.g. 23 In terms of electric charge, 1 Faraday = 96500 C (coulombs) = 6 x 10 electrons If you have 2.5 moles of the ionic aluminium oxide (Al2O3) you have ... 3+ 2o 2 x 2.5 = 5 moles of aluminium ions (Al ) and 3 x 2.5 = 7.5 mol of oxide ions (O ) When you write ANY balanced chemical equation, the balancing numbers, including the un-written 1, are the reacting molar ratio of reactants and products.

Extra Advanced Level Chemistry Questions - more suitable for Advanced AS-A2 students which can be completely tackled after ALSO studying section 9 on the molar volume of gases and ANSWERS to QA7.1 QA7.1 This question involves using the mole concept and the Avogadro Constant in a variety of situations. The Avogadro Constant = 6.02 x 10

23

-1

mol

3

The molar volume for gases is 24dm at 298K/101.3kPa. Atomic masses: Al = 27, O = 16, H = 1, Cl = 35.5, Ne = 20, Na = 23, Mg = 24.3, C = 12 Where appropriate assume the temperature is 298K and the pressure 101.3kPa. Calculate .... (a) how many oxide ions in 2g of aluminium oxide? (b) how many molecules in 3g of hydrogen? 3

(c) how many molecules in 1.2 cm of oxygen? (d) how many molecules of chlorine in 3g? (e) how many individual particles in 10g of neon? (f) the volume of hydrogen formed when 0.2g of sodium reacts with water. (g) the volume of hydrogen formed when 2g of magnesium reacts with excess acid. (h) the volume of carbon dioxide formed when the following react with excess acid (1) 0.76g of sodium carbonate (2) 0.76g sodium hydrogencarbonate

3

(i) the volume of hydrogen formed when excess zinc is added to 50 cm of hydrochloric acid, concentration 0.2 mol -3 dm . 3

(j) the volume of carbon dioxide formed when excess calcium carbonate is added to 75 cm of 0.05 mol dm hydrochloric acid.

-3

SECTION 8. USING MOLES TO DEDUCE EMPIRICAL/MOLECULAR FORMULA OF A COMPOUND/MOLECULE STARTING WITH REACTING MASSES OR % COMPOSITION BY MASS

8. Using moles to calculate empirical formula deduce molecular formula of a compound/molecule starting with reacting masses or % composition by mass The basis of this method, is that a mole of defined species has the same number of defined 'particles' in it. So calculating the mole ratio of atoms in the combination, gives you the actual atomic ratio in the compound. This ratio is then expressed in the simplest whole number atomic ratio, which is the empirical formula. (it means the formula 'from experiment', see examples 1 and 2). (see section 3. for some simpler examples not using moles) However, there is one problem to resolve for covalent molecular compounds. The molecular formula is the summary of all the atoms in one individual molecule - so the molecular formula might not be the same as the empirical formula! In order to deduce the molecular formula from the empirical formula, you ALSO need to know the molecular mass of the molecule from another data source. See example 3, which also illustrates the calculation using % element composition AND the method is no different for more than two elements! The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound. Here, the word 'empirical' means derived from experimental data. Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule. The molecular formula and empirical formula can be different or the same. They are the same if the molecular formula cannot be simplified on a whole number basis. Examples where molecular formula = empirical formula e.g. sodium sulfate Na2SO4 and propane C3H8 You can simplify the atomic ratios 2 : 1 : 4 or 3 : 8 to smaller whole number (integer) ratios Examples of where molecular formula and empirical formula are different e.g. butane molecular formula C4H10, empirical formula C2H5 numerically, the empirical formula of butane is 'half' of its molecular formula 4 : 10 ==> 2 : 5 glucose molecular formula C6H12O6, empirical formula CH2O 1

numerically, the empirical formula of glucose is ' /6th' of the molecular formula 6 : 12 : 6 ==> 1 : 2 : 1 Where the empirical formula and molecular formula are different, you need extra information to deduce the molecular formula from the empirical formula (examples set out further on down the page).

Simple empirical formula calculations NOT using moles were covered in section 5. Examples of where the empirical formula is the same as the molecular formula ... water H2O, methane CH4, pentane C5H12 (these molecular formula cannot be 'simplified') Examples of where the molecular formula is different from the empirical formula in () ... e.g. ethane C2H6 (CH3), phosphorus(V) oxide P4O10 (P2O5), benzene C6H6 (CH) Three examples are set out below to illustrate all the situations and how to calculate an empirical formula and a molecular formula from reacting masses or % element composition by mass. For non-molecular compounds (i.e. inorganic ionic compounds), the empirical formula is usually, but not always, quoted as the compound formula. The relative atomic masses of the elements (Ar) are given in the tabular format method of solving the problem.

Empirical formula calculation Example 8.1: The compound formed between sodium and sulfur 1.15g of sodium reacted with 0.8g of sulphur. Calculate the empirical formula of sodium sulphide. You convert the masses to moles i.e. mass in g divided by the relative atomic mass. Since one mole of any defined substance contains the same number of particles (e.g. atoms), it means that the atomic mole ratio is also the actual ratio of atoms in the compound. The ratio is then expressed as the simplest whole ratio from which the empirical formula is derived. Apart from a 1, other numbers e.g. 2, 3 etc. should be seen as subscripts in the empirical formula. RATIOS ...

Sodium Na (Ar = 23.0)

Sulphur S (Ar = 32.0)

Comments and tips

masses

1.15g

0.80g

not the real atom ratio

moles (mass in g / Ar)

1.15 / 23 = 0.05 mol

0.8 / 32 = 0.025 mol

0.05 / 0.025 = 2

0.025 / 0.025 = 1

or 0.05 x 40 = 2

or 0.025 x 40 = 1

atom ratio = simplest whole number mole ratio by trial and error

can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio

therefore the simplest integer ratio = 2 : 1, so empirical formula for sodium sulphide = Na2S

Empirical formula calculation Example 8.2 The empirical formula of aluminium oxide 1.35g of aluminium was heated in oxygen until there was no further gain in weight. The white oxide ash formed weighed 2.55g. Deduce the empirical formula of aluminium oxide. Note: to get the mass of oxygen reacting, all you have to do is to subtract the mass of metal from the mass of the oxide formed. RATIOS ...

Aluminium Al (Ar = 27.0)

Oxygen O (Ar = 16.0)

Comments and tips

masses

1.35g

2.55 - 1.35 = 1.2g

not the real atom ratio

moles (mass in g / Ar)

1.35 / 27 = 0.05 mol

1.2 / 16 = 0.075 mol

can now divide by smallest

atom ratio = simplest whole number mole ratio by trial and error

0.05 / 0.05 = 1

0.075 / 0.05 = 1.5

(then x 2 = 2)

(then x 2 = 3)

or 0.05 x 40 = 2

or 0.075 x 40 = 3

ratio number or scale up by x factor to get simplest whole number ratio

therefore the simplest integer ratio is 2 : 3, so empirical formula for aluminium oxide = Al2O3

Empirical formula calculation Example 8.3 for an oxide of iron. 1.448g of iron was heated in air in a crucible until no further gain in weight was observed. The final mass of the iron oxide was found to be 2.001g Calculate the empirical formula of the iron oxide. Atomic masses: Fe = 56 and O = 16 The mass of oxygen combined with the iron is deduced by subtracting the original mass of iron from final total mass of iron oxide. RATIOS ...

iron (Ar = 56.0)

Oxygen O (Ar = 16.0)

Comments and tips

masses

1.448g

2.001 - 1.448 = 0.553g

not the real atom ratio

moles (mass in g / Ar)

1.448 / 56 = 0.0259 mol

0.553 / 16 = 0.0346 mol

0.0259 / 0.0259 = 1

0.0346 / 0.0259 = 1.336

(then x 3 = 3)

(then x 3 = 4.008)

3

~4

you arrive at this

stage by trial and error!

atom ratio = simplest whole number mole ratio by trial and error

can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio, in this case you have to make a reasonable judgement as to the values of the integers

therefore the simplest integer ratio = 3 : 4, so empirical formula for the iron oxide = Fe3O4

Empirical formula and molecular formula calculation Example 8.4 for a hydrocarbon compound On analysis a hydrocarbon was found to consist of 81.8% carbon and 18.2% hydrogen. Molecular ion measurements in a mass spectrometer show that the hydrocarbon has a molecular mass of 44. Treat the percentages as if they were masses in grams, and it all works out fine. RATIOS ...

carbon (Ar = 12.0)

hydrogen O (Ar = 1.0)

Comments and tips

masses

81.8

18.2

not the real atom ratio

moles (mass in g / Ar)

atom ratio = simplest whole number mole ratio by trial and error

81.8 / 12 = 6.817

18.2 / 1 = 18.2

6.817/6.817 = 1.0

18.2/6.817 = 2.67

1.0 x 2 = 2

2.670 x 2 = 5.34

1.0 x 3 = 3.0

2.670 x 3 = 8.01 ~8.0

you arrive at this

stage by trial and error!

can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio in this case you have to make a reasonable judgement as to the values of the integers

therefore the simplest integer ratio = 3 : 8, so empirical formula for the hydrocarbon = C3H8 The empirical formula mass = (3 x 12) + 8 = 44 This equals the molecular mass, therefore the molecular formula is also C3H8

Empirical formula and molecular formula calculation Example 8.5 for another hydrocarbon compound On analysis a hydrocarbon was found to consist of 83.72% carbon and 16.28% hydrogen. Molecular ion measurements in a mass spectrometer show that the hydrocarbon has a molecular mass of 86. Again, treat the percentages as if they were masses in grams, and it all works out fine. RATIOS ...

carbon (Ar = 12.0)

hydrogen O (Ar = 1.0)

Comments and tips

masses

83.72

16.28

not the real atom ratio

moles (mass in g / Ar)

83.72 / 12 = 6.977

16.28 / 1 = 16.28

atom ratio = simplest whole number mole ratio by trial and error

6.977/6.977 = 1.0

16.28/6.977 = 2.333

1.0 x 2 = 2.0

2.333 x 2 = 4.667

1.0 x 3 = 3.0

2.333 x 3 = 7.00

you arrive at this

stage by trial and error!

can now divide by smallest ratio number or scale up by x factor to get simplest whole number ratio, a bit fiddly to find the ratio on this one to deduce the empirical formula

therefore the simplest integer ratio = 3 : 7, so empirical formula for the hydrocarbon = C3H7 The empirical formula mass = (3 x 12) + 7 = 43 BUT the molecular mass is double this, so the molecular formula must be double the empirical formula Therefore the molecular formula is C6H14 of which there are many structural isomers!

Empirical formula and molecular formula calculation Example 8.6 for a carbohydrate sugar compound A carbohydrate compound e.g. a sugar, was found on analysis to contain 40.00% carbon, 6.67% hydrogen and 53.33% oxygen. The molecular mass was 150. From the information calculate the empirical formula and deduce the molecular formula.

RATIOS ...

Carbon (Ar = 12.0)

Hydrogen (Ar = 1.0)

Oxygen (Ar = 16.0)

Comments and tips

masses

40.00

6.67

53.33

just think of it as based on 100g

molar ratio (mass in g / Ar)

40.00 / 12 = 3.333 mol

6.67 / 1 = 6.67 mol

53.33 / 16.0 = 3.333 mol

atom ratio = simplest whole number mole ratio by trial and error

can now divide by smallest ratio number 3.333/3.333 = 1.0

6.67/3.333 ~2.0

3.333/3.333 = 1.0

therefore the simplest integer ratio = 1 : 2 : 1, so empirical formula for the sugar = CH2O The empirical formula mass = 12 + 2 + 16 = 30 BUT the molecular mass is 5 x this (150/30), so the molecular formula must be 5 x the empirical formula Therefore the molecular formula is C5H10O , a pentose, of which there are many structural isomers!

Empirical formula and molecular formula calculation Example 8.7 for a chloroalkane compound A chlorinated hydrocarbon compound when analysed, consisted of 24.24% carbon, 4.04% hydrogen, 71.72% chlorine. The molecular mass was found to be 99 from another experiment. Deduce the empirical and molecular formula. You can 'treat' the %'s as if they were grams, and it all works out like examples 1 and 2, i.e. based on a total mass of 100g. RATIOS ...

Carbon (Ar = 12)

Hydrogen (Ar = 1)

Chlorine (Ar = 35.5)

Comments and tips

Reacting mass or % mass

24.24

4.04

71.72

just think of it as based on 100g

molar ratio (mass in g / Ar)

24.24 / 12 = 2.02 mol

4.04 / 1 = 4.04 mol

71.72 / 35.5 = 2.02 mol

atom ratio = simplest whole number mole ratio by trial and error

can now divide by smallest ratio number 2.02 / 2.02 = 1

4.04 / 2.02 = 2

2.02 / 2.02 = 1

therefore the simplest atomic ratio = 1 : 2 : 1, so empirical formula for the chlorinated hydrocarbon = CH2Cl BUT the molecular mass is 99, and the empirical formula mass is 49.5 (12+2+35.5) AND 99 / 49.5 = 2, and so the molecular formula must be 2 x CH2Cl = C2H4Cl2 two possible structures, which cannot be distinguished by the data or calculation above 1,2-dichloroethane and 1,1-dichloroethane

SECTION 9. MOLES AND MOLAR GAS VOLUME AND AVOGADRO'S LAW

9. The molar gas volume in calculations, moles, gas volumes and Avogadro's Law 

 



 





Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules. o So the volumes have equal moles of separate particles in them. o One mole of any gas (or the formula mass in g), at the same temperature and pressure occupies the same volume . 3 3 o o This is 24dm (24 litres) or 24000 cm , at room temperature of 25 C/298K and normal pressure of 101.3 kPa or 1 atmosphere (both = RTP). 3 o o The molar volume for s.t.p is 22.4 dm (22.4 litres) at 0 C and 1atmosphere pressure. o o Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25 C/298K is usually considered the standard temperature (RTP). Some handy relationships for substance Z below: moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol) o mass of Z in g = moles of Z x atomic or formula mass of Z o atomic or formula mass of Z = mass of Z / moles of Z o 1 mole = formula mass of Z in g. gas volume of Z = moles of Z x volume of 1 mole o rearranging this equation gives ... o moles of Z = gas volume of Z / volume of 1 mole o The latter form of the equation can be used to calculate molecular mass from experimental data because  moles = mass / molecular mass  molecular mass = mass / moles  so, if you know the mass of a gas and its volume, you can work out moles of gas and then work out molecular mass.  This has been done experimentally in the past, but these days, molecular mass is readily done very accurately in a mass spectrometer. o Note (i): In the following examples, assume you are dealing with room temperature and pressure i.e. 25 C and 1 atmosphere pressure. Note (ii): o Apart from solving the problems using the mole concept (method (a) below, and reading any equations involved in a 'molar way' ... o It is also possible to solve them without using the mole concept (method (b) below). You still use the molar volume itself, but you think of it as the volume occupied by the formula mass of the gas in g and never think about moles! Molar gas volume calculation Example 9.1 o What is the volume of 3.5g of hydrogen? [Ar(H) = 1] o common thinking: hydrogen exists as H2 molecules, so Mr(H2) = 2, so 1 mole or formula mass in g = 2g o method (a)  so moles of hydrogen = 3.5/2 = 1.75 mol H2 3 3  so volume H2 = mol H2 x molar volume = 1.75 x 24 = 42 dm (or 42000 cm ) o method (b): 3  2g occupies 24 dm , so scaling up for the volume of hydrogen ... 3 3  3.5 g will have a volume of 3.5/2 x 24 = 42 dm (or 42000 cm ) Molar gas volume calculation Example 9.2 o Given the equation o MgCO3(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) +CO2(g) 3 o What mass of magnesium carbonate is needed to make 6 dm of carbon dioxide? [Ar's: Mg = 24, C = 12, O = 16, H =1 and S = 32] o method (a): 3 3  since 1 mole = 24 dm , 6 dm is equal to 6/24 = 0.25 mol of gas 3  From the equation, 1 mole of MgCO3 produces 1 mole of CO2, which occupies a volume of 24 dm .  so 0.25 moles of MgCO3 is need to make 0.25 mol of CO2  formula mass of MgCO3 = 24 + 12 + 3x16 = 84,  so required mass of MgCO3 = mol x formula mass = 0.25 x 84 = 21g o method (b):  converting the equation into the required reacting masses ..  formula masses: MgCO3 = 84 (from above), CO2 = 12 + 2x16 = 44  MgCO3 : CO2 equation ratio is 1 : 1  so 84g of MgCO3 will form 44g of CO2 3  44g of CO2 will occupy 24dm 3  so scaling down, 6 dm of CO2 will have a mass of 44 x 8/24 = 11g











 if 84g MgCO3 ==> 44g of CO2, then ...  21g MgCO3 ==> 11g of CO2 by solving the ratio, scaling down by factor of 4 Molar gas volume calculation Example 9.3 3 o 6g of a hydrocarbon gas had a volume of 4.8 dm . Calculate its molecular mass. o method (a): 3  1 mole = 24 dm , so moles of gas = 4.8/24 = 0.2 mol  molecular mass = mass in g / moles of gas  Mr = 6 / 0.2 = 30  i.e. if 6g = 0.2 mol, 1 mol must be equal to 30g by scaling up o method (b): 3  6g occupies a volume of 4.8 dm 3  the formula mass in g occupies 24 dm 3  so scaling up the 6g in 4.8 dm 3  there will be 6 x 24/4.8 = 30g in 24 dm  so the molecular or formula mass = 30 Molar gas volume calculation Example 9.4 o Given the equation ... (and Ar's Ca = 40, H = 1, Cl = 35.5) o Ca(s) + 2HCl(aq) ==> CaCl2(aq) + H2(g) o What volume of hydrogen is formed when ...  (i) 3g of calcium is dissolved in excess hydrochloric acid?  (ii) 0.25 moles of hydrochloric acid reacts with calcium? o (i) method (a):  3g Ca = 3/40 = 0.075 mol Ca  from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H 2  so 0.075 mol Ca produces 0.075 mol H2 3 3  so volume H2 = 0.075 x 24 = 1.8 dm (or 1800 cm ) o (i) method (b):  from equation 1 Ca ==> 1 H2 means 40g ==> 2g  so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H2 3  2g H2 has a volume 24 dm , so scaling down ... 3 3  0.15g H2 has a volume of (0.15/2) x 24 = 1.8 dm (or 1800 cm ) o (ii) method (a) only:  from equation: 2 moles HCl ==> 1 mole H2 (mole ratio 2:1) 3  so 0.25 mol HCl ==> 0.125 mol H2, volume 1 mole gas = 24 dm 3  so volume H2 = 0.125 x 24 = 3 dm Molar gas volume calculation Example 9.5 o Given the equation ... (and Ar's Mg = 24, H = 1, Cl = 35.5) o Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) 3 o How much magnesium is needed to make 300 cm of hydrogen gas? o method (a) 3 3  300 cm = 300/24000 = 0.0125 mol H2 (since 1 mol of any gas = 24000 cm )  from the equation 1 mole Mg ==> 1 mole H2  so 0.0125 mole Mg needed to make 0.0125 mol H2  so mass of Mg = mole Mg x Ar(Mg)  so mass Mg needed = 0.0125 x 24 = 0.3g o method (b)  reaction ratio in equation is 1 Mg ==> 1 H2,  so reacting mass ratio is 24g Mg ==> 2g H2, 3  2g H2 has a volume of 24000 cm (volume of formula mass in g)  so scaling down: mass Mg needed = 24 x (300/24000) = 0.3g Molar gas volume calculation Example 9.6 o A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g. Calculate the moles, mass and volume of carbon dioxide formed when it is thermally decomposed in the oven. Assume room temperature for the purpose of the calculation.  2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(g)  Formula mass of NaHCO3 is 23+1+12+(3x16) = 84 = 84g/mole  Formula mass of CO2 = 12+(2x16) = 44 = 44g/mole (not needed by this method) 3  or a molar gas volume of 24000 cm at RTP (definitely needed for this method)  In the equation 2 moles of NaHCO3 give 1 mole of CO2 (2:1 mole ratio in equation)  Moles NaHCO3 = 4.2/84 = 0.05 moles ==> 0.05/2 = 0.025 mol CO2 on decomposition.  Mass = moles x formula mass, so mass CO2 = 0.025 x 44 = 1.1g CO2 3  Volume = moles x molar volume = 0.025 x 24000 = 600 cm of CO2 Molar gas volume calculation Example 9.7 o What volume of carbon dioxide is formed at RTP when 5g of carbon is burned?  C(s) + O2(g) ==> CO2(g)  1 mole carbon gives 1 mole of carbon dioxide, atomic mass of carbon = 12  moles = mass / atomic mass, moles carbon = moles carbon dioxide = 5/12 = 0.417 mol

3



 1 mole of gas at RTP occupies 24 dm 3  so 0.417 mol occupies a volume of 0.417 x 24 = 10.0 dm Molar gas volume calculation Example 9.8 o What volume of carbon dioxide gas is formed at RTP if 1Kg of propane gas fuel is burned?  C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)  1 mole of propane gas gives 3 moles of carbon dioxide gas on complete combustion  1 kg = 1000g, atomic masses: C = 12, H =1  Mr(propane) = (3 x 12) + (8 x 1) = 44  moles = mass in g / molecular mass, therefore moles propane = 1000/44 = 22.73 mol  from equation molar ratio: moles carbon dioxide = 3 x moles of propane  mol propane = 3 x 22.73 = 68.18 mol 3  1 mole of gas at RTP occupies a volume of 24 dm 3  so 68.18 mol of gas occupies a volume of 68.18 x 24 = 1636 dm

SECTION 10. REACTING GAS VOLUME RATIOS, AVOGADRO'S LAW AND GAY-LUSSAC'S LAW

10. Reacting gas volume ratios of reactants or products (Avogadro's Law, Gay-Lussac's Law)

In the diagram above, if the volume on the left syringe is twice that of the gas volume in the right, then there are twice as many moles or actual molecules in the left-hand gas syringe. 





REACTING GAS VOLUMES and MOLE RATIO o Historically Gay-Lussac's Law of volumes states that 'gases combine with each other in simple proportions by volume'. o Avogadro's Law states that 'equal volumes of gases at the same temperature and pressure contain the same number of molecules' or moles of gas.  Note the mention of the mole concept - in these sorts of calculations you are essentially reading the equation as a mole ratio. o This means the molecule ratio of the equation or the relative moles of reactants and products automatically gives us the gas volumes ratio of reactants and products, if all the gas volumes are measured at the same temperature and pressure. o These calculations only apply to gaseous reactants or products AND if they are all at the same temperature and pressure. o The balanced equation can be read/interpreted in terms of either ...  (i) a gas volume ratio, obviously for gaseous species only (g), AND at the same temperature and pressure.  or (ii) a mole ratio, which applies to anything in the equation, whether (g), (l) or (s).  Note: If you have to convert from moles to volume or volume to moles, you need to know 3 o the molar volume at that temperature and pressure e.g. 24 dm (litres) at 25 C (298K) and 1 atm (101 kPa) pressure. 3  i.e. if the volume is in dm (litres) at ~ room temperature and pressure  moles of gas = Vgas/24 or Vgas = 24 x moles of gas  If the gas volume is given in cm3, then dm3 = V/1000 Reacting gas volume ratio calculation Example 10.1 o Given the equation: HCl(g) + NH3(g) ==> NH4Cl(s) o 1 mole hydrogen chloride gas combines with 1 mole of ammonia gas to give 1 mole of ammonium chloride solid. o 1 volume of hydrogen chloride will react with 1 volume of ammonia to form solid ammonium chloride 3 3 o e.g. 25cm + 25cm ==> solid product (no gas formed) 3 3 o or 400dm + 400 dm ==> solid product etc. 3 3 o so, if 50 cm HCl reacts, you can predict 50 cm of NH3 will react etc. etc. ! o The note the Reacting gas volume ratio calculation Example 10.2 o Given the equation: N2(g) + 3H2(g) ==> 2NH3(g) o 1 mole of nitrogen gas combines with 3 mols of hydrogen gas to form 2 mol of a ammonia gas.









o 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of ammonia 3 o e.g. what volume of hydrogen reacts with 50 cm nitrogen and what volume of ammonia will be formed? o The ratio is 1 : 3 ==> 2, so you multiply equation ratio numbers by 50 giving ... 3 3 3 o 50 cm nitrogen + 150 cm hydrogen (3 x 50) ==> 100 cm of ammonia (2 x 50) Reacting gas volume ratio calculation Example 10.3 o Given the equation: C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l) o Reading the balanced equation in terms of moles (or mole ratio) ... o 1 mole of propane gas reacts with 5 mols of oxygen gas to form 3 moles of carbon dioxide gas and 4 mols of liquid water. 3 o (a) What volume of oxygen is required to burn 25cm of propane, C3H8.  Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5 for burning the fuel propane. 3  so actual ratio is 25 : 5x25, so 125cm oxygen is needed. 3 o (b) What volume of carbon dioxide is formed if 5dm of propane is burned?  Theoretical reactant-product volume ratio is C3H8 : CO2 is 1 : 3 3  so actual ratio is 5 : 3x5, so 15dm carbon dioxide is formed. 1 3 o (c) What volume of air ( /5th oxygen) is required to burn propane at the rate of 2dm per minute in a gas fire?  Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5 3  so actual ratio is 2 : 5x2, so 10dm oxygen per minute is needed, 1 3  therefore, since air is only /5th O2, 5 x 10 = 50dm of air per minute is required Reacting gas volume ratio calculation Example 10.4 o Given the equation: 2H2(g) + O2(g) ==> 2H2O(l) 3 o o If 40 dm of hydrogen, (at 25 C and 1 atm pressure) were burned completely ...  a) What volume of pure oxygen is required for complete combustion?  From the balanced equation the reacting gas volume ratio is 2 : 1 for H 2 to O2 3  Therefore 20 dm of pure oxygen is required (40 : 20 is a ratio of 2 : 1).  b) What volume of air is required if air is ~20% oxygen? 1  ~20% is ~ /5, therefore you need five times more air than pure oxygen 3  Therefore volume of air needed = 5 x 20 = 100 dm of air  c) What mass of water is formed?  The easiest way to solve this problem is to think of the water as being formed as a gas-vapour.  The theoretical gas volume ratio of reactant hydrogen to product water is 1 : 1  Therefore, prior to condensation at room temperature and pressure, 3 40 dm of water vapour is formed. 3  1 mole of gas occupies 24 dm , and the relative molar mass of water is 18 g/mol  (atomic masses H = 1, O = 16, so Mr(H2O) = 1 + 1 + 16 = 18).  Therefore moles of water formed = 40/24 = 1.666 moles  Since moles = mass / formula mass  mass = moles x formula mass  mass water formed = 1.666 x 18 = 30g of H2O Reacting gas volume ratio calculation Example 10.5 3 3 o It was found that exactly 10 cm of bromine vapour (Br2(g)) combined with exactly 30 cm chlorine gas (Cl2(g)) to form a bromine-chlorine compound BrClx. o a) From the reacting gas volume ratio, what must be the value of x? and hence write the formula of the compound.  Since both reactants have the same formula, i.e. both diatomic molecules, the ratio of bromine to chlorine atoms in the compound must be 1 : 3 because the reacting gas volume ratio is 1 : 3  Therefore x must be 3, and the formula must be BrCl3 o b) Write a balanced equation to show the formation of BrClx  The reacting gas volume ratio is 1 : 3, therefore we can write with certainty that 1 mole (or molecule) of bromine reacts with 3 moles (or molecules) of chlorine, and balancing the symbol equation, results in two moles (or molecules) of the bromine-chlorine compound being formed.  Br2(g) + 3Cl2(g) ==> 2BrCl3(g) Reacting gas volume ratio calculation Example 10.6 o If 0.25 moles of ammonia is decomposed,  (a) how many moles of nitrogen and ammonia will be formed? o  (b) what volume of nitrogen and hydrogen will be formed at 25 C and 1 atm pressure?  (a) moles of products  First set out the balanced equation (which may be given in the question)  2NH3(g) ==> N2(g) + 3H2(g)  From the equation the mole ratio of reactants to products is 2 ==> 1 : 3  Therefore 0.25 mol ammonia ==> 3  0.125 mol nitrogen (0.25/2) : 0.375 mol hydrogen ( /2 x 0.25)  (b) volumes of products  First, convert the moles of ammonia into a gas volume  Vgas = mol gas x molar volume

3



 VNH3 = 0.25 x 24 = 6.0 dm  From the mole ratio OR reacting volume ratio (its all the same!) 2 ==> 1 : 3 3 3 3 3  6 dm NH3 ===> 3 dm N2 (6/2 for nitrogen) : 9 dm H2 ( /2 x 6.0 for hydrogen) Reacting gas volume ratio calculation Example 10.7

SECTION 11. MAKING UP SOLUTIONS - CONCENTRATION & MOLARITY

11. Molarity, volumes and the concentration of solutions

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Revise section 7. moles and mass before proceeding in this section 11 and eventually you may need to be familiar with the use of the apparatus illustrated above, some of which gives great accuracy when dealing with solutions and some do not. It is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution. o So we need a standard way of comparing the concentrations of solutions in. o The more you dissolve in a given volume of solvent, or the smaller the volume you dissolve a given amount of solute in, the more concentrated the solution. The concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre (reminder mole formula triangle on the right), -3 3 with units mol dm (or mol/dm )and this is called molarity, sometimes denoted in shorthand as M. 3 3 3 3 o Note: 1dm = 1 litre = 1000ml = 1000 cm , so dividing cm /1000 gives dm , which 3 is handy to know since most volumetric laboratory apparatus is calibrated in cm (or ml), but solution 3 concentrations are usually quoted in molarity, that is mol/dm (mol/litre). 3 o Concentration is also expressed in a 'non-molar' format of mass per volume e.g. g/dm Equal volumes of solution of the same molar concentration contain the same number of moles of solute i.e. the same number of particles as given by the chemical formula. You need to be able to calculate o the number of moles or mass of substance in an aqueous solution of given volume and concentration o the concentration of an aqueous solution given the amount of substance and volume of water, for this you use the equation .... (reminder molarity formula triangle on the right), 3 o (1a) molarity (concentration) of Z = moles of Z / volume in dm  you need to be able to rearrange this equation ... therefore ... 3  (1b) moles = molarity (concentration) x volume in dm and ... 3  (1c) volume in dm = moles / molarity (concentration) o You may also need to know that ... 3  (2) molarity x formula mass of solute = solute concentration in g/dm 3  and dividing this by 1000 gives the concentration in g/cm , and 3 3  (3) concentration in g/dm / formula mass = molarity in mol/dm

 both equations (2) and (3) result from equations (1) and (4), work it out for yourself. and don't forget by now you should know:  (4) moles Z = mass Z / formula mass of Z  (5) 1 mole = formula mass in grams  (reminder molarity formula triangle top right) Molarity calculation Example 11.1 3 3 o What mass of sodium hydroxide (NaOH) is needed to make up 500 cm (0.500 dm ) of a 0.500 mol -3 dm (0.5M) solution? [Ar's: Na = 23, O = 16, H = 1] o 1 mole of NaOH = 23 + 16 + 1 = 40g 3 o molarity = moles / volume, so mol needed = molarity x volume in dm o mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH o therefore mass = mol x formula mass o = 0.25 x 40 = 10g NaOH required Molarity calculation Example 11.2 3 -3 o (a) How many moles of H2SO4 are there in 250cm of a 0.800 mol dm (0.8M) sulphuric acid solution? o (b) What mass of acid is in this solution? [Ar's: H = 1, S = 32, O = 16] 3  (a) molarity = moles / volume in dm , rearranging equation for the sulfuric acid 3  mol H2SO4 = molarity H2SO4 x volume of H2SO4 in dm  mol H2SO4 = 0.800 x 250/1000 = 0.200 mol H2SO4  (b) mass = moles x formula mass  formula mass of H2SO4 = 2 + 32 + (4x16) = 98  0.2 mol H2SO4 x 98 = 19.6g of H2SO4 Molarity calculation Example 11.3 3 o 5.95g of potassium bromide was dissolved in 400cm of water. Calculate its molarity. [Ar's: K = 39, Br = 80] o moles = mass / formula mass, (KBr = 39 + 80 = 119) o mol KBr = 5.95/119 = 0.050 mol 3 3 o 400 cm = 400/1000 = 0.400 dm o molarity = moles of solute / volume of solution o molarity of KBr solution = 0.050/0.400 = 0.125M Molarity calculation Example 11.4 3 3 o What is the concentration of sodium chloride (NaCl) in g/dm and g/cm in a 1.50 molar solution? o At. masses: Na = 23, Cl = 35.5, formula mass NaCl = 23 + 35.5 = 58.5 3 o since mass = mol x formula mass, for 1 dm 3 o concentration = 1.5 x 58.5 = 87.8 g/dm , and 3 o concentration = 87.75 / 1000 = 0.0878 g/cm Molarity calculation Example 11.5 3 o A solution of calcium sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm of water. 3 3 3 o Calculate the concentration in (a) g/dm , (b) g/cm and (c) mol/dm . 3 3 3  (a) concentration = 0.500/2.00 = 0.250 g/dm , then since 1dm = 1000 cm 3  (b) concentration = 0.250/1000 = 0.00025 g/cm  (c) At. masses: Ca = 40, S = 32, O = 64, f. mass CaSO4 = 40 + 32 + (4 x 16) = 136  moles CaSO4 = 0.5 / 136 = 0.00368 mol 3  concentration CaSO4 = 0.00368 / 2 = 0.00184 mol/dm o











SECTION 14. % ACTUAL AND THEORETICAL YIELD, DILUTION CALCULATIONS, ATOM ECONOMY, VOLUMETRIC TITRATION APPARATUS, WATER OF CRYSTALLISATION, HOW MUCH REACTANT IS NEEDED?

14.1 Percentage purity of a chemical reaction product  

Purity is very important e.g. for analytical standards in laboratories or pharmaceutical products where impurities could have dangerous side effects in a drug or medicine. However in any chemical process it is almost impossible to get 100.00% purity and so samples are always analysed in industry to monitor the quality of the product. % purity is the percentage of the material which is the actually desired chemical in a sample of it. MASS of USEFUL PRODUCT PERCENT PURITY = 100 x

-----------------------------------------------------in TOTAL MASS of SAMPLE





Example 14.1.1 o A 12.00g sample of a crystallised pharmaceutical product was found to contain 11.57g of the active drug. o Calculate the % purity of the sample of the drug. o % purity = actual amount of desired material x 100 / total amount of material o % purity = 11.57 x 100 / 12 = 96.4% (to 1dp) o Example 14.1.2 o Sodium chloride was prepared by neutralising sodium hydroxide solution with dilute hydrochloric acid. The solution was evaporated to crystallise the salt. o The salt is required to be completely anhydrous, that is, not containing any water. o o The prepared salt was analysed for water by heating a sample in an oven at 110 C to measure the evaporation of any residual water. o The following results were obtained and from them calculate the % purity of the salt. o Mass of evaporating dish empty = 51.32g. o Mass of impure salt + dish = 56.47g o Mass of dish + salt after heating = 56.15g o Therefore the mass of original salt = 56.47 - 51.32 = 5.15g o and the mass of pure salt remaining = 56.15 - 51.32 = 4.83g o % salt purity = 4.83 x 100 / 5.15 = 93.8% (to 1dp)

14.2a Percentage yield of the product of a reaction 

The % yield of a reaction is the percentage of the product obtained compared to the theoretical maximum (predicted) yield calculated from the balanced equation. o You get the predicted maximum yield from a reacting mass calculation (see examples further down). ACTUAL YIELD (e.g. in grams) PERCENTAGE YIELD =

100 x ----------------------------------------------------PREDICTED YIELD (g)





In carrying out a chemical preparation, the aim is to work carefully and recover as much of the desired reaction product as you can, and as pure as is possible and practicable. o Despite the law of conservation of mass, i.e. no atoms lost or gained, in real chemical preparations things cannot work out completely according to chemical theory, often for quite simple, physical or sometimes chemical reasons, o and it doesn't matter with its a small scale school laboratory preparation of a large scale industrial manufacturing process, the percent yield is never 100%. So, in any chemical process, it is almost impossible to get 100% of the product because of several reasons:

1.

The reaction might not be 100% completed because it is reversible reaction and an equilibrium is established (note the sign in the equation below. Both reactants and products co-exist in the same reaction mixtures (solutions or gases) i.e. the reaction can never go to completion.  A good example of this is the preparation of an ester  ethanoic acid + ethanol ethyl ethanoate + water

 + + H2O 2. You always get losses of the desired product as it is separated from the reaction mixture by filtration, distillation, crystallisation or whatever method is required.  e.g. bits of solid or droplets are left behind on the sides of the apparatus or reactor vessel.  Small amounts of liquid will be left in distillation units or solid particles on the surface of filtration units.  You cannot avoid losing traces of product in all stages of the manufacturing process. 3. Some of the reactants may react in another way to give a different product to the one you want (so-called by products).  By products are very common in organic chemistry reactions  (i) A + B ==> C + D  (i) The main reaction to give the main desired products A + B  (ii) A + B ==> E + F  (ii) A con-current reaction, maybe just involving a few % of the reactants to give the minor, and often undesirable, by products of E + F.  Sometimes the by products can be separated as a useful product and sold to help the economics of the chemical process. 4. The aim is to work carefully and recover as much of the desired reaction product, and as pure as is possible and practicable  % yield = actual amount of desired chemical obtained x 100 / maximum theoretical amount formed o If the reaction doesn't work the yield is zero or 0%. o If the reaction works perfectly and you obtain all the product, the yield is 100%, BUT this never happens in reality (as already discussed above).  Example 14.2a.1 o Magnesium metal dissolves in hydrochloric acid to form magnesium chloride. o Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) o Atomic masses : Mg = 24 and Cl = 35.5, and formula mass MgCl2 = 24 + (2 x 35.5) = 95 o (a) What is the maximum theoretical mass of anhydrous magnesium chloride which can be made from 12g of magnesium?  Reacting mass ratio calculation from the balanced equation:  1 Mg ==> 1 MgCl2, so 24g ==> 95g or 12g ==> 47.5g MgCl2 o (b) If only 47.0g of purified magnesium chloride was obtained after crystallising the salt from the solution and heating it to drive off the water of crystallisation, what is the % yield from the salt preparation?  % yield = actual amount obtained x 100 / maximum theoretical amount possible  % yield = 47.0 x 100 / 47.5 = 98.9% (to 1dp) o More examples of % yield and atom economy calculations in section 6. o  Example 14.2a.2 o 2.8g of iron was heated with excess sulphur to form iron sulphide. o Fe + S ==> FeS o The excess sulphur was dissolved in a solvent and the iron sulphide filtered off, washed with clean solvent and dried. o If 4.1g of purified iron sulphide was finally obtained, what was the % yield of the reaction? o 1st a reacting mass calculation of the maximum amount of FeS that can be formed:  Relative atomic/formula masses: Fe =56, FeS = 56 + 32 = 88  This means 56g Fe ==> 88g FeS, or by ratio, 2.8g Fe ==> 4.4g FeS  because 2.8 is 1/20th of 56, so theoretically you can get 1/20th of 88g of FeS. o 2nd the % yield calculation itself.  % yield = actual amount obtained x 100 / maximum theoretical amount possible  % yield = 4.1 x 100 / 4.4 = 93.2% (to 1dp) o More examples of % yield and atom economy calculations in section 6. o  Example 14.2a.3 o (a) Theoretically how much iron can be obtained from 1000 tonne of pure haematite ore, formula Fe2O3 in a blast furnace? o If we assume the iron(III) oxide ore (haematite) is reduced by carbon monoxide, the equation is: o Fe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g) o (atomic masses: Fe = 56, O = 16) o For every Fe2O3 ==> 2Fe can be extracted, formula mass of ore = (2 x 56) + (3 x 56) = 160 o Therefore reacting mass ratio is: 160 ==> 112 (from 2 x 56)

o o



so, solving the ratio, 1000 ==> 112 / 160 = 700 tonne copper = max. can be extracted (b) If in reality, only 670 tonne of iron is produced what is the % yield of the overall blast furnace process? o % yield = actual yield x 100 / theoretical yield o % yield = 670 x 100 / 700 = 95.7% o In other words, 4.3% of the iron is lost in waste e.g. in the slag. o More examples of % yield and atom economy calculations in section 6. o Example 14.2a.4 o Given the atomic masses: Mg = 24 and O = 16, o and the reaction between magnesium to form magnesium oxide is given by the symbol equation o 2Mg(s) + O2(g) ==> 2MgO(s) o (a) What mass of magnesium oxide can be made from 1g of magnesium?  2Mg ==> 2MgO  in terms of reacting masses (2 x 24) ==> {2 x (24 +16)}  so 48g Mg ==> 80g MgO (or 24g ==> 40g, its all the same)  therefore solving the ratio  1g Mg ==> w g MgO, using the ratio 48 : 80  w = 1 x 80 / 48 = 1.67g MgO o (b) Suppose the % yield in the reaction is 80%. That is only 80% of the magnesium oxide formed is actually recovered as useful product. How much magnesium needs to be burned to make 30g of magnesium oxide?  This is a bit tricky and needs to done in two stages and can be set out in several ways.  Now 48g Mg ==> 80g MgO (or any ratio mentioned above)  so y g Mg ==> 30g MgO  y = 30 x 48 / 80 = 18g Mg  BUT you only get back 80% of the MgO formed,  so therefore you need to take more of the magnesium than theoretically calculated above.  Therefore for practical purposes you need to take, NOT 18g Mg, BUT ...  ... since you only get 80/100 ths of the product ...  ... you need to use 100/80 ths of the reactants in the first place  therefore Mg needed = 18g x 100 / 80 = 22.5g Mg  CHECK: reacting mass calculation + % yield calculation CHECK:  22.5 Mg ==> z MgO, z = 22.5 x 80 / 48 = 37.5g MgO,  but you only get 80% of this,  so you actually get 37.5 x 80 / 100 = 30g  This means in principle that if you only get x% yield ...  ... you need to take 100/x quantities of reactants to compensate for the losses.

14.2b The Atom economy of a chemical reaction The atom economy of a reaction is a theoretical measure of the amount of starting materials that ends up as 'desired' reaction product. The greater the atom economy of a reaction, the more 'efficient' or 'economic' it is likely to be, though this is a gross simplification when complex and costly chemical synthesis are looked at. Quite simply, the larger the atom economy of a reaction, the less waste products are produced. It can be defined numerically in words in several ways, all of which amount to the same theoretical % number! You can do the calculation in any mass units you want, or non at all by simply using the atomic/formula masses of the reactants and products as appropriate , and I suggest you just think like that. The formula to calculate atom economy can be written in several different ways and they are all equivalent to each other because of the law of conservation of mass e.g. MASS of desired USEFUL PRODUCT ATOM ECONOMY = 100 x

-----------------------------------------------------------TOTAL MASS of REACTANTS

MASS of desired USEFUL PRODUCT ATOM ECONOMY = 100 x

-----------------------------------------------------------TOTAL MASS of PRODUCTS

TOTAL FORMULA MASSES of USEFUL PRODUCT ATOM ECONOMY = 100 x

----------------------------------------------------------------------------------TOTAL FORMULA MASS of REACTANTS

TOTAL FORMULA MASSES of USEFUL PRODUCT ATOM ECONOMY = 100 x

----------------------------------------------------------------------------------TOTAL FORMULA MASS of PRODUCTS

THEY ALL GIVE THE SAME ANSWER! Example 14.2b.1 This is illustrated by using the blast furnace reaction from example 14.2a.3 above. Fe2O3(s) + 3CO(g) ===> 2Fe(l) + 3CO2(g) Using the atomic masses of Fe = 56, C = 12, O = 16, we can calculate the atom economy for extracting iron. the reaction equation can be expressed in terms of theoretical reacting mass units [(2 x 56) + (3 x 16)] + [3 x (12 + 16)] ===> [2 x 56] + [3 x (12 + 16 + 16)] [160 of Fe2O3] + [84 of CO] ===> [112 of Fe] + [132 of CO2] so there are a total of 112 mass units of the useful/desired product iron, Fe out of a total mass of reactants or products of 160 + 84 = 112 + 132 = 244. Therefore the atom economy = 100 x 112 / 244 = 45.9% Note: It doesn't matter whether you use the total mass of reactants or the total mass products in the calculations, they are the same from the law of conservation of mass Example 14.2b.2 The fermentation of sugar to make ethanol ('alcohol') e.g. glucose (sugar) == enzyme ==> ethanol + carbon dioxide C6H12O6(aq) ==> 2C2H5OH(aq) + 2CO2(g) atomic masses: C = 12, H = 1 and O = 16 formula mass of glucose reactant = 180 (6x12 + 12x1 + 6x16) formula mass of ethanol = 46 (2x12 + 5x1 + 1x16 + 1x1) relative mass of desired useful product = 2 x 46 = 92 Atom economy = 100 x 92/180 = 51.1%

14.3 Dilution of solutions calculations calculating dilutions - volumes involved etc.

   







In conjunction with this page it be important to study o Calculations Part 11 Introducing Molarity, volumes and the concentration of solutions o before tackling this section and note the triangle of relationships you need to know! It is important to know how to accurately dilute a more concentrated solution to a specified solution of lower concentration. It involves a bit of logic using ratios of volumes. The diagram above illustrates some of the apparatus that might be used when dealing with solutions. Example 14.3.1 3 A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm . How would you 3 3 prepare 100 cm of a 0.1 mol/dm solution to do a titration of an acid? o The required concentration is 1/10th of the original solution. 3 3 3 o To make 1dm (1000 cm ) of the diluted solution you would take 100 cm of the original solution and mix 3 with 900 cm of water. 3 o The total volume is 1dm but only 1/10th as much sodium hydroxide in this diluted solution, so the 3 concentration is 1/10th, 0.1 mol/dm . 3 3 3 o To make only 100 cm of the diluted solution you would dilute 10cm by mixing it with 90 cm of water. o How to do this in practice is described at the end of Example 14.3.2 below and a variety of accurate/'less inaccurate' apparatus is illustrated above. Example 14.3.2 3 3 o Given a stock solution of sodium chloride of 2.0 mol/dm , how would you prepare 250cm of a 0.5 3 mol/dm solution? 3 3 o The required 0.5 mol/dm concentration is 1/4 of the original concentration of 2.0 mol/dm . 3 3 3 o To make 1dm (1000 cm ) of a 0.5 mol/dm3 solution you would take 250 cm of the stock solution and add 3 750 cm of water. 3 3 o Therefore to make only 250 cm of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm of 3 the stock solution plus 187.5 cm of pure water. o This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker. o It can be done much more accurately by using a burette or a pipette to measure out the stock solution 3 directly into a 250 cm graduated-volumetric flask. o Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper on at the same time! Example 14.3.3

In the analytical laboratory of a pharmaceutical company a laboratory assistant 3 -2 -3 was asked to make 250 cm of a 2.0 x 10 mol dm (0.02M) solution of paracetamol (C8H9NO2). o (a) How much paracetamol should the laboratory assistant weigh out to make up the solution?  Atomic masses: C = 12, H = 1, N = 14, O = 16  method (i): Mr(paracetamol) = (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151 3  1000 cm of 1.0 molar solution needs 151g

3

o

o

 

-2

-2

 1000 cm of 2.0 x 10 molar solution needs 151 x 2.0 x 10 /1 = 3.02g  (this is just scaling down the ratio from 151g : 1.0 molar) 3  Therefore to make 250 cm of the solution you need 3.02 x 250/1000 = 0.755 g method (ii): Mr(paracetamol) = 151 3  moles = molarity x volume in dm -2 -3  mol paracetamol required = 2.0 x 10 x 250/1000 = 5.0 x 10 (0.005) -3  mass = mol x Mr = 5.0 x 10 x 151 = 0.755 g -2 3 (b) Using the 2.0 x 10 molar stock solution, what volume of it should be added to a 100cm 3 -3 -3 volumetric flask to make 100 cm of a 5.0 x 10 mol dm (0.005M) solution? -2 -3  The ratio of the two molarities is stock/diluted = 2.0 x 10 /5.0 x 10 = 4.0 or a dilution factor of 1/4 (0.02/0.005). 3 1 -2 3  Therefore 25 cm ( /4 of 100) of the 2.0 x 10 molar solution is added to the 100 cm volumetric 3 -3 -3 flask prior to making it up to 100 cm with pure water to give the 5.0 x 10 mol dm (0.005M) solution. There are more questions involving molarity in section 12. on dilution

section 7. introducing molarity and

Example 14.3.4 -3 o You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm (conc. ammonia! ~18M!) 3 o (a) What volume of the conc. ammonia is needed to make up 1dm of 1.0 molar ammonia solution?  Method (i) using simple ratio argument.  The conc. ammonia must be diluted by a factor of 1.0/17.9 to give a 1.0 molar solution. 3 3  Therefore you need (1.0/17.9) x 1000 cm = 55.9 cm of the conc. ammonia. 3 3 3 -3  If the 55.9 cm of conc. ammonia is diluted to 1000 cm (1 dm ) you will have a 1.0 mol dm (1M) solution.  Method (ii) using molar concentration equation - a much better method that suits any kind of dilution calculation involving molarity. 3 3  molarity = mol / volume (dm ), therefore mol = molarity x volume in dm 3  Therefore you need 1.0 x 1.0 = 1.0 moles of ammonia to make 1 dm of 1.0M dilute ammonia.  Volume = mol / molarity 3 3  Volume of conc. ammonia needed = 1.0 / 17.9 = 0.0559 dm (55.9 cm ) of the conc. ammonia is required, 3 -3  and, if this is diluted to 1 dm , it will give you a 1.0 mol dm dilute ammonia solution. 3 o (b) What volume of conc. ammonia is needed to make 5 dm of a 1.5 molar solution? 3 3  molarity = mol / volume (dm ), therefore mol = molarity x volume in dm 3  Therefore you need 1.5 x 5 = 7.5 moles of ammonia to make 5 dm of 1.5M dilute ammonia.  Volume (of conc. ammonia needed) = mol / molarity 3 3  Volume of conc. ammonia needed = 7.5 / 17.9 = 0.419 dm (419 cm ) of the conc. ammonia is required, 3 -3  and, if this is diluted to 5 dm , it will give you a 1.5 mol dm dilute ammonia solution.

14.4 Water of crystallisation in a crystallised salt 





Example 14.4.1: Calculate the % of water in hydrated magnesium sulphate MgSO 4.7H2O salt crystals o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1 o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246 o 7 x 18 = 126 is the mass of water o so % water = 126 x 100 / 246 = 51.2% Example 14.4.2 The % water of crystallisation and the formula of the salt are calculated as follows: o Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O, (x unknown) was gently heated in a crucible until the mass remaining was 4.00g. This is the white anhydrous copper(II) sulphate. o The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g o The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36% o [ Ar's Cu=64, S=32, O=16, H=1 ] o The mass ratio of CuSO4 : H2O is 4.00 : 2.25 o To convert from mass ratio to mole ratio, you divide by the molecular mass of each 'species' o CuSO4 = 64 + 32 + (4x18) = 160 and H2O = 1+1+16 = 18 o The mole ratio of CuSO4 : H2O is 4.00/160 : 2.25/18 o which is 0.025 : 0.125 or 1 : 5, so the formula of the hydrated salt is CuSO4.5H2O Example 14.4.3 How to calculate the theoretical % of water in a hydrated salt o eg magnesium sulphate MgSO4.7H2O salt crystals



o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1 o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246 o 7 x 18 = 126 is the mass of water o so % water = 126 x 100 / 246 = 51.2% Example 14.4.4 Determination and calculation of salt formula containing 'water of crystallisation'. o Some salts, when crystallised from aqueous solution, incorporate water molecules into the structure. This is known as 'water of crystallisation', and the 'hydrated' form of the compound. o e.g. magnesium sulphate MgSO4.7H2O. The formula can be determined by a simple experiment (see the copper sulphate example below). o A known mass of the hydrated salt is gently heated in a crucible until no further water is driven off and the weight remains constant despite further heating.  The mass of the anhydrous salt left is measured.  The original mass of hydrated salt and the mass of the anhydrous salt residue can be worked out from the various weighings. o The % water of crystallisation and the formula of the salt are calculated as follows:  Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O, (x unknown) was gently heated in a crucible until the mass remaining was a constant4.00g.  When the mass on subsequent weighings stays constant, you know all the water of crystallisation has driven off by the heat.  This is the white anhydrous copper(II) sulphate.  The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g  The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%  [ Ar values: Cu=64, S=32, O=16, H=1 ]  The mass ratio of CuSO4 : H2O is 4.00 : 2.25  To convert from mass ratio to mole ratio, you divide by the molecular mass of each 'species'  CuSO4 = 64 + 32 + (4x18) = 160 and H2O = 1+1+16 = 18  The mole ratio of CuSO4 : H2O is 4.00/160 : 2.25/18  which is 0.025 : 0.125 or 1 : 5, so the formula of the hydrated salt is CuSO4.5H2O

14.5 Calculation of quantities required for a chemical reaction and a brief mention of % yield and atom economy 

Example 1. How much iron and sulfur do you need to heat together to make 20.0 g of iron sulfide o Atomic masses: Fe = 56 and S = 32 o Balanced equation: Fe + S ==> FeS o This is essentially a reacting mass calculation Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

comments

Fe + S

==>

FeS

You need the balanced chemical equation and in this case its very simple.

56g + 32g

==>

88g

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

56/88 = 0.6364g + 32/88 = 0.3636

==>

88g/88 = 1.00g

divide by 88 to scale down to 1g of FeS product

20 x 0.6364 = 12.7g + 20 x 0.3636 = 7.3g

20 x 100 = 20.0g

==>

then scale by a factor of 20 (for 20g of FeS)

Therefore to make 20g of iron sulfide you need 12.7g of iron and 7.3g of sulfur.



o Example 2. How much iron do you need to make 100.00g of iron(III) chloride by passing excess chlorine gas over heated iron filings? o Atomic masses: Fe = 56 and Cl = 35.5, formula mass of FeCl3 = 162.5 (56 + 3x35.5) o Balanced equation: 2Fe(s) + 3Cl2(g) ==> 2FeCl3(s) o This is essentially a reacting mass calculation Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product

2Fe

==>

comments

2FeCl3

You need the balanced chemical equation BUT only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation

2 x 56 = 112g

==>

2 x 162.5 = 325g

basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses

112/325 = 0.3446g

==>

325/325 = 1.00g

divide by 325 to scale down to 1g of FeCl3 product

100 x 0.3446 = 34.46g

==>

100 x 1 = 100.00g

then scale by a factor of 100 to make 100g of FeCl3

Therefore to make 100g of iron(III) chloride you need 34.46g of iron

 

o Example 3 A more complex example based on a salt preparation Suppose you want to make 50g of the blue hydrated copper(II) sulphate crystals o All of section 14.5 is based on this quantity of 50g of the familiar blue crystals.  The blue crystals contain water of crystallisation, which must be taken into account in doing the calculation.  The preparation is briefly described in the GCSE Acids, Bases and Salts Notes. o Your reactants are dilute sulphuric acid and the black solid copper(II) oxide. o You can use copper(II) carbonate, but this is not a pure simple compound and the predictive nature of the calculations will not be as good. o copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water o (i) CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l) o on crystallisation you get the blue hydrated crystals of formula CuSO4.5H2O o so strictly speaking, after evaporation-crystallisation the overall equation is o (ii) CuO(s) + H2SO4(aq) + 4H2O(l) ==> CuSO4.5H2O(s) (formula of the blue crystals) o o o o o o

How much copper(II) oxide is needed? A 'non-moles' calculation first of all, involving a reacting mass calculation. The crucial change overall is CuO ==> CuSO4.5H2O (Note: In reacting mass calculations you can often ignore other reactant/product masses) Atomic masses: Cu = 64, S = 32, H = 1, O = 16 Formula masses are for: CuO = 64 + 16 = 80, CuSO4 = 64 + 32 + (4 x 16) = 160

o o o o o o o o o

o

and CuSO4.5H2O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250 The crucial reacting mass ratio is: 80 ==> 250 since formula ratio is 1:1 in the equation. 1 Therefore, theoretically, to make 50g of the crystals ( /5th of 250), 1 you need /5th of 80g of copper(II) oxide, and 80/5 = 16g of copper(II) oxide is required. However, in reality, things are not so simple because the method involves adding excess copper(II) oxide to the dilute sulphuric acid. (see salt preparation method (b)) So in practice you would need to use more of the CuO to get anything like 50g of the salt crystals. There is another way to calculate the quantities required based on the acid. -3 How much dilute sulphuric acid (of concentration 1 mol dm ) is required?  Mol = mass in g / formula mass,  so moles of CuSO4.5H2O required = 50/250 = 0.2 mol (see basics of moles) 1  Therefore 0.2 mol of H2SO4 is required ( /5th mol), since the mole ratio CuO : H2SO4 is 1 : 1 in the equation. 3 3 -3  1dm (1000 cm ) of a 1 mol dm solution of contains 1 mole (by definition - see molarity page) 3  Therefore 1/5th of 1dm is required to provide 1/5th of mole of the sulfuric acid. 3 -3  so 200 cm of 1 mol dm dilute sulphuric acid is required, 3 -3  or 100 cm of 2M dilute sulphuric acid. (2M is old fashioned notation for 2 mol dm still seen on many laboratory bottles!)  You would then add copper(II) oxide in small amounts until no more dissolves in the warm-hot acid and the excess black powder is filtered off. There is no need to weigh out an exact amount of copper oxide. 3  If you want just 25g of copper sulfate crystals you would use 100 cm of 1 molar sulfuric acid, or 3 50cm of 2 molar sulfuric acid.  BUT REMEMBER, in practice, you will NOT get a 100% yield, see calculation below. Suppose after carrying out the preparation you finally crystallise 34g of pure the blue crystals of CuSO4.5H2O after weighing the dry product.  What is the 'atom economy' of the preparation? (you need to refer to equations (i)/(ii) at the start of section 14.5  Atom economy = useful theoretical products x 100/mass of all reactants  based on equation (i) Atom economy = mass CuSO4 x 100 / (mass CuO + H2SO4)  = 160 x 100 / (80 + 98) = 16000/178 = 89.9%  based on equation (ii) the atom economy is 100% if you include water as a 'reactant', can you see why?  What is the % yield? i.e. comparing what you actually get with the maximum possible, i.e. a 'reality check'!  % yield = mass of product obtained x 100 / theoretical mass from the equation  % yield = 34 x 100 / 50 = 68%

END OF FORMULAE, EQUATIONS AND AMOUNTS OF SUBSTANCE

ENERGETICS ADVANCED LEVEL ENERGETICS–THERMOCHEMISTRY – ENTHALPIES OF REACTION, FORMATION & COMBUSTION

1.1 Advanced Introduction to Enthalpy (Energy) Changes in Chemical Reactions

1.1a I have ASSUMED you have studied the GCSE notes on the basics of chemical energy changes in what you might call a lower level introduction which bridges GCSE and AS level and you are completely ok in interpreting enthalpy level and activation energy diagrams such as ...

... and can clearly distinguish between enthalpy change and activation energy and the activation energy change with a catalyst does NOT change the enthalpy value of the reaction (activation energy will be rarely mentioned until the end of Part 3), but all important ideas from the GCSE page will be re-studied in their advanced level context, but I would still study the GCSE page first! REMEMBER - all chemical changes are accompanied by energy changes or energy transfers, many of which can be directly measured, or, theoretically calculated from known values.

1.1b Enthalpy Changes and Thermochemistry Some important initial definitions and examples: The system: The reactants and products of the reaction being studied i.e. the contents of the calorimeter. The surroundings: The means the rest of the 'world' including the i.e. a copper calorimeter, the surrounding air etc. etc. Enthalpy H: The heat energy content of a substance. This cannot be determined absolutely but enthalpy changes for a chemical reaction can be measured directly or indirectly from theoretical calculations using known enthalpy values. Enthalpy change ΔH: The net heat energy transferred to a system from the surroundings or from the surroundings to a system at constant pressure. The Greek letter delta Δ in maths implies a change, in this case a net heat energy change. -1

ΔH = Hfinal - Hinitial (the units of delta H are kJ mol ) or ΔH = ∑Hproducts - ∑Hreactants

θ

θ

θ

or ΔH (reaction) = ∑ΔH f(products) - ∑ΔH f(reactants) The Greek letter delta Δ implies 'change in' .... The Greek letter ∑ implies 'sum of' ....' θ

ΔH f denotes a standard enthalpy of formation - which is explained further down. Exothermic reaction A reaction in which heat energy is given out from the system to the surroundings i.e. the enthalpy of the reacting system decreases and the temperature of the system and surroundings rises. This means Hreactants > Hproducts so that ΔH is negative (-ve). The enthalpy of the reaction system is decreasing. Example: All combustion reactions are exothermic e.g. CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l) ΔH = -890 kJmol

-1

3

i.e. the figure of 890 kJ released refers to the complete combustion of 1 mole of gaseous methane (24 dm ), using exactly 2 3 3 moles of gaseous oxygen (48 dm ) to form exactly 1 mole of gaseous carbon dioxide (24 dm ) and 2 moles of liquid water. o These values refer to 298K (25 C and 1 atm/101 kPa) Note some general points (which apply to all exothermic or endothermic changes, physical or chemical changes): (i) All enthalpy values must be quoted with referenced to the ambient/assumed temperature and pressure of the system undergoing the physical or chemical change. o

The usual standard reference conditions are 298K (25 C and 1 atm/101 kPa), and other criteria may apply e.g. 1 molar solution if applicable. (ii) Not only the molar quantities must clearly indicated BUT the physical states of all the substances must be clearly stated too. This is a convenient point to make the point about the importance of state symbols via the combustion of hydrogen. eg 1

-1

H2(g) + /2O2(g) ==> H2O(l) ΔH = -285.9 kJ mol , but for 1

H2(g) + /2O2(g) ==> H2O(g) ΔH = -241.8 kJ mol

-1

If the water forms remains as steam/vapour/gas, then 44.1 kJ less heat energy is released to the surroundings, because condensation is an exothermic process (g ==> l) and forming liquid water releases an extra 44.1 kJ. The -285.9 (~-286) kJ -1 mol is the usual value for the enthalpy of combustion of hydrogen you will encounter in your studies because at the standard temperature of 298K water is a liquid in its normal stable state. (iii) This sort of combustion reaction can be measured in a calorimeter (see section 1.3). BUT, however the enthalpy change is measured, all equations should be read in molar terms when dealing with enthalpy values i.e. a delta H value goes with a specific equation. -1

(iv) Enthalpy change values are usually quoted in kJ mol , but take care in their interpretation because you must know what equation goes with the ΔH value! eg the enthalpy of combustion usually refers to the complete combustion of one mole of the combustible material as for water above, BUT if you double the equation you must also double the enthalpy values for that equation -1

2H2(g) + O2(g) ==> 2H2O(l) ΔH = 2 x -285.9 = 571.8 kJ mol

Endothermic reaction

A reaction in which the system takes in or absorbs heat energy from the surroundings i.e. the enthalpy of the system increases and the temperature of the system and surroundings falls OR the system must be heated to initiate the reaction and provide the heat absorbed. This means Hproducts > Hreactants so that ΔH is positive (+ve). The enthalpy of the reaction system is increasing. Example: The thermal decomposition of calcium carbonate CaCO3(s) ==> CaO(s) + CO2(g) ΔH = +179 kJmol

-1

i.e. 179 kJ of heat energy must be absorbed to decompose 1 mole of solid calcium carbonate into 1 mole of solid calcium oxide and 1 mole of gaseous carbon dioxide. Mr(CaCO3) = 100, so 17.9 kJ of heat energy is absorbed in decomposing 10g o of limestone. This reaction requires an experimental temperature of 800-1000 C to achieve an appreciable rate of reaction and cannot be studied quantitatively in the laboratory. However it can be theoretically calculated from known enthalpy change values by means of a Hess's Law cycle calculation.

The two diagrams below illustrate how exothermic (left) and endothermic (right) reactions are specified on an enthalpy level diagram.

Standard conditions Standard conditions for referencing enthalpy values are essential for communicating accurate data throughout the scientific community. It means values measured/calculated in one laboratory/research team can be used in another scientific establishment anywhere!, OR checked for accuracy by any other scientists. In this way accurate enthalpy data can be built up and through time validated and perhaps more accurately measured with technological developments and theoretical calculations become more reliable. This means the reactants/products start/finish at a specified temperature, pressure and concentration whatever the 'temporary' temperature change in the reaction - which is required to calculate the enthalpy change. The net energy change is based on the products returning to the same temperature and pressure that the reactants started o o at. The most frequently used standard conditions are a temperature of 298 K/25 C (K = 273 + C) and a pressure of 1 -3 atm/101 kPa and a concentration of 1.00 mol dm . The use of standard conditions enables a database of delta H change to be assembled from which you can do theoretical calculations (see section 1.2 using Hess's Law). Strictly speaking the standard conditions should be indicated in terms of the standard temperature and the reactants involved and standard delta H values are denoted with the Greek letter theta (θ). By using data based on standard. agreed and defined conditions, then the data can be used universally by any laboratory around the world and also allows scientists to check each others experimental results.

o

Its pertinent here to consider the question - how can you have an standard enthalpy of combustion at 25 C when the o flame temperature is perhaps peaking at over 1000 C !!! The answer applies to all enthalpy changes what-so-ever! The enthalpy change represents the heat energy change needed to restore the products to the temperature of the reactants o at the start e.g. room temperature/25 C. Standard Enthalpy of Reaction ΔHr/react/reaction is the enthalpy change (heat absorbed/released, endothermic/exothermic) o when molar quantities of reactants as stated in an equation react under standard conditions (i.e. 298K/25 C, 1 atm/101kPa) Examples (i) NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(aq) (exothermic) ΔH

θ r,298

-1

= -57.1 kJ mol (can also be described as an 'enthalpy of neutralisation')

(ii) CaCO3(s) ==> CaO(s) + CO2(g) (endothermic) ΔH

θ r,298

-1

= +179 kJ mol (can also be described as an 'enthalpy of thermal decomposition')

Standard Enthalpy of Formation ΔHf/form/formation is the enthalpy change when 1 mole of compound is formed from its o constituent elements with both the compound and elements in their standard states ('normal stable states) i.e. at 298K/25 C, 1 atm/101kPa (a) It may be endothermic or exothermic (b) Any accompanying equation should involve the formation of 1 mole of the compound o

The standard state is the most stable state at the standard temperature and pressure e.g. at 298K/25 C and 1 atm/101kPa e.g. H2(g) H2O(l) C(s) O2(g), C3H8(g) C8H18(l) C24H50(s) CO2(g) CH3CH2OH(l) etc. Examples (i) C(s) + 2H2(g) ==> CH4(g) ΔH

θ f,298(methane)

(ii) 2C(s) + 2H2(g) ==> C2H4(g) ΔH

θ f,298(ethene)

= -74.9 kJ mol

= +52.3 kJ mol

1

(iii) 2C(s) + 3H2(g) + /2O2(g) ==> CH3CH2OH(l) ΔH 1

(iv) /2N2(g) + O2(g) ==> NO2(g) ΔH

-1

θ f,298(nitrogen

-1

θ f,298(ethanol)

= -278 kJ mol

dioxide) = +33.9 kJ mol

-1

-1

Note (a) The values can be positive/endothermic or negative/exothermic. (b) The enthalpy of formation of elements in their standard stable states is arbitrarily assigned a value of zero. This definition, together with experimental values of enthalpy changes allows a body of enthalpy change data to be accumulated and extended via theoretical calculations. Standard Enthalpy of Combustion ΔHc/comb/combustion is the enthalpy change when 1 mole of a fuel (or any combustible o material) is completely burned in oxygen (or air containing oxygen) equated to standard conditions (298K/25 C, 1 atm/101kPa). You should ensure just 1 mole of fuel appears in the equation to accompany the delta H value which is always negative i.e. always exothermic. Examples (i) C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l) ΔH

θ c,298K(propane)

(ii) CH3COOH(l) + 2O2(g) ==> 2CO2(g) + 2H2O(l) ΔH

= -2219 kJ mol

θ c,298K(ethanoic

-1

acid) = -876 kJ mol

-1

In the calculations explained below just the subscripted letters r/f/c will be used for brevity and a temperature of 298K and a constant pressure 1atm assumed unless otherwise stated. There is more the enthalpies of combustion of alkanes and alcohols in section 1.4a Standard enthalpy of neutralisation is the energy released when unit molar quantities of acids and alkalis completely neutralise each other at 298K (pressure effects are insignificant for reactions only involving liquids/solutions/solids) (i) NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) ΔH

θ neutralisation

= -57.1 kJ mol

(ii) Ba(OH)2(aq) + 2HNO3(aq) ==> Ba(NO3)2(aq) + 2H2O(l) ΔH 1

1

θ neutralisation

(iii) /2Ba(OH)2(aq) + HNO3(aq) ==> /2Ba(NO3)2(aq) + H2O(l) ΔH

-1

= -116.4 kJ mol

θ neutralisation

-1

= -58.2 kJ mol

-1

Note! It looks as if the enthalpy of neutralisation of barium hydroxide is approximately double that of sodium hydroxide ie ~ twice as exothermic! Well yes it is! and no it isn't! Yes - ~twice as much energy is released per mole of soluble base/alkali. No - however, on the basis of heat released per mole of water formed, they are actually very similar. In other words, which value you quote, depends on which point you want to make. Yet another example of carefully qualifying enthalpy values with respect to the context. More on enthalpies of neutralisation Bond Enthalpy ('bond energy') This is the average energy absorbed to break 1 mole of a specified bond when all species involved are in the gaseous state. e.g. for (i) H2(g) ==> 2H(g) ΔH = +436 kJ mol

-1

for the H-H bond

or for (ii) CH3CH2Br(g) ==> CH3CH2(g) + Br(g) ΔH = +276 kJ mol

-1

for the C-Br bond

It is always endothermic and the reverse process - bond formation, is always exothermic. In many cases the values are averaged from a variety of 'molecular' situations. More on this in the bond enthalpy section.

Some examples of points made on this page with reference to an enthalpy level change diagram

General points: Arrows pointing downwards represent exothermic changes and arrows pointing upwards represent endothermic changes 1. The energy released when 1 mole of aluminium oxide is formed. -1

The ΔH value of -1669 kJ mol corresponds to the very exothermic enthalpy of formation of Al2O3 or the enthalpy of the complete combustion of two moles of Al. The very exothermicity of the reaction suggests, and correctly, that aluminium oxide is a very stable compound - it is o thermally stable to at least 2500 C. 2. The endothermic enthalpy of formation of gold(III) oxide o

It is a compound not readily formed and it decomposes on heating at ~150 C, so contrast this thermal instability with that of aluminium oxide. 3. This is a much more complex enthalpy level diagram involving hydrogen, chlorine and hydrogen chloride. -1

The +436 kJmol represents the bond enthalpy for splitting hydrogen molecules into hydrogen atoms. -1

The +242 kJmol is the bond energy of chlorine molecules. -1

The -184 kJmol is the enthalpy of formation of hydrogen chloride gas. -1

The very exothermic -862 kJmol is the energy released theoretically when two moles of hydrogen chloride are formed directly from hydrogen and chlorine atoms. The latter indicates that the H-Cl bond enthalpy is +862/2 = 431 kJmol

-1

END OF ENERGETICS

Atomic Structure and the Periodic Table Advanced Level Inorganic Chemistry Periodic Table Revision Notes

1. A few snippets of the past and continuing history of the Periodic Table Not all scientists are mentioned who perhaps should be, but I've tried to pick out a few 'highlight' and added some footnotes on what was happening in terms of the development of the detailed knowledge of the structure of atoms, so essential to the modern interpretation of the Periodic Table. Its a good 'advanced' example of how science works i.e. the relationship between experimental data and theories to account for it, questions posed, questions answered, leading to more comprehensive and accurate theories developing. 1.1 The early classification of Antoine Lavoisier of 1789 Antoine Lavoisier's 1789 classification of substances into four 'element' groups acid-making elements

gas-like elements

metallic elements

earthy elements

sulphur

light

cobalt, mercury, tin

lime (calcium oxide)

phosphorus

caloric (heat)

copper, nickel, iron,

magnesia (magnesium oxide)

charcoal (carbon)

oxygen

gold, lead, silver, zinc

barytes (barium sulphate)

azote (nitrogen)

manganese, tungsten

argilla (aluminium oxide)

hydrogen

platina (platinum)

silex (silicon dioxide)

     

The understanding that an element as a unique atomic 'building block' which could not be split into simpler substances and compound is a chemical combination of two or more elements were not at all understood at the time of Lavoisier. 'light' and 'caloric' (heat), were considered 'substances' and the last 'scientific' vestige of the elements of 'earth, fire, air and water' which had there conceptual origin in the Greek civilisation of 2300-2800 years ago. However, Lavoisier was correct on a few things e.g. the elements sulphur, phosphorus and carbon and correctly described their oxides as acidic e.g. dissolved in water turned litmus turns red. Many metallic elements, were correctly identified though I doubt if they were pure though! What he described as the 'earthy elements' are of course compounds, a chemical combination of a metal plus oxygen or sulfur (both O and S in case of barium). He didn't have very high temperature smelting technology, or a reactive metal from electrolysis (came in about 1806 onwards)' to 'separate' the elements in some way e.g. he couldn't extract a reactive metal! In other words, at this time, the wrong 'classification' was due to a lack of chemical technology as much as lack of knowledge. o Atomic structure history note: You can see from the 1789 'table' Lavoisier and his contemporaries did not have the experiment techniques, data or theoretical framework to clearly distinguish between 'elements' and 'compounds'. It was only in 1808 Dalton proposed his atomic theory based on experimental data and produced the first list of 'atomic weights', which we now call relative atomic masses.

1.2 The 1829 work of Johann Döbereiner



 

Johann Döbereiner noted that certain elements seemed to occur as 'triads' of similar elements e.g. o (i) lithium, sodium and potassium o (ii) calcium, strontium and barium o (iii) chlorine, bromine and iodine

Döbereiner was amongst the first scientists to recognise the 'group' idea of chemically very similar elements. Three groups he 'recognised' were (i) Group 1 Alkali Metals, (ii) Group 2 Alkaline Earth Metals, (iii) Group 7 Halogens. o Atomic structure history note: The physical and chemical likeness of the three members of these 'triad groups' should be evident and it was based purely on observation, however Döbereiner and contemporaries where unaware of the atomic and molecular nature of these elements e.g. the atomic nature of the metals (M atoms) and the molecular nature of the Halogens (X 2 diatomic molecules). In fact the concept of a 'molecule' was first realised by Avogadro in 1811 but it took 50 years before the genius of his experimental work and intuition was fully realised.

1.3 The work of John Newlands 1864 Newland's 'Law of Octaves' (his 'Periodic Table' of 1864) H

Li

Ga

B

C

N

O

F

Na

Mg

Al

Si

P

S

Cl

K

Ca

Cr

Ti

Mn

Fe

Co, Ni

Cu

Zn

Y

In

As

Se

Br

Rb

Sr

Ce, La

Zr

Di, Mo

Ro, Ru

Pd

Ag

Cd

U

Sn

Sb

Te

I

Cs

Ba, V

Ta

W

Nb

Au

Pt, Ir

Tl

Pb

Th

Hg

Bi

Cs

  

 

Newlands recognised that every 7 elements, the 8th seemed to be very similar to the 1st of the previous 7 when laid out in a 'periodic' manner and he was one of the first scientists to derive a 'Periodic Table' from the available knowledge. e.g. his 'table' consists of almost completely genuine elements (Di was a mix of two elements), classified roughly into groups of similar elements and a real recognition of 'periodicity' He also recognised that the 'groups' had more than 3 elements (not just 'triads'), and was correct to mix up metals and non-metals in same group e.g. in 5th column there is carbon, silicon, tin (Sn) from what we know call Group 4. However, indium is in group 3 but Ti, Zr have a valency of 4, like Group 4 elements and do form part of vertical column in what we know call the Transition Metal series Other correct 'patterns' if not precise are recognisable in terms of the modern Periodic Table e.g. half of column 2 is Group 1, half of column 3 is Group 2, half of column 5 is Group 4, half of column 6 is Group 5, half of column 7 is Group 6. If we put his column 1 as column 7, it would seem a better match of today! Although none of his vertical column groups match completely but the basic pattern of the modern periodic table was emerging. However column's 1 and 7 do seem particularly mixed up compared to the modern periodic table. o Atomic structure history note: A good wedge of history at this point!  The Greeks Leucippus and Democritus ~500-400 BC wondered what was the result of continually dividing a substance i.e. what was the end product or smallest bit i.e. what was left that was indivisible - the word atom/atomic is from Greek adjective atomos meaning 'not divisible'.









They considered that matter is made of atoms that are too small to be see and cannot be divided into smaller particles. They speculated that there was empty space between solid atoms and that atoms were the same throughout a cross section and atoms could have different sizes, shapes and masses.  These were brilliant ideas for their time and such concepts were the result of excellent intuitive thinking BUT the famous and much more eminent and revered philosopher Aristotle, didn't think much of their theory, and so atomic theory never developed for nearly 2000 years!  Its worth commenting further on the Greeks. Although brilliant in intellectual discourse on many subjects and legendary mathematicians, they were NOT very good at science. Most Greek intellectuals did not consider doing experiments to test out theories as very important, and therefore over 2000 years ago they actually rejected the principal methods by which we today practice science! However, the Greeks idea of atoms was not completely forgotten and later revived by Boyle and Newton but with little progress.  Robert Boyle (1627-1691) in his book 'The Sceptical Chemist' talked about tiny identical particles that were indivisible but could be joined in various ways to make 'compounds'. But, in 1808, Dalton (1766-1844) proposed his atomic theory that all matter was made up of tiny hard particles called atoms and the different types of atoms (elements) combined together to give all the different substances of the physical world.  His theory included the idea that atoms in an element are all the same.  Dalton considered that a compound is made by joining at least two different elements to form a compound,  and atoms do not change themselves in a reaction but from the original reactants they rearrange to form the products.  This is real progress! Most of his ideas were correct except the 'indivisibility' of the atom! but it would take nearly another 60 years before the idea of 'atomic structure' would take shape from experimental results.  He also produced the first list of 'atomic weights' (we now call relative atomic masses) on a scale based on hydrogen which was given the arbitrary value of 1 since it was lightest element known, and, as it happens, correctly so.  Dalton also devised symbols for the different elements, but his 'picturesque' symbols were not universally adopted and today's elements letter symbols were introduced and promoted by the chemist Jons Berzelius in 1811. In 1876 Goldstein and Jean Perrin in 1895 (1870-1942) passed a high-voltage electrical discharge through various gases and discovered beams of negatively charge particles where formed.  They where called cathode rays and, where in fact, what we now know as negative electrons (but he didn't know this!).  The electrons were emanating from the negative electrode and being accelerated towards the positive anode.  They were unaware that positive ions were also produced and beamed in the opposite direction.  Up till then, it was just assumed that matter consisted of Daltons 'atoms' i.e. particles that could not be broken down into smaller particles, so did not have any meaningful structure but just combined in various ways to make different compounds.  This was the real start of research into 'atomic structure', especially as it was soon found later on that a stream of positive particles was travelling in the opposite direction to the 'negative electrons'!  Goldstein's and Perrin's experiments also provided the experimental basis for the development of the mass spectrograph by Aston - what we know now as a mass spectrometer.

1.4 Dmitri Mendeleev's Periodic Table, Lothar Meyer's Graphs of 1869   

Mendeleev (Russian chemist) first published his 'Periodic Table' work simultaneously in 1869 with the work of Lothar Meyer (German chemist) who looked at the physical properties of all known elements. Lothar Meyer noted 'periodic' trend patterns e.g. peaks and troughs when melting or boiling points, specific heat and atomic volume values were plotted against 'atomic weight' - what we now call relative atomic mass. My modern versions of Lothar Meyer's graphs are shown on a separate pages, plus others and now the properties are plotted against atomic/proton number and I've managed to collect most data upto element 96.

o o

Elements Z = 1 to 20 covering Periods 1-3 and start of Period 4 Elements Z = 1 to 38 covering Periods 1-4 and start of Period 5 Elements Z = 1 to 96 covering Periods 1-6 and start of Period 7 The atomic volume graph is shown below clearly showing the 'periodic' highest volumes for the alkali metals - the least dense of the elements in liquid or solid form.

My modern version of Lothar-Meyer's 'atomic volume' curve and below one of Mendeleev's early versions of the Periodic Table



  

Mendeleev laid out all the known elements in order of 'atomic weight' (what we know call relative atomic mass, Ar) except for several examples like tellurium (Te, Ar = 127.60) and iodine (I, Ar = 126.90) whose order he reversed because chemically they seemed to be in the wrong vertical column! Smart thinking! o Argon (Ar, Ar = 39.95) and potassium (K, Ar = 39.10) is the 2nd example, but that was not a problem for chemists at the time, because the Group 0 Noble Gases hadn't been discovered by then! o These 'anomalies' in the order of 'atomic weights' are explained by the existence of isotopes which were discovered ~1916 and the neutron finally characterised in 1932. o Isotopes of elements are atoms of the same proton number with different numbers of neutrons, hence atoms of the same element with different mass numbers. 39 40 o The most abundant stable isotope of potassium is K, and that of argon is Ar, hence the anomaly. 127 120 130 o Naturally occurring iodine is 100% I, but tellurium has a range of isotopic masses from Te to Te but more the heavier isotopes are more abundant than the lighter isotopes. By 1869, Mendeleev and Lothar Meyer had an advantage over Newlands (1864) because by then there was an increased number of known elements and hence 'groups' of similar elements were becoming more clearly defined. Mendeleev used a double column approach which is NOT incorrect, i.e. a sort of group xA and xB classification. This is due to the 'insert' of transition metals, some of whom show chemical similarities to the vertical 'groups'. We now recognise theses dual columns as His 'presentation' was sufficiently accurate to predict missing elements and their properties * e.g. germanium (Ge) below silicon (Si) and above tin (Sn) in Group IV and Mendeleev is rightly called

the 'father of the modern Periodic Table'. o Atomic structure history notes: In 1897 Wien and J J Thompson measured the charge mass ratio of the 'particles' of the cathode rays (electrons) and also showed that the smallest positively charged particle was obtained from hydrogen gas. This 'smallest particle' we now know is the proton. o Thompson ~1897 proposed his 'plum pudding' theory based on the growing evidence that atoms where themselves composed of even small more fundamental particles and the mass and charge of the proton and electron. Thompson envisaged a plumb pudding atom consisting of a positively charged 'pudding' with just enough lighter negatively charged electrons embedded in it to produce a neutral atom. The positive balancing the negative was correct but the relative size and nature of the nucleus were not. o Between 1910-1914 Millikan established the value of the electric charge on an electron in his famous 'oil drop' experiments, hence the mass of the electron could be calculated. o From 1902-1910 Rutherford, Geiger and Marsden and others used alpha particle scattering experiments (GCSE-AS atomic structure notes) to establish the concept of the nucleus and were even able to make an estimate of the value of its positive charge (which we now know equals the atomic/proton number). Even at that stage it was recognised that this positive nuclear charge bore some relationship to the order of the elements, as given by 'atomic weights', which Mendeleev and others were using to construct their periodic table. o Experimentally the 'atomic number' of an element was established by Chadwick in 1920 from beta particle scattering experiments (an atoms electrons deflecting the bombarding beta particle electrons) and from the X-ray spectra results of Moseley in 1913. Moseley showed that when atoms were bombarded with cathode rays (electrons) X-rays where produced. It was found that the square root of the highest energy emission line (called the K alpha line, Kα) gave a linear plot with the apparent atomic number. However the plot of √K α against atomic weight (relative atomic mass) gave a zig-zag plot. Therefore finally establishing that the really important 'chemical identity number' was the charge on the nucleus, i.e. what we know as the atomic/proton number and this would be the crucial number for ordering the elements, ultimately into the modern periodic table. o However, there was still the problem of why the atomic mass and atomic number where different i.e. in the case of the lighter elements, the atomic weight was often about twice the atomic number. In 1919 Aston developed a cathode ray tube i.e. like those used by Wien and Thompson etc. into a 'mass spectrograph', which we now know as a mass spectrometer GCSE-AS atomic structure notes. This showed that atoms of the same element had different masses but there was no experimental evidence that they had different atomic numbers (which of course they didn't). In 1920 Rutherford suggested there might be a 'missing' neutral particle and in 1932 Chadwick discovered the neutron by bombarding beryllium atoms with alpha particles which produced a beam of neutrons 9 4 12 1  4Be + 2He ==> 6C + 0n  Incidentally, the neutrons are unaffected by electrical and magnetic fields and not directly 'observed', they were primarily detected because they produced a beam of protons on collision with molecules of the hydrocarbon wax by a sort of snooker ball collision effect. The protons are readily detected and characterised (mass 1, charge +10 and their formation linked to the presence of a neutral particle of the same mass (neutron mass 1, charge 0). o Once the nature of the neutron was finally deduced by Chadwick, it completely explained the nature of isotopes and backed up the ideas from Moseley's work that the fundamentally important number that characterises an element is its atomic number and NOT the atomic mass.

1.5 A modern version Periodic Table based on the electronic structure of atoms The electronic basis of the periodic table is explained in Part 2. s block

Pd

1 2 3

3d to 6d blocks of Transition Metals (Periods 4 to 7), note p block 1 10 that the 1st (d ) and 10th (d ) are NOT true transition Gp1 Gp2 Gp3/13 Gp4/14 Gp5/15 Gp6/16 Gp7/17 Gp0/18 elements. 1H 3Li

Note: (i) H does not readily fit into any group, (ii) He not strictly a 'p' element but does belong in Gp 0/18 4Be

11Na 12Mg

The full IUPAC modern Periodic Table of Elements (ZSymbol, z = atomic or proton number)

2He

5B

6C

7N

8O

9F

10Ne

13Al

14Si

15P

16S

17Cl

18Ar

4

19K 20Ca

21Sc

22Ti

5

37Rb 38Sr

39Y

40Zr

6

55Cs 56Ba

*57-71

72Hf

7

87Fr 88Ra

*89-103

23V

24Cr 25Mn 26Fe 27Co

31Ga

32Ge

33As

34Se

35Br

36Kr

41Nb 42Mo 43Tc 44Ru 45Rh 46Pd 47Ag 48Cd

49In

50Sn

51Sb

52Te

53I

54Xe

73Ta

79Au 80Hg

81Tl

82Pb

83Bi

84Po

85At

86Rn

104Rf 105Db 106Sg 107Bh 108Hs 109Mt 110Ds 111Rg 112Cn

113?

114Fl

115?

116Lv

117?

118?

*57La

58Ce

59Pr

66Dy

67Ho

68Er

69Tm

70Yb

71Lu

*89Ac

90Th

91Pa

98Cf

99Es

100Fm

101Md

102No

103Lr

74W

75Re 76Os

77Ir

28Ni

78Pt

29Cu

30Zn

60Nd 61Pm 62Sm 63Eu 64Gd 65Tb 92U

93Np 94Pu 95Am 96Cm 97Bk

Gp 1 Alkali Metals

*Horizontal insert in Period 6 of the Lanthanide Metal Series (Lanthanides/Lanthanoids) Z=57 to 71 including the 4f-block series. *Horizontal insert in Period 7 of the Actinide Series of Metals Gp 2 Alkaline Earth (Actinides/Actinoids) Z=89-103 including the 5f-block series. Metals 1. Using 0 to denote the Group number of the Noble Gases is historic i.e. when its valency was considered zero since no compounds were known. However, from 1961 stable compounds Gp 7/17 Halogens of xenon have been synthesised exhibiting up to the maximum possible valency of 8 e.g. in XeO4. Gp 0/18 Noble Gases 2. Because of the horizontal series of elements e.g. like the Sc to Zn block (10 elements), Groups 3 to 7 and 0 can also be numbered as Groups 13 to 18 to fit in with the maximum number of vertical columns of elements in periods 4 and 5 (18 elements per period 4 and Take note of the four period 5). points on the right 3. This means that 21Sc to 30Zn can be now considered as the top elements in the vertical Groups 3 to 12. 4. I'm afraid this can make things confusing, but there it is, classification is still in progress and the notation Group 1 to 18 seems due to become universal.

 



 

With increasing knowledge of the elements of the Periodic Table it is now laid out in order of atomic (proton) number. Due to isotopic masses, the relative atomic mass does go 'up/down' occasionally (there is no obvious 'nuclear' rule that accounts for this, at least at GCSE/GCE level!). BUT chemically Te is like S and Se etc. and I is like Cl and Br etc. and so are placed in their correct 'chemically similar family' group and this is now backed up by modern knowledge of the electron structure of atoms. We now know the electronic structure of elements and can understand sub-levels and the 'rules' in electron structure e.g. 2 in shell 1 (period 1, 2 elements H to He), 8 in shell 2 (period 2, 8 elements Li to Ne), there is a sub-level which allows an extra 10 elements (the transition metals) in period 4 (18 elements, K to Kr). this also explains the sorting out of Mendeleev's A and B double columns in a group. The periods are complete now that we know about Noble Gases. The use and function of the Periodic Table will never cease! Newly 'man-made' elements are being synthesised. In the 1940's Glenn Seaborg was part of a research team developing the materials required to produce the first atomic bombs dropped on Hiroshima and Nagasaki. He specialised in separating all the substances made in the first nuclear reactors and helped discover the series of 'nuclear synthesised' elements beyond the naturally occurring limit of uranium (92U). From element 93 to 113 are now known, so the structure of the bottom part of the periodic table will continue to grow. There is plenty of scope for present day, and future Mendeleev's!!!! (will you be one of em'?). o Atomic structure history note: From 1913 onwards the electron structure of atoms was gradually being understood and paralleling the developing knowledge of the structure of the nucleus and its importance in determining which element an atom was i.e. the atomic/proton number. The Bohr theory of the hydrogen spectrum (see section 2.6) postulated that the electrons surrounding the positive nucleus could only exist in specific energy levels and that any electron level change must involve a specific input/output of energy the quanta e.g. a photon of light or X-rays etc. o In the 1920's and 1930's scientist-mathematicians like Heisenberg and Schrödinger were developing the mathematical equations known as wave mechanics. These mathematical theories describe the detailed behaviour of electrons, and out of these equations come the four quantum numbers from which are derived the set of rules we use to assign electrons in their respective levels (see section 2.2), which ultimately determines the chemistry of an element.

1.6 Where did elements come from originally? Where do we get the elements from? Where did elements come from originally? It all starts in the STARS! 

The ultimate origin of all elements is the nuclear reactions that go on when stars are formed from inter-stellar dust and gas forming a huge combined mass due to gravity, and then 'chunks' of a star cool down to form planets. The heaviest elements are formed in nuclear fusion reactions when stars self-destruct in super-nova explosions. o The nucleosynthesis of light elements up to Z = 26 (Fe, iron) occurs in stars formed from the condensation of hydrogen and helium atoms. o Eventually, as the mass increases, the force of gravity causes such compression that the temperatures rise considerably at high matter densities and nuclear reactions begin. o Up to Z = 26 nuclei, they are usually formed energy releasing fusion processes or the decay of unstable nuclei o There are hundreds of possible nuclear transformations possible, so, below, I've chosen some examples of possible nuclear reactions, whose products fit in with the isotopes, mass numbers, relative atomic masses etc. which A level chemistry students are likely to come across ... A o ... in the nuclear equations, for the nuclide symbol ZX, A = mass number, Z = atomic number, X = element symbol  and gamma photons, and neutrinos, often formed in nuclear changes, are NOT shown for simplicity. 1 1 2 1 o H + 1 1H ==> 1H + 0n 7  The formation of hydrogen-2 (deuterium) occurs slowly at a temperature of 1.5 x 10 K e.g. in our o Sun (~15 million C). 2 1 3 o 1H + 1H ==> 2He  This is one way the next heaviest element, helium can be formed. 3 3 4 1 o 2He + 2He ==> 2He + 2 1H  From helium-3, the formation of helium-4, the most common isotope of helium we find on earth.  From helium-4, by what is known as the alpha process, a succession of heavier elements can be synthesised in subsequent nuclear reactions ... 4 8 o 2 2He ==> 4Be  beryllium-8 is very unstable, but is readily converted on collision with another helium-4 nucleus to give stable carbon-12 (and ultimately life on earth!) 8 4 12 o 4Be + 2He ==> 6C 8 o  These sorts of nuclear reactions need star temperatures of 1 x 10 K (~100 million C) 12 4 16 o 6C + 2He ==> 8O  From carbon-12 you can get oxygen-16, the most common oxygen isotope on earth today. 16 4 20 o 8O + 2He ==> 10Ne  From oxygen-16 you can get neon-20, the most common neon isotope on earth today. 20 4 24 o 10Ne + 2He ==> 12Mg  From neon-20 you can get magnesium-24, the most common magnesium isotope on earth today. 24 4 28 o 10Mg + 2He ==> 14Si  From magnesium-24 you can get silicon-28, the most common silicon isotope on earth today. 28 4 32 o 14Si + 2He ==> 16S  From silicon-28 you can get sulfur-32, the most common sulphur isotope on earth today. 32 4 36 o 16S + 2He ==> 18 Ar  From sulphur-32 you can get argon-36, the most common argon isotope on earth today 36 4 40 o 18 Ar + 2He ==> 20Ca  From argon-36 you can get calcium-40, the most common calcium isotope on earth today.

o o



You can see from the Periodic Table of relative atomic masses how the alpha-process ('helium burning' has produced the values for C, O, Ne, Mg, Si, S, Ar and Ca from the principal isotope of multiples of four mass units. o There are lots of other possibilities involving H and He nuclei and particularly complicated nuclear fusion cycle involving carbon nuclei e.g. the six step cycle ... 12 1 13  6C + 1H ==> 7N 13 13 0  7N ==> cC + +e 13 1 14  6C + 1H ==> 7N 14 1 15  7N + 1H ==> 8O 15 15 0  8O ==> 7N + +e (decay of oxygen-15 by positron emission) 17 1 12 4  7N + 1H ==> 6C + 2He  You can also see how other isotopes of an element can be formed and in the cycle carbon-12 is reformed to continue these particular nucleosynthesis pathways.  There is a good analogy here with auto-catalytic cycles in chemistry e.g. the removal of ozone by chlorine atoms. o The heavier elements beyond iron i.e. Z > 26 Co cobalt onwards must be formed by energy absorbing processes including neutron capture e.g. the formation of technetium from molybdenum 98 1 99  42Mo + 0n ==> 42Mo  neutron absorbed by molybdenum-98 nucleus to give an unstable Mo nucleus 99 99 0  42Mo ==> 43Tc + -e  Mo-98 nucleus decays by beta particle emission to give an atom of technetium, with a higher atomic number.  Similarly, gallium can be formed from zinc, i.e. again forming an element of higher atomic number ... 68 1 69 69 69 0  followed by 30Zn + 0n ==> 30Zn 30Zn ==> 31Ga + -e o So you can see that these nuclear fusion, neutron or proton capture, nuclear decay etc., can over time, gradually produce all the heavier elements up to element 92 uranium, the last of our naturally occurring elements. 238 o Even though small amounts of 92U are eventually formed, it requires the highest of temperature e.g. in a super-nova explosion of giant stars a lot bigger than our sun! o Some examples of nuclear fusion reactions to form heavier elements are quoted in Part 3.4 Where do heavier elements come from? All the elements from atomic numbers 1-92 (H-U) naturally occur on Earth, though some are very unstableradioactive and decay to form more nuclear stable elements. o Many isotopes of elements after lead, 82Pb are unstable. o After uranium, 92U, the vast majority of the isotopes of the elements of atomic number 93+ are inherently unstable. o They will not have survived even if they were formed billions of years ago in the Sun, and retained or formed in the initial 'spin-off' material that formed the 'very early' Earth. o However, the advent of nuclear reactors has enabled up to kg quantities of e.g. plutonium, 94Pu (used in nuclear reactors and weapons) and americium, 95Am (used in smoke alarms) to be produced. o Cyclotrons, particle bombardment linear accelerators, have enabled 'super-heavy' elements up to Z = 118? to be 'synthesised', but only a few atoms at a time (The Russia-US space race seems to have been partly replaced by 'who can synthesize the biggest atom'!). o One things for certain, the Periodic Table still keeps growing with newly synthesised elements!

Where, and how, do we get the elements from the earth? 





Everything around you is made up of the elements of the periodic table, BUT most are chemically combined with other elements in the form of many naturally occurring compounds e.g. o hydrogen and oxygen in water, sodium and chlorine in sodium chloride ('common salt'), iron, oxygen and carbon as iron carbonate, carbon and oxygen as carbon dioxide etc. etc.! Therefore, most elements can only be obtained by some kind of chemical process to separate or extract an element from a compound e.g. o Less reactive metals are obtained by reduction of their oxides with carbon and more reactive metals are extracted by electrolysis of their chlorides or oxides (see GCSE/IGCSE/AS notes on Metal Extraction) o Non-metals are obtained by a variety of means e.g. chlorine is obtained by electrolysis of sodium chloride solution (see GCSE/IGCSE notes on Group 7 The Halogens). However some elements never occur as compounds or they occur in their elemental form as well as in compounds e.g. o The Group 0 Noble Gases are so unreactive they are only present in the atmosphere as individual atoms. Since air is a mixture, these gases are separated from air by a physical method of separation by distillation of liquified air. The elements oxygen and nitrogen are obtained from air at the same time, which is far more convenient than trying to get them from compounds like oxides and nitrates etc. o Gold/platinum is are the least reactive metals and are usually found 'native' as the yellow/silver elemental metal. o Relatively unreactive metals like copper and silver can also be found in their elemental form in mineral deposits as well as in metal ores containing compounds like copper carbonate, copper sulphide and silver sulphide. o The non-metal sulphur is found combined with oxygen and a metal in compounds known as sulphates, but it can occur as relatively pure sulphur in yellow mineral beds of the element.

ATOMIC STRUCTURE

1. The Structure of Atoms – three fundamental particles WHAT ARE ATOMS? and WHAT DO WE MEAN BY FUNDAMENTAL PARTICLES? (sub–atomic particles) Compounds are formed when two or more elements are chemically combined to form a new substance in a reaction which is not easily reversed ie its difficult to separate a compound back into its constituent elements. However, what are elements made up of? How and why do elements bond together? In order to answer these questions we must look a bit deeper into the fundamental structure of matter, that is everything around you! Atoms are the smallest particles of matter whose properties we study in Chemistry. Every element or compound is comprised of atoms. All the atoms are the same in the structure of an element (ignoring isotopes) and two or more different atoms/elements must be present in a compound. Initially, once the concept of an atom was established, it was assumed that atoms were indestructible and not divisible into smaller particles, but merely combined in different proportions to give the range of compounds we know about. th

th

However from experiments done in the late 19 and early 20 century it was deduced that atoms are made up of three fundamental or sub–atomic particles called protons, neutrons and electrons, which are listed below with their relative masses and electrical charges. WHAT ARE THE CHARACTERISTIC PROPERTIES OF THESE SUB–ATOMIC PARTICLES?

WHAT IS THE NUCLEUS? WHAT ARE NUCLEONS? The three fundamental particles of which atoms are composed The table gives the relative mass and electric charge of the three sub–atomic particles known as the proton, neutron and electron Sub–atomic particle

Relative mass

Electric charge

Comments

Proton

1

+1 (+ positive)

In the nucleus, a nucleon

Neutron

1

0 (zero)

In the nucleus, a nucleon

–1 (– negative)

NOT a nucleon. Electrons are arranged in energy levels or shells in orbit around the nucleus

Electron

1

/1850 or 0.00055

1

You can think of the mass of an electron as about /2000th of the mass of a proton or neutron, so, a pretty small mass BUT they occupy most of the space of atom!!!

What can we say about 'A Portrait of an Atom'? – an image of what you can't see!

However this diagram, which is based on the Bohr model of atomic structure, although more realistic in terms of the real size of the nucleus compared to the atom as a whole, it is not convenient to give a brief diagrammatic picture of the composition of an atom like the style of the diagram below. The central nucleus of protons and neutrons (most of mass) is extremely small even compared to the size of an atom. The rest of the 'almost empty space' of an atom is occupied by the negative electrons, held by, and moving around the positive nucleus in their energy levels or 'shells'. The electrons are also pretty tiny in mass too, compared to a proton or neutron. Bohr theorised the negative electrons can only exist in certain specific energy levels (shells) held in place by the positive nucleus. All of these theories must, and have been, backed up by repeated and varied experiments. As each new experiment was/is done, it must support the current theory or the theory needs to be modified to take into account new discoveries. Some of these important experiments are described further down the page. Even new experimental findings written up in research papers should be thoroughly peer reviewed, that is checked by scientists of at least equal academic ranking to the researchers. That's how science works! The number of protons in the nucleus of an atom decides what element that atom is. e.g. if the atom has 3 protons in the nucleus, it cannot be anything except lithium! Elements consist of one type of atom only.

Some more concise and handy styles to show the atomic composition of the same lithium atom  



 

  

 

What sub–atomic particles make up atoms? What is their mass and charge? The diagram above of a 'portrait of an atom' gives some idea on the structure of an atom (sometimes called the Bohr Atomic Model), it also includes some important definitions and notation used to describe atomic structure o The three fundamental particles you need to know are ... o proton: particle mass = 1, electric charge = +1, the charged particle in the nucleus o neutron: particle mass = 1, charge = 0, uncharged particle in the nucleus o electron: particle mass = 1/1850 ~1/2000, electric charge = –1, NOT in nucleus but exist in electronic energy levels around the nucleus (a sort of orbit, often described as a shell, see later). o The nucleus of protons and neutrons is tiny, even compared to the tiny atom!  So most of the volume of an atom is where the electrons are, the diameter of the nucleus of protons plus neutrons is about a ten thousandth of the diameter of an atom!  Since the nucleus is composed of positive protons and neutral neutrons, the nucleus itself must be positive.  A neutral atom carries no overall charge because the number of positive protons equals the number of negative electrons, and this information is given by the atomic/proton number. Protons and neutrons are the 'nucleons' or 'sub–atomic' particles present in the minute positive nucleus and the negative electrons are held by the positive protons in 'orbits' called energy levels or shells. o Although the nucleus must be positive because of the positive protons (neutrons are neutral) an individual atom is neutral because the number of electrons equals the number of protons – so the charges 'cancel out'. o If electrons are removed from an atom you get a positive ion and if electrons are added to an atom you get a negative ion.  An ion, by definition, cannot be neutral. Some important evidence for this 'picture' is obtained from alpha particle scattering experiments (see Appendix 1). The atomic number (Z) is the number of protons in the nucleus and is also known as the proton number of the particular element. o Each element has its own atomic number, so all the atoms of a particular element have the same atomic number. It is the proton/atomic number (Z) that determines the number of electrons an element has, its specific electron structure and therefore the specific identity of a particular element in terms of its physical and chemical properties. It cannot be overemphasised that it is the electronic structure that determines the chemical character of an element, hence the proton/atomic number determines everything about a particular element The mass number (A) is also known as the nucleon number, that is the number of particles in the nucleus of a particular atom–isotope (notes on isotopes – definition and examples). o Therefore the mass/nucleon number = sum of the protons plus neutrons in the nucleus. o Since the mass of a proton or neutron equals 1, the mass number equals the mass of an atom to the nearest whole number. o Relative Atomic Mass is dealt with on a separate calculation page The neutron number (N) = mass number – proton/atomic number In a neutral atom the number of protons (+) equals the number of electrons (–), that is the number of positive charges is equal to the number of negative charges. o If not, and the atom has an overall surplus or deficiency electrical charge, and is the resulting electrically charged particle called an ion e.g. +  the positive sodium ion Na (11 protons, 10 electrons, excess positive protons)





 or the negative chloride ion Cl (17 protons, 18 electrons, excess negative electrons)  for more details and examples see ionic bonding notes. In the example above for lithium–7, the nuclide notation states that o before the chemical symbol of the element Li o the top left number = nucleon/mass number = 7 o and the bottom left number = proton/atomic number = 3 o Similarly for ... o

, atom of hydrogen–1, symbol H, mass 1, just 1 proton and NO neutrons (only atom with no neutrons) , atom of helium–4, symbol He, mass 4, 2 protons, 4 – 2 = 2 neutrons

o

  

o , atom of sodium, symbol Na, mass 23, 11 protons, 23 – 11 = 12 neutrons The electrons are arranged in specific energy levels according to a set of rules (dealt with in section 3). This description of an atom consisting of the relatively minute nucleus of protons and neutrons surrounded by electrons in particular shells or energy levels is sometimes referred to as the Bohr Model of the atom, after the great Danish scientist Niels Bohr (1885–1962), one of the brilliant founders of modern atomic theory. Other examples of interpreting the nuclide notation and definition reminders: o Top left number is the nucleon number or mass number (A = sum of protons + neutrons = nucleons) o Bottom left number is the atomic number or proton number (Z = protons in nucleus) o Electrons = protons if the atom is electrically neutral i.e. NOT an ion. o The neutron number N = A – Z i.e. mass/nucleon number – atomic/proton number 

Iron atom (isotope iron–56), mass 56, 26 protons, 30 neutrons (56 – 26), 26 electrons



Cobalt atom (isotope cobalt–59), mass 59, 27 protons, 32 neutrons (59 – 27), 27 electrons



Californium atom (isotope californium–246), mass 246, 98 protons, 148 neutrons (246 – 98), 98 electrons So, at this point we had better explain, slightly belatedly, what isotopes are!



2. ISOTOPES WHAT ARE ISOTOPES? ARE THEY IMPORTANT? 

     



Isotopes are atoms of the same element with different numbers of neutrons and therefore different masses (different nucleon/mass numbers). o This gives each isotope of a particular element a different mass or nucleon number, but, being the same element they have the same atomic or proton number. o They are also chemically identical, because they have the same number of electrons, hence the same electron structure. o Study the diagrams of the isotopes of carbon further down the page. o Relative Isotopic Mass is dealt with on a separate calculation page The phrase 'heavier' or 'lighter' isotope means 'bigger' or 'smaller' mass number for a particular element. There are small physical differences between the isotopes e.g. the heavier isotope has a greater density or boiling point, the lighter the isotope the faster it diffuses. However, because they have the same number of protons (proton/atomic number) isotopes of a particular element have the same electronic structure and identical chemistry. Examples of isotopes are illustrated and described below. Caution Note: Do NOT assume the word isotope means the atom it is radioactive, this depends on the stability of the nucleus i.e. unstable atoms (radioactive) might be referred to as radioisotopes. Many isotopes are extremely stable in the nuclear sense and NOT radioactive i.e. most of the atoms that make up you and the world around you! hydrogen–1, hydrogen–2, and hydrogen–3 are the three isotopes of hydrogen with mass numbers of 1, 2 and 3, with 0, 1 and 2 neutrons respectively. All have 1 proton, since all are hydrogen! Hydrogen–1

is the most common, there is a trace of hydrogen–2 (sometimes called deuterium) naturally but hydrogen–3 (sometime called tritium) is very unstable and is used in atomic bombs – nuclear fusion weapons. 1 2 3 o They are sometimes denoted more simply as H, H and H since the chemical symbol H means hydrogen and therefore must have only one proton. 

3

4

and or He and He, are the two isotopes of helium with mass numbers of 3 and 4, with 1 and 2 neutrons respectively but both have 2 protons. Helium–3 is formed in the Sun by the initial nuclear fusion process. Helium–4 is also formed in the Sun and as a product of radioactive alpha decay of an unstable nucleus. o An alpha particle is a helium nucleus (mass 4, charge +2) and if it picks up two electrons it becomes a stable atoms of the gas helium. For more details see

Radioactivity Revision Notes Part 4



and or Na and Na, are the two isotopes of sodium with mass numbers of 23 and 24, with 12 and 13 neutrons respectively but both have 11 protons in the nucleus and 11 surrounding electrons. Sodium–23 is quite stable e.g. in common salt (NaCl, sodium chloride) but sodium–24 is a radio–isotope and is a gamma emitter used in medicine as a radioactive tracer e.g. to examine organs and the blood system.



and are the two nuclear symbols for the two most common and stable isotopes of the element chlorine. They both have 17 protons in the nucleus and 35–17 = 18 and 37–17 = 20 neutrons respectively (and both have 17 surrounding electrons).



and are the two nuclide symbols for the two most common and stable isotopes of the element bromine. They both have 35 protons in the nucleus and 79–35 = 44 neutrons and 81–35 = 46 neutrons respectively. By coincidence, there are almost exactly 50% of each isotope present in naturally occurring bromine. The three known isotopes of carbon



23

24

o

o o isotope

nuclide symbol protons neutrons electrons

% abundance

o o o

carbon–12

12 6C

6

6

6

98.9%, stable

carbon–13

13 6C

6

7

6

1.1%, stable

carbon–14

14 6C

6

8

6

trace, unstable radioactive

The table of information on the three isotopes of carbon is illustrated by the diagrams above it. Now is an appropriate point to introduce the concept and definition of relative atomic mass (Ar), which is required for very accurate quantitative chemistry calculations. The relative atomic mass of an element is the average mass of all the isotopes present compared to 12 1/12th of the mass of a carbon–12 atom ( C = 12.00000 amu i.e. the standard).  When you average the masses of the isotopes of carbon, taking into account their relative abundance (%), you arrive at a relative atomic mass of carbon of 12.011, Ar(C) = 12.011, though at this academic level 12.0 is accurate enough!  





See also chemical calculations on how to calculate relative atomic mass I've put this calculation on its own page because there is plenty on atomic structure already on this page!  Anything on this page relevant to the calculation of RAM is repeated on the page. Knowledge of isotopes is important in modern science. o Radioactive isotopes are used in medicine to trace aspects of body chemistry due to their radioactive emissions, and in chemical synthesis as tracers to follow how a reaction sequence occurs. o Radioactive isotopes are used in radiotherapy to kill malignant cancer cells.  For lots more details see the RADIOACTIVITY NOTES DO NOT CONFUSE ISOTOPES and ALLOTROPES – see Appendix 3.

3. The Electronic Structure of Atoms – rules to be learned WHAT DO WE MEAN BY Electron configuration, electronic structure of atoms – arrangement in shells or energy levels? 

  

 

The electrons are arranged in energy levels or shells around the nucleus and with 'orbits' on average increasing in distance from the nucleus. o The lowest energy levels are always filled first, you can think of the lower the shell, the nearer the nucleus, and numbered 1, 2, 3 etc. as the shell gets further from the nucleus. Each electron in an atom is in a particular energy level (or shell) and the electrons must occupy the lowest available energy level (or shell) available nearest the nucleus. When the level is full, the next electron goes into the next highest level (shell) available. There are rules to learn about the maximum number of electrons allowed in each shell and you have to be able to work out the arrangements for the first 20 elements (for GCSE students, upto at least 36 for Advanced level students). st o The 1 shell can contain a maximum of 2 electrons (electrons 1–2) o The 2nd shell can contain a maximum of 8 electrons (electrons 3–10) o The 3rd shell also has a maximum of 8 electrons (electrons 11–18) th th th o The 19 and 20 electrons go into the 4 shell, (required limit of GCSE knowledge). o Remember the total electrons to be arranged equals the atomic/proton number for a neutral atom. If you know the atomic (proton) number, you know it equals the number of electrons in a neutral atom, you then apply the rules to work out the electron arrangement (configuration). For elements 1 to 20 the electron arrangements/configurations are written out in the following manner: o Note that each number represents the number of electrons in a particular shell, dots or commas are used to separate the numbers of electrons in each shell. They are written out in order of increasing average distance from the positive nucleus which holds these negative electrons in their energy levels (shells). o The electron configurations are summarised below with reference to the periods of the periodic table and in order of increasing atomic number.  For more see the Periodic Table and Electron Structure notes below. o Period 1 – elements 1 to 2 (2 elements)  the electron arrangement is written out simply as 1 or 2 o Period 2 – elements 3 to 10 (8 elements)

o o

o

o o

o

 electron arrangements of 2.1 to 2.8 (since 1st shell is full with 2 electrons i.e. the first number) Period 3 – elements 11 to 18 (8 elements)  denoted by 2.8.1 to 2.8.8 (1st,2nd full shells with 2,8 electrons) Period 4 – first two elements 19 to 20  written out as 2.8.8.1 and 2.8.8.2 (1st,2nd,3rd full shells with 2,8,8 electrons)  Reminder – this is as far as GCSE students need to know, after that things get more complicated, BUT only for advanced level students!  For example, after element 18, the 3rd shell can hold a maximum of 18 electrons! The above is summarised in the diagram below

The electron shell arrangements are quoted in numbers e.g. 2.4 for C (carbon) but you need to be able to draw electron diagrams showing the electronic structure of the atom.  Some examples are given below and GCSE/IGCSE/O level students need to be able to work and draw the electronic structures of the first 20 elements.  You should notice that the number of shells used equals the period number of the element in the periodic table.  They can be all worked by the 'shell filling' rules described above. For the rest of Period 4 and other Periods you need a more advanced electron configuration system upto at least Z=36 using s, p, d and f orbital notation BUT for advanced level chemistry students only!

Examples: diagram, symbol or name of element (Atomic Number = number of protons and the number of electrons in a neutral atom), shorthand electron arrangement and a diagram to help you follow the numbers.

Filling 1st shell, electron level 1

Filling 2nd shell, electron level 2

2 elements only, Period 1 of the Periodic Table

to

to

3 of the 8 elements of Period 2

Filling 3rd shell, electron level 3

to

The first 2 elements of the 4th shell

3 of the 8 elements of Period 3

to Kr [2.8.18.8], start of Period 4

Only the first 2 of the 18 elements of Period 4 are shown above, the rule for 3rd shell changes from element 21 Sc onwards (studied at Advanced level, so GCSE students don't worry!) A few more 'snappy' examples – given atomic number, work out electron configuration (abbreviated to e.c.) Z = 3 e.c. 2.1 or Z = 7 e.c. = 2.5 or Z = 14 e.c. = 2.8.4 or Z = 19 e.c. = 2.8.8.1 etc. upto Z = 20

4. Which electron arrangements are stable and which are not? Both atoms and ions are considered    





WHY ARE SOME ELECTRON ARRANGEMENTS ARE MORE STABLE THAN OTHERS? WHICH ELECTRON ARRANGEMENTS ARE THE MOST STABLE AND WHICH ELECTRON ARRANGEMENTS THE LEAST STABLE? HOW DO ELECTRON ARRANGEMENTS RELATE TO THE REACTIVITY OF CHEMICAL ELEMENTS? When an atom has its outer level full to the maximum number of electrons allowed, the atom is particularly stable electronically and very unreactive. o This is the situation with the Noble Gases: He is [2], neon is [2,8] and argon is [2,8,8] etc. o There atoms are the most reluctant to lose, share or gain electrons in any sort of chemical interaction because they are so electronically stable. o For all elements most of their chemistry is about what outer electrons do or don't! o [2], [2,8] and [2,8,8] etc. are known as the 'stable Noble Gas arrangements', and the atoms of other elements try to attain this sort of electron structure when reacting to become more stable. o More details on Electron configuration notes for Advanced Level Chemistry Students The most reactive metals have just one outer electron. o These are the Group 1 Alkali Metals, lithium [2,1], sodium [2,8,1], potassium [2,8,8,1] o With one outer shell electron, they have one more electron than a stable Noble Gas electron structure. o So, they readily lose the outer electron when they chemically react to try to form (if possible) one of the stable Noble Gas electron arrangements – which is why atoms react in the first place! o When Group 1 Alkali Metal atoms lose an electron they form a positive ion because the positive proton number doesn't change, but with one negative electron lost, there is a surplus of one + charge e.g.  sodium atom ==> sodium ion +  Na ==> Na  is [2.8.1] ==> [2.8] electronically  in fundamental particles [11p + 11e] ==> [11p + 10e]  IONS are atoms or group of atoms which carry an overall electrical charge i.e. not electrically neutral. The most reactive non–metals are just one electron short of a full outer shell. o These are the Group 7 Halogens, namely fluorine [2,7], chlorine [2,8,7] etc.

o

 

These atoms are one electron short of a stable full outer shell and seek an 8th outer electron to become electronically stable – yet again, this is why atoms react! o They readily gain an outer electron, when they chemically react, to form one of the stable Noble Gas electron arrangements either by sharing electrons (in a covalent bond) or by electron transfer forming a singly charged negative ion (ionic bonding) e.g.  chlorine atom ==> chloride ion –  Cl ==> Cl  is [2.8.7] ==> [2.8.8] electronically  in fundamental particles [17p + 17e] ==> [17p + 18e]  the positive proton number of Cl doesn't change but the chloride ion carries one extra negative electron to give the surplus charge of a single – on the ion. EXTRA NOTE ON 'ATOMIC' NOTATION – representation of isotopes of ions Nuclide notation and ions (interpretation required for advanced level students only)

o

sodium–24 isotope ion, 11 protons, 13 neutrons, 10 electrons (one electron lost to form a positive ion)

o

sodium–23 isotope ion, 11, protons, 12 neutrons, 10 electrons (one electron lost to form a positive ion)

o



isotope sulfur–32 in the form of the sulfide ion, 16 protons, 16 neutrons, 18 electrons (two electrons gained to form the double charged negative ion) For more on electron structure and chemical changes and compound formation see ...



o GCSE/IGCSE/AS notes on CHEMICAL BONDING and for more on metal and non–metal reactivity see o

GCSE/IGCSE notes REACTIVITY SERIES of METALS

o

GCSE/IGCSE notes Group 1 ALKALI METALS

o

GCSE/IGCSE notes GROUP 7 HALOGENS

5. The Periodic Table and Electronic Structure – more patterns! Selected Elements of the Periodic Table are shown below with atomic number and chemical symbol. HOW DOES AN ELEMENT'S ELECTRON ARRANGEMENT RELATE TO ITS POSITION IN THE PERIODIC TABLE?



The elements are laid out in order of Atomic Number – that is the number of protons in the nucleus.



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It is important to realise that the 'chemical structure' of the periodic Table (shown above), that is the chemical similarity of vertical groups 'like' elements (apart from the Noble Gases), was known well before the electronic structure of atoms was understood. o In other words the elements are laid out in vertical columns (groups) and horizontal rows (periods) so that chemically (usually) VERY similar elements appear under each other – and there is a very good electronic structure reason for this! o However, it wasn't understood why they behaved in the same way chemically e.g. similar compound formulae and reactions etc. nor was it understood at first why Noble Gases were so unreactive towards other elements. o BUT, once the electronic structure of atoms was understood, 'electronic' theories could then be applied to explain the chemical similarity of elements in a vertical Group of the Periodic Table. Originally they were laid out in order of ' relative atomic mass' (the old term was 'atomic weight'). This is not correct for some elements now that we know their detailed atomic structure in terms of protons, neutrons and electrons, and of course, their chemical and physical properties. For example: Argon (at. no. 18, electrons 2,8,8) has a relative atomic mass of 40. Potassium (at. no. 19, electrons 2,8,8,1) has a relative atomic mass of 39. BUT Argon, in terms of its physical, chemical and electronic properties is clearly a Noble Gas in Group 0. Likewise, potassium is clearly an Alkali Metal in Group 1. Hydrogen, 1, H, does not readily fit into any group A Group is a vertical column of chemically and physically similar elements e.g. o Group 1 The Alkali Metals (Li, Na, K etc.) with one outer electron (one more than a Noble Gas structure), o Group 7 The Halogens (F, Cl, Br, I etc.) with seven outer electrons (one short of a Noble Gas arrangement) o and Group 0 The Noble Gases (He, Ne, Ar etc.). The group number equals the number of electrons in the outer shell (e.g. chlorine's electron arrangement is 2.8.7, the second element down Group 7 on period 3). A Period is a horizontal row of elements with a variety of properties (left to right goes from metallic to non– metallic elements. All the elements use the same number of electron shells which equals the period number (e.g. sodium's electron arrangement 2.8.1, the first element in Period 3). The ten elements Sc to Zn are called the Transition Metals Series and form part of a period between Group 2 and Group 3 from Period 4 onwards. Below are the electron arrangements for elements 1 to 20 set out in Periodic Table format (Hydrogen and The Transition metals etc. have been omitted). When you move down to the next period you start to fill in the next shell according to the maximum electrons in a shell rule (see previous section). NOTE: In the most modern periodic table notation Groups 3–7 and 0 are numbered Groups 3 to 18.

o o o o o o

The first element in a period has one outer electron (e.g. sodium Na 2.8.1), and the last element has a full outer shell (e.g. argon Ar 2.8.8) Apart from hydrogen (H, 1) and helium (He, 2) the last electron number is the group number (in the old notation) and the number of shells used is equal to the Period number. The periodicity of elements i.e. the repetition of very chemically similar elements in a group is due to the repetition of a the same outer electron structure – check out the last number from element 3 onwards. More GCSE/IGCSE notes on the Periodic Table and the

electronic explanations of chemical bonding–formulae

Advanced Level Chemistry – electron configurations/arrangements and the Periodic Table

APPENDIX 1. The history of the atom concept and the Alpha Particle Scattering Experiment (some of the ideas described here go above GCSE/IGCSE level!) The Greeks Leucippus and Democritus ~400 BC wondered what was the result of continually dividing a substance i.e. what was the end product or smallest bit i.e. what was left that was indivisible – the word atomic is from Greek adjective meaning 'not divisible'. The Greeks idea was not forgotten and later revived by Boyle and Newton but with little progress. However in 1808 Dalton proposed his atomic theory that all matter was made up of tiny hard particles/spheres called atoms and the different types of atoms (elements) combined together to give all the different substances of the physical world (all which of course is true, except for the 'hard solid indivisible spheres'!). He also produced the first list of 'atomic weights' (we now call relative atomic masses) on a scale based on hydrogen – given the arbitrary value of 1 since it was lightest element known, and, as it happens, correctly so. J J Thompson Around 1897 proposed his 'plum pudding' theory based on the growing evidence that atoms were themselves composed of even small more fundamental particles and the mass and charge of the proton and electron i.e. atoms were not hard indivisible spheres. His experiments had shown that atoms contained small negatively charged particle called electrons. From Thompson envisaged a plumb pudding atom consisting of a positively charged 'pudding' with just enough lighter negatively charged electrons embedded in it to produce a neutral atom. The idea of positive particles balancing the negative particles was correct but the relative size and nature of the nucleus were not. Ernest Rutherford, Hans Geiger and Ernest Marsden (the latter two were students of Rutherford at Cambridge University) conducted alpha particle scattering experiments (1902–1910, and described in detail below). These experiments established (i) minute nature of the nucleus even compared to the size of an atom (ii) the nucleus was positive and the positive charge varied from element to element. The Rutherford and Geiger–Marsden scattering experiments (1902–1910). When alpha particle beams are fired on very thin layers of metals (e.g. very fine gold leaf) some rather surprising results were by scientists of the early 20th century. o

By using a 360 charged particle detection system it was found that ... 3. most particles passed through un–deflected (as if there was nothing there!) 2. a small proportion were deflected slightly (so there was something there!) o

1. about 1 in 20,000 were 'bounced' back through an angle of over 90 , in other words were reflected backwards, a totally unexpected result. So, whatever was there was substantial, positively charged to cause the repulsion 'bounce', BUT not very big! From a detailed mathematical analysis of the scattering results, the only 'model' which could account for the pattern was an atom of ... ... mainly empty space (why most alpha particles passed through!), ... a positive centre (the nucleus) causing deflection (like charges repel, alpha particles are positively charged and so were being repelled by the 'later to be discovered' positive protons in the nucleus), ... a tiny dense centre of similar or greater charge or mass to an alpha particle (which we now call the nucleus), and this is the modern picture of the 'nuclear atom'. So an atom is quite well represented by the Bohr model of the atom diagram near the top of this page were we started! Bohr's suggestion that the negative electrons can only exist in certain specific energy levels (shells) held in place by

the positive nucleus complimented the Rutherford model of the atom to gives a reasonably complete picture of an atom (at least for this academic level!). Earlier theories of atomic structure, e.g. the 'plum pudding' model in which 'protons' and 'electrons' were scattered or arranged evenly across the atom, were superceded by this model. It was the only model that could explain the scattering of the high speed alpha particles by a small dense and positive atomic centre. Later experiments showed that the outer bits could be knocked off atoms and these had a very tiny mass and a negative charge, in other words the electron! Moseley studied the X–rays emitted by highly energised–ionised atoms and from the X–ray spectra of elements (the K alpha line, Kα) he was able to deduce the electric charge of the nucleus which we now know is equal to the atomic number of protons in the nucleus. Moseley showed that when atoms were bombarded with cathode rays (electrons) X–rays where produced. It was found that the square root of the highest energy emission line (called the K alpha line, K α) gave a linear plot with the apparent atomic number. However the plot of √Kα against atomic weight (relative atomic mass) gave a zig–zag plot. However, there was still the problem of why the atomic mass and atomic number where different i.e. in the case of the lighter elements, the atomic weight was often about twice the atomic number. In 1919 Aston developed a cathode ray tube i.e. like those used by Wien and Thompson etc. into a 'mass spectrograph', which we now know as a mass spectrometer GCSE–AS atomic structure notes. This showed that atoms of the same element had different masses but there was no experimental evidence that they had different atomic numbers (which of course they didn't). In 1920 Rutherford suggested there might be a 'missing' neutral particle and in 1932 Chadwick discovered the neutron by bombarding beryllium atoms with alpha particles which produced a beam of neutrons. It was not until 1932 that the nature of the neutron was finally deduced by Chadwick and this completely explained the nature of isotopes and backed up the ideas from Moseley's work that the fundamentally important number that characterises an element is its atomic number and NOT the atomic mass. See section 2. Radioactivity Notes page on other experiments with mixed particle beams and their separation.

Appendix 2. Atomic structure diagrams – some variations! e.g. for the element lithium

7 3Li

consisting of three protons and four neutrons

Appendix 3. Allotropes – don't confuse with isotopes!

WHAT ARE ALLOTROPES? As explained above, Isotopes are atoms of the same element with different masses due to different numbers of neutrons in the nucleus. Same protons and electrons. e.g. atomic number 6 = 6 protons = carbon, but there can be 6, 7 or 8 neutrons giving isotopes of carbon–12, 13 or 14. Oxygen atoms usually form 'stable' O2 oxygen molecules (also called dioxygen), BUT they can form an unstable molecule O3 ozone (also called trioxygen). The mass of the oxygen atoms in each of the molecules is mainly 16 (99.8%), and about 0.2% of two other stable isotopes of masses 17 and 18. Whatever isotope or isotopes make up the molecule, it doesn't affect the molecular structure or the respective chemistry of the O 2 or O3 molecules. However, what sometimes confuses the issue is the fact that oxygen O2 and ozone O3 are examples of allotropes. Allotropes are defined as different forms of the same element in the same physical state. The different physical allotropic forms arise from different arrangements of the atoms and molecules of the element and in the case of solids, different crystalline allotropes. They are usually chemically similar but always physically different in some way e.g. O2 (oxygen, dioxygen) and O3 (ozone, trioxygen) are both gases but have different densities, boiling points etc. Graphite, diamond and buckminsterfullerene are all solid allotropes of the element carbon and have significantly different physical and in some ways chemical properties! (details on bonding page) Rhombic and monoclinic sulphur have different geometrical crystal structures, that is different ways of packing the sulphur atoms (which are actually both made up of different packing arrangements of S8 ring molecules). They have different solubilities and melting points. There is also a 3rd unstable allotrope of sulfur called plastic sulphur made by pouring boiling molten sulphur into cold water which forms a black plastic material consisting of chains of sulphur atoms –S– S–S–S–S– etc.. It doesn't matter which isotopes make up the structure of any of an element's allotropes described above, so to summarise by one example ... oxygen–16, 17 or 18 are isotopes of oxygen with different nuclear structures due to different numbers of neutrons, and O2 and O3 are different molecular structures of the same element in the same physical state and are called allotropes irrespective of the isotopes that make up the molecules.

Part 2 Electronic Structure, Spectroscopy & Ionisation Energies Sections 2.1 The modern Periodic Table and 2.2 Electron configurations

2.1 The modern version of the Periodic Table is based on the electronic structure of atoms   



With our knowledge of atomic structure the modern Periodic Table is now laid out in order of atomic/proton number (Z) and any apparent anomalies sorted out. The atomic/proton number of the nucleus (Z) decides which element the atom is, the number of electrons surrounding the nucleus of a neutral atom and hence the element's chemistry which is based on the electron configuration. The full Periodic Table (Z = 1 to 112) is shown in section 2.4 with the element symbol, atomic/proton number (Z) and another version of the Periodic Table (Z = 1 to 56) showing the electron configuration which is introduced and explained in the next section 2.2. o In these notes the convention Z = atomic/proton number is used extensively, its a handy shorthand. Due to isotopic mass variations and their nuclear stability, the relative atomic mass does sometimes go 'up/down' as you proceed through the Periodic Table.

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o See Part 1 Mendeleev's Periodic table work. The use and function of the Periodic Table will never cease! Newly 'man–made' elements, beyond uranium (Z=92), are being 'synthesised' in nuclear reactors and cyclotrons. See GCSE/IGCSE nuclear reactions and radioactivity pages We now know the electronic structure of elements and can understand how the electrons are arranged in principal and sub–electronic levels and the 'quantum rules' of electron structure are understood. This knowledge now allows us to understand why the Periodic Table makes sense in terms of the known chemistry of the elements, and their subsequent classification, prior to the discovery and understanding of the significance of the sub–atomic particles, particularly the proton and electron and their 'arrangement' in an atom. Mendeleev and his contemporaries central ideas on classifying elements, despite some errors and omissions (i.e. not discovered), are now fully vindicated by our knowledge of the electronic structure of atoms. Mendeleev's powerful intuition on 'element patterns' was brought to full fruition by Rutherford and his contemporaries in discovering the secrets of the atom and quantum physicists elucidating the 'quantum patterns' of how multi–electron systems function. For the simplified version of expressing electronic arrangements up to atomic number 20 and the relationship of the element in the Periodic Table, see the GCSE/IGCSE Atomic Structure Notes. o Its not a bad idea to revise the basics before getting stuck into the advanced stuff!

2.2 Orbitals and the electronic structure of the atoms The details required by different pre–university syllabuses as regards background theory and orbital knowledge seems to vary quite a lot, so I've done by best to cater for all of them. If you wish to go straight to working out the s, p, d electron configuration of an element, click here! 

How to use the advanced s, p, d (f) notation for the electron configuration/arrangement of atoms/ions is outlined below, but no knowledge of quantum mechanics is required, but you do need to know how to work out electron arrangements from the rules and a little knowledge of the shape of orbitals wouldn't go amiss! You do NOT need to know the origin of the rules or know all about the four quantum numbers, BUT I can't stand pulling rules out of a hat, so I have given a little theoretical introduction, if can't stand that, tough!  To accurately describe an electron in an atom requires four quantum numbers which arise from solutions to the elaborate mathematical equations of quantum mechanics, which describe the exceedingly complex wave behaviour of electrons. o These four quantum numbers arise from solutions to the complex equations which describe the wave and quantised behaviour of electrons surrounding the nucleus. 1 o The first three quantum numbers have 0 or +/– integer values and the fourth one is +/– /2)  The Pauli exclusion principle states that no electron in an atom can have the same four quantum numbers, i.e. at least one must differ from electron to electron for a single atom.  The four quantum numbers are: 1. The principal quantum energy level number n or shell (n = 1,2, 3 ...), often just referred to as 'the level'. It is important to think of this as the principal energy level, i.e. the principal quantum level an electron can occupy. 2. The subsidiary/azimuthal/angular quantum number, l, this defines the 'spatial' type of sub–shell orbital, (l = 0 to n–1). often just referred to as 'the sub–level or more specifically the s/p/d/f sub–level' (see orbital diagrams later). Again, it is important to think of this as a sub–energy level of an electron.  For s orbital (l = 0), p orbital (l = 1), d orbital (l = 2) diagrams below, and for the f orbital (l = 3).  For a given principal quantum number the order of energy of the sub–level is s < p < d < f. 3. The magnetic or spatial orientation (of the orbital) quantum number, m, in terms of x,y,z axis (m = –l ... 0 ... l)  where l = the azimuthal quantum number 2. above and allows for each principal quantum level n, one s orbital for n = 1, 2, 3 etc., three p orbitals per for n = 2, 3, 4 etc., 5 d orbitals for n =3, 4, 5 etc. and seven orbitals for n = 4, 5, 6 etc.  See the orientation of the three p type orbitals and the five d type orbitals. 1 1 4. The electron's spin, s, which has the value of + /2 or – /2 and can be envisaged as the electron spinning clockwise/anti–clockwise in a full individual orbital.  Electrons possess spin and if an orbital is filled then the pair of electrons must have opposite spins (spin–paired).  This due to Pauli exclusion principle, which states that no electron can have the same four quantum numbers, since the other three quantum numbers would be the same for a specific orbital, it is the 1 spin quantum number which will differ (+/– /2).  The principal quantum electronic energy levels (n) can be split into sub–levels denoted by s, b, d and f depending on the number of electrons in the 'system'.  The 'space' in which the electron exists with its particular quantum level energy is called the atomic orbital and each type, s, p, d or f has its own particular 'shape' or 'shapes'.

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Each individual atomic orbital can 'hold' a maximum of two electrons. s, p and d orbital diagrams. o Orbital diagram notes: 1. The diagrams are NOT to scale and are somewhat simplified. 2. These are from theoretical calculations based on the probability functions of the peculiar behaviour of electrons from the deep realms of quantum mechanics! Don't worry about it! 3. These mathematical functions giving rise to an electron probability distribution e.g. illustrated by the pictures below of s, p and d orbitals. 4. They only give a very approximate representation of electron density. 5. Each orbital, that is the space a particular quantum level occupies, can hold a maximum of two electrons of opposite 1 spin quantum number (+/– /2) 6. Quantum physicists would say that these picture are not real, its all matrix mathematics really, BUT chemists like pictures, and pictures can often help students understand difficult concepts and most importantly, use the concepts to describe chemical systems and predict properties of atoms and molecules etc.

o   o

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o

o

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s atomic orbital s orbitals have a spherical shell shape and the faint dark blue circle represents in cross–section, the region of maximum electron density. Only one s orbital exists for each principal quantum number denoted by 1s, 2s, 3s etc.

p orbitals o p orbitals are pairs of 'dumb–bells' aligned along the x, y and z axis at 90 to each other. There are three p orbitals for each principal quantum number from 2 onwards denoted by 2p, 3p and 4p etc.  e.g. 2p can be composed of 2px, 2py and 2pz if all three orbitals for a particular principal quantum number are occupied.  If a p sub–shell is full it holds a maximum of 3 x 2 = 6 electrons.  There is no 1p because quantum rules do not allow this.

*

d atomic orbitals  d orbitals have complex shapes, I say no more except their relative alignment is important in explaining the origin of colour in transition metal complexes.  There are five d orbitals for each principal quantum number from 3 onwards denoted by 3d, 4d, 5d etc.  If a d sub–shell is full it contains a maximum of 5 x 2 = 10 electrons.  There are no 1d or 2d quantum levels, the quantum rules do not permit these. o f orbitals – orbital shapes not relevant at this level, the first is the 4f level and there are 7 orbitals holding a maximum of 7 x 2 = 14 electrons if the sub–shell is full. Don't worry too much about all the 'quantum' details above, the important features to appreciate are described below. To sum up 'numerically' from the quantum number rules, for the principal quantum number n ... o Each atomic orbital can hold a maximum of two electrons. o For each principal quantum level n, the following rules apply ... o for n = 1, there is just one sub–shell: 1s, maximum of 2 electrons, o for n = 2 there are two sub–shells: 1 x 2s atomic orbital and 3 x 2p orbitals, maximum of 2 + 6 = 8 electrons,

o

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for n = 3 there are three sub–shells: 1 x 3s,3 x 3p orbitals and 5 x 3d orbitals, maximum of 2 + 6 + 10 = 18 electrons, o for n = 4 there are four sub–shells: 1 x 4s,3 x 4p orbitals, 5 x 4d orbitals and 7 x 4f orbitals, maximum of 2 + 6 + 10 + 14 = 32 electrons. o However the order of filling is not this simple (see below, with visual diagrammatic help). How do we work out electron the arrangement of an atom? The arrangement of electrons in the shells and orbitals is called the electronic configuration or electron arrangement/structure and is written out in a particular sequence. 3 The orbital electrons are denoted in the form of e.g. 2p 3 o means there are three electrons (super–script number ) o in the p sub–shell (the lower case letter) o and in the second principal quantum level/shell (prefix number 2). The quantum levels and associated orbitals are filled according to the Aufbau Principle which states that an electron goes into the lowest available energy level providing the following 'sub–rules' are obeyed. o The Pauli exclusion principle states that no two electrons can have the same four quantum numbers. o Hund's Rule of maximum multiplicity states that, as far as is possible, electrons will occupy orbitals so that they have parallel spins. This means if a set of sub–shell orbitals of the same energy level e.g. a 2p or 3d set, each orbital will be singly occupied before pairing (to minimise electron repulsion within a single atomic orbital, i.e. a lower energy state than paired electron orbitals and unoccupied orbitals. The orbitals are filled in a definite order to produce the system of lowest energy and any electron will go into the lowest available energy level. o The order of 'filling' for an electron configuration is shown in the diagram below. o It uses is a simple diagrammatic convention to show an atomic orbital as a box. o Electrons are shown as half–arrows (up/down to represent the different spin quantum number s), see the 2nd diagram.

o The order of filling (up to atomic number Z = 36, H to Kr) is 1s 2s 2p 3s 3p 4s 3d 4p, up to a total 36 electrons from Z = 1 to 36 i.e. the order of increasing energy of the subshell or energy sub–level. o Note the 'quirk' in order for filling the 3d sub–shell energy level (see also the diagram below). o Until atomic number 21 (Sc) is reached, the 3d level is too high in energy and the electrons go into the 4s level and then the 3d level is filled from Sc to Zn. o This, and other 'quirks' I'm afraid, are a feature of the quantum complexity of multi–electron systems, so just learn the rules and get on with life! After Z=30, the 'filling' of the 4p level begins with Ga (Z=31) and finishes with Kr (Z=36). After Z=36, and up to Z=56, so after 4p the filling order is, 5s 4d 5p 6s, thus completing period and starting period 6 (and also repeating the pattern of filling in period 4 including a 2nd block of metals, the 4d block. 2 2 6 2 6 3 2 o The diagram for vanadium (Z=23), 1s 2s 2p 3s 3p 3d 4s is shown below.

o o

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* Just a thought experiment do the following ...  'Empty' the 3d level of electron arrows and you get the diagram for calcium (Z = 20).  Fill up completely the 3d and 4p boxes with arrows and you get krypton (Z = 36) The table in Part 2.3 shows how they are written out up to Z = 56 and a few others and note the orbital order when writing out. They are written out in strict order of principal quantum number 1, 2, 3 etc. and each principal quantum number is followed by the s, p or d sub–levels etc., and this is irrespective of the order of filling, i.e. when writing out the configuration, you ignore the 3d filling 'quirk' described above. Also in the table, some are written out in box diagram format, each box represents an orbital with a maximum of two electrons of opposite spin (shown by the opposing arrows). o Note the electrons only pair up when all sub–orbitals are filled separately with a single electron (this minimises electron pair repulsion within an orbital). Elements with one or two outer s electrons, and no outer p or d electrons etc., are called s–block elements (Groups 1 and 2). Elements with at least one outer p electron are called p–block elements (Groups 3 to 8/0). Elements where the highest available d sub–shell is being filled are called d–block elements (*Transition Metals) and similarly elements where the highest available f sub–shell is being filled are called f–block elements (the Lanthanides and Actinides). o * Sc–Zn is the 3d block, BUT true transition elements form at least one chemically stable ion with a partly filled sub–shell of d electrons. 3+ 0 2+ 10 2  Sc only forms Sc [Ar]3d , and Zn only forms Zn [Ar]3d 4s , so the true 3d–block transition metals are from Ti to Cu.  Can you spot the other electronic 'quirks' for chromium and copper?  Explanation: It would appear that a half–filled 3d subshell (Cr) or a full 3d sub–shell (Cu) is a tad more stable than a full 4s level. Quantum theory dictates that electrons can only have certain specific 'quantised' energies and any electronic level change requires a specific energy change i.e. energy quanta absorbed or energy quanta emitted. o Any electron will occupy the lowest available energy level according to the Aufbau principle (previously described). o The order of 'filling' up to atomic number 56 from the lowest to highest quantum level is ...  1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s o Writing out electron configurations for atoms o BUT first, how to work out the electron arrangement from the atomic/proton number, o AND then how to write out the electron configuration. o To work out an electron arrangement for an atom, you start with the atomic number, then 'fill in' the levels and sub–levels according to the rule.  The electron configuration is written out in order of,  firstly, the principal quantum energy level  then within this level in s, p, d, f order  and the total number of electrons in each sub–energy level is shown as a super– script. o Example 1. sodium, Na, Z = 11 o 1s filled (2e) 9e's left, 2s filled (2e's) 7e left, 2p filled (6e's) 1e left, last electron goes into the 3s level. o According to the notation rule this is written as ... 2 2 6 1 o 1s 2s 2p 3s (2.8.1 in simplified shell notation) o Example 2. vanadium, V, Z = 23 o 1s filled (2e's) 21e's left, 2s filled (2e's) 19e left, 2p filled (6e's) 13e's left, 3s filled (2e's) 11e's left, 3p filled (6e's) 5e's left, 4s filled (2e's) 3e's left, last 3e's go into 3d level.

o o o o o o

According to the notation rule this is written as ... 2 2 6 2 6 3 2 1s 2s 2p 3s 3p 3d 4s (2.8.11.2 in simplified shell notation) Example 3. bromine, Br, Z = 35 2 2 6 2 6 Filling in the first 18e's as described in example 2. will give an argon structure (1s 2s 2p 3s 3p ), which can be abbreviated to [Ar], the next 2e's go into the 4s level (15e's left), the next 10e's go into the 3d level, the final 5e's go into the 4p level. 10 2 5 [Ar]3d 4s 4p (2.8.18.7 in simplified notation) Note the use of 'noble gas notation' as an abbreviation for all the filled inner sub–shells making up the equivalent of noble gas electron arrangement, and will not include the 'outer electrons').

Part 3. Survey of Period 1: hydrogen H to helium He

3.1 Survey of Period 1: H to He (2 elements, Z = 1 to 2)

Z = 1. Hydrogen H not in a Group   

The structure of the element: o Non-metal existing as a diatomic molecule, H2, with a single covalent bond. Physical properties: o o o Colourless gas, less dense than air; mpt -259 C (14K), bpt -253 C (20K). o Poor conductor of heat/electricity in any physical state. Group, electron configuration and oxidation states: 1 o Not in any group strictly speaking*; e.c. [1] or 1s o Oxidation states: (+1) e.g. in HCl, H2O and (-1) with electropositive metals e.g. sodium hydride, NaH or + Na H . o * It has been displayed at the top of Gp 1 Alkali Metals or Gp 7 (17) Halogen in the Periodic Table in the past, for the following reasons .. +  Gp 1: Forms H ion and in alkali metal vapour there is a tiny fraction of M2 molecules.  BUT, you can hardly argue it has any real metallic properties physically since it is a nonmetallic gas and chemically it doesn't/can't react with water like alkali metals to form an alkali and hydrogen!  When ignited it will react rapidly with oxygen/chlorine if initiation energy is supplied e.g. heat/uv light but the resulting compounds are very different,.  Water, H2O, is a covalent neutral molecule as opposed to the ionic group 1 oxide + 2 (M )2O , which forms a strong alkali in water.  Hydrogen chloride is an acidic covalent molecule dissolving in water to form a + strongly acid solution, whereas group 1 metal chlorides are ionic, M Cl , and dissolve to form neutral solutions with water.  Gp 7(17): Hydrogen is non metallic in character and occurs as the gaseous H2 molecule, being colourless fits in with increase in colour intensity down the Halogen group. It similarly chemically combines with a valency of 1, forming covalent simple molecular compounds with non-metals e.g. hydrogen chloride, HCl, methane, CH4 etc. as do halogens. It also forms ionic compounds with the + most electropositive metals e.g. sodium hydride, Na H where the hydride ion parallels the halide ion + and sodium hydride exist as white crystalline solid just like sodium chloride, Na Cl ('salt').  However, it hardly matches the halogen in reactivity e.g. it aught to be more reactive than fluorine and more importantly, its maximum oxidation state is +1 and doesn't have the extensive chemistry of halogens below fluorine. Fluorine only has an oxidation state of -1 in compounds and has no +1 compound, since it has the highest electronegativity of any element (4.0, hydrogen is 2.2).  From chlorine downwards, all the halogens exhibit a huge range of compounds with oxidation states of +1, +3, +5 and +7 (max. possible) and even +4 ox. state oxides. The oxides are acidic and the oxyanions acts as strong oxidising agents. Hydrogen oxide is water, is neutral and acts as relatively weak oxidising agent.









 



So, although it physically and chemically it fits in a bit more with Group 7/17 than Group 1, it does not really fit convincingly in any group, and many modern periodic tables show it on its own at the top on period 1. I'm afraid I'm irritated when its put at the top of Group 1 or 7 in the periodic table. Reaction of element with oxygen: o Explosive reaction when ignited in an air mixture, a jet of the gas burns with very pale blue flame to form the colourless covalent liquid molecule, water.  2H2(g) + O2(g) ==> 2H2O(l) Reaction of element with chlorine: o Explosive if the mixture is heated or subjected to uv light. The colourless covalent gas hydrogen chloride is formed.  H2(g) + Cl2(g) ==> 2HCl(g) Reaction of hydrogen with metals o Hydrogen will combine with alkali metals to form a series of Group 1 hydrides  H2(g) + 2M(s/l) ==> 2MH(s)  These are white ionic compounds.  Hydrides are formed with other reactive metals but their chemical structure is more complex, with different degrees of ionic/covalent character. Reaction of chloride with water: o Hydrogen chloride gas dissolves to form the strong hydrochloric acid (100% ionised). +  HCl(g) + H2O(l) ==> H3O (aq) + Cl (aq) Other comments: o In chemistry, its 'proton form' is the basis of the concept of Bronsted-Lowry acids and bases. o An acid is a proton donor e.g. hydrogen chloride. o A base is a proton acceptor e.g. ammonia. +  NH3(g) + HCl(g) ==> NH4 Cl (s)  The hydrogen donates a proton to the ammonia molecule. o Section 3.4 shows how the heavier elements are built up from hydrogen isotopes via nuclear reactions. Links to other pages on this site: o Preparation of gas : Test for gas : Acid-Base theory : o Use in ammonia-fertiliser production or margarine production :

3.2 Z = 2. Helium He in Group 0/18     



The structure of the element: o Non-metal existing as single atoms, He, sometimes described as 'monatomic molecules'. Physical properties: o o o Colourless gas, less dense than air; mpt -272 C (1K), bpt -269 C (4K) o Poor conductor of heat/electricity in any physical state. Group, electron configuration (and oxidation states): 2 o Gp0/18 Noble gas; e.c. [2] or 1s ; no stable oxidation states (other than 0 for the element itself!) so no compounds! Reaction with anything: o None! Far too stable electron configuration which also explains why it prefers to exist as single atoms. Other comments: o Last element in the period, as the outer principal quantum level 1 (shell 1) is full to the maximum number of electrons allowed, conferring extra chemical stability on the atom. -1 -1 o The 1st ionisation energy of helium (2372 kJmol ) is nearly twice that of hydrogen (1312 kJmol ) because the effective nuclear attraction charge is doubled for the same 1s quantum level (+1 to +2). Links to other pages on site: o GCSE notes on Group 0 Noble Gases o Advanced Level Chemistry Notes on Group 0/18 Noble Gases - Helium

3.3 Summary of Period 1: H to He (elements 1 to 2)   

Maximum of two elements in period 1 as there is a maximum of two electrons allowed in the 1st principal quantum level (1s) and only an s-orbital energy sub-level is allowed. Both elements are non-metallic gases, but a considerable difference in chemical reactivity! Hydrogen, as outlined above, has quite an extensive chemistry ie combines with all non-metals except the noble gases and can also combine with some reactive metals eg the group 1 alkali metals.



Helium, with one full shell only (outer = inner !) has the highest ionisation energy of any element and is chemically the most stable and least reactive of any element in the periodic table and has no meaningful chemistry.

3.4 Where do the heavier elements come from?    

There are 92 naturally occurring elements on planet Earth. All the 91 naturally occurring elements after hydrogen up to 92U, uranium, were formed in stars by nuclear fusion reactions. Extremely high temperatures are needed in stars to give the atomic nuclei enough kinetic energy to overcome the very powerful positive nucleus-nucleus repulsion forces and fuse together in a 'fruitful collision'. Examples of stellar nuclear fusion reactions building up the heavier elements from hydrogen and helium are shown below. key:

A ZX,

A = mass number, Z = atomic/proton number, X = element symbol From hydrogen, helium is formed e.g. the sequence ...

1 1H

+

1 0n

2

==> 1H (hydrogen-2, 'heavy hydrogen', 'deuterium') 2 1H

+

1 0n

3

==> 1H (hydrogen-3, 'tritium')

3 1H 3 2He

1

+ 0n ==>

4

2He

+

1 1H

3

==>

2He

1

+ 0n

(the most common isotope of helium)

From helium the heavier elements are formed as bigger and bigger nuclei fuse together. 4

8

e.g. 2 2He ==> [ 4Be] which is highly unstable and rapidly changes, on impact with a 3rd helium nucleus, into a carbon nucleus, 8

[ 4Be] +

4

2He

==>

12 6C

and from carbon-12, oxygen-16 and neon-20 are formed e.g. via 4 2He

2

12 6C

+

12 6C

==>

==>

20 10Ne

16 8O

+

4 2He

238

and so on, until even small amounts of 92U are eventually formed but require the highest of temperature e.g. in a super-nova explosion of giant stars a lot bigger than our sun! Many isotopes of elements after lead, 82Pb are unstable. After uranium, 92U, the vast majority of the isotopes of the elements of atomic number 93+ are inherently unstable. They will not have survived even if they were formed billions of years ago in the Sun, and retained or formed in the initial 'spin-off' material that formed the 'very early' Earth. However, the advent of nuclear reactors has enabled up to kg quantities of e.g. plutonium, 94Pu (used in nuclear reactors and weapons) and americium, 95Am (used in smoke alarms) to be produced. Cyclotrons, particle bombardment linear accelerators, have enabled 'super-heavy' elements up to Z = 118? to be 'synthesised', but only a few atoms at a time (The Russia-US space race seems to have been partly replaced by 'who can synthesize the biggest atom'.

Part 4. Revising Survey of Period 2 Li to Ne

Part 4. Survey of Period 2: Li across to Ne (8 elements, Z = 3 to 10) 4.1 Survey of the individual elements Li, Be, B, C, N, O, F and Ne

Z = 3 lithium Li in Group 1 Alkali Metals      

  

 

The structure of the element: o Giant lattice metallic structure of immobile positive metal ions surrounded by a 'sea' of freely moving mobile electrons (so-called delocalised electrons). Physical properties: o o o Relatively soft silvery solid, less dense than water; mpt 181 C; bpt 1347 C; good conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 1 o Gp1 Alkali Metal; e.c. 2,1 or 1s 2s ; (+1 only) e.g. LiCl, Li2O etc. Reaction of element with oxygen: + 2o Burns when heated in air to form the ionic white solid, lithium oxide, (Li )2O .  4Li(s) + O2(g) ==> 2Li2O(s) Reaction of oxide with water: o It is a basic oxide, readily dissolving/reacting to form an alkaline solution of lithium hydroxide of pH 13-14.  Li2O(s) + H2O(l) ==> 2LiOH(aq) Reaction of oxide or hydroxide with common mineral acids: o Li2O behaves as a basic oxide dissolving to form the chloride, sulphate and nitrate salt in the relevant dilute acid. o The hydroxide MOH is a strong base and alkali (since a soluble base) and similarly forms salts. o Li2O(s) + 2HCl(aq) ==> 2LiCl(aq) + H2O(l)  LiOH(aq) + HCl(aq) ==> LiCl(aq) + H2O(l) o Li2O(s) + H2SO4(aq) ==> Li2SO4(aq) + H2O(l)  2LiOH(aq) + H2SO4(aq) ==> Li2SO4(aq) + 2H2O(l) o Li2O(s) + 2HNO3(aq) ==> 2LiNO3(aq) + H2O(l)  LiOH(aq) + HNO3(aq) ==> LiNO3(aq) + H2O(l) o In all cases the ionic equations are + +  for M2O: Li2O(s) + 2H (aq) ==> 2Li (aq) + H2O(l) +  for MOH: OH (aq) + 2H (aq) ==> H2O(l) Reaction of oxide with strong bases/alkalis: o None, lithium oxide is ONLY basic. Reaction of element with chlorine: + o Burns when heated in chlorine to form white powder/colourless crystals of ionic lithium chloride, Li Cl .  2Li(s) + Cl2(g) ==> 2LiCl(s) Reaction of chloride with water: o The salt readily dissolves forming a neutral solution of hydrated lithium and chloride ions (~pH 7). +  LiCl(s) + aq ==> Li (aq) + Cl (aq) +  or LiCl(s) + 4H2O(l) ==> [Li(H2O)4] (aq) + Cl (aq)  The chloride ion is such a weak base it shows no chemical interaction with water and the lithium ion shows virtually no acidic character, so lithium chloride solution is neutral ~pH 7. Reaction of element with water: o The metal reacts quite quickly forming hydrogen gas and alkaline lithium hydroxide, pH 13-14.  2Li(s) + 2H2O(l) ==> 2LiOH(aq) + H2(g) Other comments:

Z = 4 Beryllium Be in Group 2 Alkaline Earth Metals    







 

 

The structure of the element: o Giant lattice metallic structure of immobile positive metal ions surrounded by a 'sea' of freely moving mobile electrons (so-called delocalised electrons). Physical properties: o o o Quite hard silvery solid; mpt 1278 C; bpt 2487 C; moderately good conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 o Gp2 Alkaline Earth Metal; e.c. 2,2 or 1s 2s ; (+2 only) e.g. BeCl2, BeO. Reaction of element with oxygen: o Forms a white powder of beryllium oxide (intermediate ionic-covalent character) when heated strongly in air.  2Be(s) + O2(g) ==> 2BeO(s) Reaction of oxide with water: o It is insoluble in water, and shows no reaction with water, but is an amphoteric oxide and reacts with acids or alkalis to form salts e.g.  with hydrochloric acid it forms beryllium chloride.  BeO(s) + 2HCl(aq) ==> BeCl2(aq) + H2O(l)  with sodium hydroxide it forms sodium beryllate(II)  BeO(s) + 2NaOH(aq) + H2O(l) ==> Na2[Be(OH)4](aq) Reaction of oxide with acids: o Behaves as a basic oxide dissolving to form the chloride, sulphate and nitrate salt in the relevant dilute acid. o BeO(s) + 2HCl(aq) ==> BeCl2(aq) + H2O(l) o BeO(s) + H2SO4(aq) ==> BeSO4(aq) + H2O(l) o BeO(s) + 2HNO3(aq) ==> Be(NO3)2(aq) + H2O(l) + 2+ o In all cases the ionic equation is: BeO(s) + 2H (aq) ==> Be (aq) + H2O(l) Reaction of oxide with strong bases/alkalis: o The oxide also behaves as an acidic oxide by dissolving in strong soluble bases to form beryllate(II) salts. o BeO(s) + 2NaOH(aq) + H2O(l) ==> Na2[Be(OH)4](aq) o e.g. forming sodium beryllate(III) with sodium hydroxide. 2o ionic equation: BeO(s) + 2OH (aq) + H2O(l) ==> [Be(OH)4] (aq) o Therefore beryllium oxide is an amphoteric oxide, because of this dual acid-base behaviour. Reaction of element with chlorine: o Forms covalent beryllium chloride when heated in chlorine.  Be(s) + Cl2(g) ==> BeCl2(s) Reaction of chloride with water: o With an excess of water, the salt tends to hydrolyse to give a gelatinous beryllium hydroxide precipitate and a hydrochloric acid solution.  BeCl2(s) + 2H2O(l) Be(OH)2(s) + 2HCl(aq) Reaction of element with water: o None with cold water. Other comments: o Beryllium shows considerable anomalous behaviour compared to the rest of Group 2 (Mg to Ra) e.g. the covalent chloride and the amphoteric oxide. It also has a maximum co-ordination number of 4, e.g. in the tetra-aqa beryllium(II) ion and the beryllate(II) ion shown above. This applies to all the period 2 elements, but for period 3 e.g. magnesium, the maximum co-ordination number is 6.

Z = 5 Boron B in Group 3/13      

The structure of the element: o Non-metal existing as a giant covalent lattice, Bn, where n is an extremely large number. Physical properties: o o o Hard high melting solid; mpt 2300 C; bpt 3659 C; poor conductor heat/electricity. Group, electron configuration (and oxidation states): 2 2 1 o Gp3; e.c. 2,3 or 1s 2s 2p ; (+3 only) e.g. B2O3 and BCl3 etc. Reaction of element with oxygen: o Reacts when heated strongly in air to form boron oxide which has a giant covalent structure.  4B(s) + 3O2(g) ==> 2B2O3(s) Reaction of oxide with water: o Insoluble, no reaction but it is a weakly acidic oxide. Reaction of oxide with acids:

  

 

o None, only acidic in acid-base behaviour. Reaction of oxide with strong bases/alkalis: o Presumably dissolves to give a solution of sodium borate. Reaction of element with chlorine: o Forms covalent liquid boron trichloride on heating in chlorine gas.  2B(s) + 3Cl2(g) ==> 2BCl3(l) Reaction of chloride with water: o It hydrolyses to form boric acid and hydrochloric acid. *  BCl3(l) + 3H2O(l) ==> B(OH)3(aq) + 3HCl(aq) *  can also be, but less accurately, written as H3BO3 Reaction of element with water: o None. Other comments: o -

Z = 6 Carbon C in Group 4/14 

  



 





 

The structure of the element: o Non-metal existing as three allotropes covalently bonded. Diamond (tetrahedral bond network) and graphite (layers of connected hexagonal rings) have giant covalent structures Cn where n is an extremely large number, and a series of large molecules (3rd allotrope) called fullerenes e.g. C60. o Bonding details and diagrams of the allotropes of carbon. Physical properties: o o o mpt 3547 C; bpt 4827 C; very hard colourless/light coloured diamond (poor conductor) or dark softish/slippery crystals of graphite (moderate conductor of heat/electricity). Group, electron configuration (and oxidation states): 2 2 2 o Gp4; e.c. 2,4 or 1s 2s 2p ; (can be +2, but usually +4) e.g. o (+2) CO, (+4) CO2 and CCl4 etc. Reaction of element with oxygen: o Burns when heated in air to form carbon dioxide gas.  C(s) + O2(g) ==> CO2(g)  In limited air/oxygen, carbon monoxide would be formed too.  2C(s) + O2(g) ==> 2CO(g) Reaction of carbon dioxide with water: o Quite soluble to form a weakly acid solution of pH 4-5. So called carbonic acid, H2CO3, does not really exist, but the dissolved carbon dioxide reacts with water to form hydrogen/oxonium ions and hydrogencarbonate ions. The equilibrium is very much on the left - hence the fizz in 'fizzy drinks'! +  CO2(aq) + 2H2O(l) H3O (aq) + HCO3 (aq) Reaction of oxide with acids: o None, only acidic in acid-base behaviour. Reaction of oxide with bases/alkalis: o It is a weakly acidic oxide dissolving sodium hydroxide solution to form sodium carbonate. o CO2(g) + 2NaOH(aq) ==> Na2CO3(aq) + H2O(l) 2o ionic equation: CO2(g) + 2OH (aq) ==> CO3 (aq) + H2O(l) o With excess of carbon dioxide, sodium hydrogencarbonate is formed. o CO2(g) + Na2CO3(aq) + H2O(l) ==> 2NaHCO3(aq) 2o ionic equation: CO2(g) + CO3 (aq) + H2O(l) ==> 2HCO3 (aq) Reaction of element with chlorine: o None directly. o Tetrachloromethane (carbon tetrachloride) is made by fully chlorinating methane in a multi-stage reaction.  CH4(g) + 4Cl2(g) ==> CCl4(l) + 4HCl(g) Reaction of chloride with water: * o None. CCl4(l) cannot readily act as a Lewis acid and accept a lone pair from a water molecule at the polar CCl bond to start the hydrolysis process. *  In the case of SiCl4, 3d orbitals can be used to accept a lone pair from water, so providing a mechanistic route for hydrolysis to occur. (compare with silicon). Reaction of element with water: o No reaction with cold water but red hot carbon reacts with steam to form carbon monoxide and hydrogen.  C(s) + H2O(g) ==> CO(g) + H2(g) Other comments: o All of organic chemistry is based on the compounds of carbon except for the oxides and carbonates.

Z = 7 Nitrogen N in Group 5/15    



 

 

 

The structure of the element: o Non-metal existing as diatomic molecule N2, with a triple covalent bond, N N. Physical properties: o o o Colourless gas; mpt -210 C; bpt -196 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 3 o Gp5; e.c. 2.5 or 1s 2s 2p ; Variety of oxidation states from -3 to +5 e.g. o NH3 (-3), N2O (+1), NO (+2), NCl3 (+3), NO2 (+4) and N2O5 and HNO3 (+5). Reaction of element with oxygen: o At high temperatures e.g. in car engines, nitrogen(II) oxide (nitrogen monoxide) is formed.  N2(g) + O2(g) ==> 2NO(g) o and the nitrogen(II) oxide rapidly reacts in air to form nitrogen(IV) oxide (nitrogen dioxide).  2NO(g) + O2(g) ==> 2NO2(g) o The theoretical highest oxide is N2O5 nitrogen(V) oxide (nitrogen pentoxide) and does exist. Reaction of oxides with water: o Nitrogen(IV) oxide dissolves to form an acidic solution of weak nitrous acid and strong nitric acid.  2NO2(g) + H2O(l) ==> HNO2(aq) + HNO3(aq) +  or 2NO2(g) + 2H2O(l) ==> HNO2(aq) + H3O (aq) + NO3 (aq) o NO and N2O are neutral oxides but nitrogen(V) oxide is strongly acidic and dissolves to form nitric acid.  N2O5(s) + H2O(l) ==> 2HNO3(aq) +  or N2O5(s) + 3H2O(l) ==> 2H3O (aq) + 2NO3 (aq) Reaction of oxides with acids: o None, only acidic (N2O3 (very unstable), NO2 and N2O5) or neutral (N2O and NO), in nature. Reaction of oxides with bases/alkalis: o Nitrogen(IV) oxide or nitrogen dioxide forms sodium nitrite and sodium nitrate with sodium hydroxide solution. o 2NO2(g) + 2NaOH(aq) ==> NaNO2(aq) + NaNO3(aq) + H2O(l) o ionic equation: 2NO2(g) + 2OH (aq) ==> NO2 (aq) + NO3 (aq) + H2O(l) o As well as being a neutralisation reaction, it is also a redox reaction, the oxidation states of oxygen (-2) and hydrogen (+1) do not change BUT the oxidation state of nitrogen changes from two at (+4) to one at (+3) and one at (+5). The simultaneous change of an element into an lower and upper oxidation sate is sometimes called disproportionation. Reaction of element with chlorine: o None, but the unstable yellow oily liquid chloride can be made indirectly. Reaction of chloride with water: o Slowly hydrolyses to form weak nitrous acid and strong hydrochloric acid.  NCl3(l) + 2H2O(l) ==> HNO2(aq) + 3HCl(aq) +  or NCl3(l) + 2H2O(l) ==> HNO2(aq) + 3H (aq) + 3Cl (aq) +  or NCl3(l) + 5H2O(l) ==> HNO2(aq) + 3H3O (aq) + 3Cl (aq) Reaction of element with water: o Slightly soluble but no reaction. Other comments: o An essential element for plants, hence need for nitrogen compounds in compost and artificial fertilisers (NPK bags!).

Z = 8 Oxygen O in Group 6/16 

 

The structure of the element: o Non-metal existing as diatomic molecule, O2, with a double covalent bond. o It has two allotropes: 'normal oxygen' O2 (dioxygen above) and the highly unstable and reactive gas ozone, O3 (trioxygen). Physical properties of the element: o o o O2 is a colourless gas; mpt -218 C; bpt -183 C; poor conductor of heat/electricity.  O3 is a pale blue gas. Group, electron configuration (and oxidation states): 2 2 4 o Gp6; e.c. 2,6 or 1s 2s 2p ; Normally (-2) e.g. H2O, CO2 etc. but can have other ox. states ...



    

o e.g. H2O2 (-1), F2O (+2). Reaction of element with oxygen: o O2 molecules won't react with themselves BUT in the upper atmosphere oxygen atoms are formed by high energy radiation/particle collision with oxygen molecules causing homolytic bond fission to produce free oxygen atom (free radicals). These combine with oxygen molecules to form ozone. Ozone can be synthesised by an electric discharge through oxygen. . .  (i) O2 = hv => 2O (ii) O + O2 ==> O3 Reaction of oxide with water, acids or bases/alkalis: Not applicable. Reaction of element with chlorine: o None, but unstable chlorine(I) oxide (chlorine monoxide) can be made indirectly and there are other chlorine oxides. (see chlorine) Reaction of chloride with water: o Slowly hydrolyses to form weak chloric(I) acid.  Cl2O(g) + H2O(l) ==> 2HClO(aq) Reaction of element with water: o Slightly soluble but no reaction. Other comments: o Formed in plant photosynthesis. Consumed in respiration.

Z = 9 Fluorine F in Group 7/17 The Halogens     

    

 

The structure of the element: o Non-metal existing as covalent diatomic molecule, F2, with a single bond. Physical properties: o o o Pale yellow gas; mpt -219 C; bpt -188 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 5 o Gp7 Halogen; e.c. 2,7 or 1s 2s 2p ; (only -1) e.g. HF, ClF, F2O (O is +2!) o An extremely reactive element and readily combines with almost every other element. Reaction of element with oxygen: o None, but oxygen difluoride, F2O, can be made indirectly. Reaction of oxide with water: o The oxide hydrolyses to form hydrofluoric acid and oxygen (it is powerful enough to oxidise water. This is an anomalous reaction for a Group 7 element due to the high oxidising power of oxygen in the +2 state in F 2O.  F2O(g) + H2O(l) ==> 2HF(aq) + O2(g) Reaction of oxide with acids: o It readily reacts with water as above. Reaction of oxide with bases/alkalis: o Presumably fluoride salt is formed and oxygen, as it oxidises water. Reaction of element with chlorine: o Can combine directly or indirectly to form ClF, ClF3, ClF5 and ClF7. o e.g. Cl2(g) + F2(g) ==> 2ClF(g) Reaction of chloride with water: o ClFx + H2O => ? Reaction of element with water: o Reacts to form hydrofluoric acid and oxygen. This is an anomalous reaction for a Group 7 element due to the high oxidising power of fluorine.  2F2(g) + 2H2O(l) ==> 4HF(aq) + O2(g) Other comments: o Links to other pages on site: o 4.2 Period 2 element trends & explanations of physical properties o 4.3 Period 2 element trends in bonding, structure, oxidation state, formulae & reactions o 6.4 Important element trends down a Group o 9 Advanced Level Chemistry Notes on Group 7 Halogens - Fluorine o KS4 Science GCSE/IGCSE Revision notes on The Halogens

Z = 10 Neon Ne in Group 0/18 The Noble Gases     

The structure of the element: o Exists as single atoms, Ne, sometimes described as 'monatomic molecules'. Physical properties: o o o Colourless gas, less dense than air; mpt -249 C, bpt -246 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 o Gp0/8 Noble gas; e.c. 2,8 or 1s 2s 2p ; no stable oxidation states (other than 0 for the element itself!) so no compounds! Reaction with anything: o None! Far too stable electron configuration. Other comments: o Last element in the period, as the outer principal quantum level (shell) is full to the maximum number of electrons, conferring extra chemical stability on the atom.

Part 4 Survey of Period 2: Li across to Ne (8 elements, Z = 3 to 10) 4.2 Period 2 trends and explanations of selected physical properties Element

Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine -1

1st ionization energy (kJ mol )

Neon

513

900

801

1086

1402

1314

1681

2081

Atomic metallic or covalent radius (pm, /1000 for nm)

152 (met)

111 (met)

88 (cov)

77 (cov)

70 (cov)

66 (cov)

64 (cov)

51 (cov)

Electronegativity (Pauling scale)

0.98

1.57

2.04

2.55

3.04

3.44

3.98

4.84

Melting Point (K)

454

1551

2573

3820

63

55

54

24

Boiling Point (K)

1620

2760

3932

5100

77

90

85

27

Relative electrical conductivity

0.150

0.250

giant covalent structure ==> small covalent molecules  The change in bonding character from ionic to covalent in the oxide follows the decreasing difference in electronegativity between that of the element and oxygen. o In terms of chemical character ... o basic oxide => amphoteric oxide => weakly acidic oxide => strongly acidic oxide o and metal basic oxides ==> metal amphoteric oxides ==> non-metal oxides o This is chemically characteristic of metallic compound ==> non-metallic compound character. o -



Reaction of element with chlorine and the structure of the chloride

(Gp 1) 2Li(s) + Cl2(g) ==> 2LiCl(s)

(Gp 2) Be(s) + Cl2(g) ==> BeCl2(s)

(Gp 3) 2B(s) + 3Cl2(g) ==> 2BCl3(l) needs high temperature

(Gp 4) carbon has no reaction with chlorine

(Gp 5) nitrogen has no reaction with chlorine

(Gp 6) oxygen has no reaction with chlorine

(Gp 7) Cl2(g) + F2(g) ==> 2ClF(g)

(Gp 0) neon has no reaction with chlorine



Reaction with chlorine and the structure of the chloride o The first three element will combine directly on heating in chlorine to give the expected chloride formula. + o Lithium gives a giant ionic lattice of LiCl or Li Cl , o Beryllium a polymeric covalent lattice of BeCl2. o Boron forms small covalent molecules of BCl3. o The others are also small covalent molecules, made indirectly, CCl4, NCl3, Cl2O o In terms of the chloride the (oxidation states) in the chlorides are o (+1) LiCl, (+2) BeCl2, (+3) BCl3, (+4) CCl4, (+3) NCl3, (-2) Cl2O, (+1) ClF, no neon chloride o So the number of atoms of chlorine combined with the Period 2 element (the valency) follows the pattern  1 2 3 4 3 2 1 0 o The overall pattern, from left to right across period 2 is ... o giant ionic lattice => polymeric covalent lattice ==> small covalent molecules. o The change in bonding character from ionic to covalent in the chloride, follows the decreasing difference in electronegativity between that of the element and chlorine, as in the case of oxides. o This is chemically characteristic of metallic ==> non-metallic element character. o -



Reaction of the chlorides with water +

(Gp 1) LiCl(s) + aq ==> Li neutral ~pH 7

-

(aq)

+ Cl (aq) just dissolves,

(Gp 2) BeCl2(s) + 2H2O(l) Be(OH)2(s) + 2HCl(aq) hydrolysis giving an acid solution

*

(Gp 3) BCl3(l) + 3H2O(l) ==> B(OH)3(aq) + 3HCl(aq) hydrolysis (Gp 4) CCl4 no reaction with water giving an acid solution (Gp 5) NCl3(l) + 2H2O(l) ==> HNO2(aq) + 3HCl(aq) hydrolysis giving an acid solution

(Gp 6) Cl2O(g) + H2O(l) ==> 2HClO(aq) hydrolysis giving an acid solution

(Gp 7) ClF + H2O ==> acidic solution ???

(Gp 0) neon has no chloride



The reaction of chlorides with water o Ionic LiCl dissolves to form a nearly neutral solution of hydrated lithium and ions. o The covalent beryllium chloride BeCl2 hydrolyses to give beryllium hydroxide and hydrochloric acid. o Boron trichloride BCl3 similarly hydrolyse to form acid solutions. o Tetrachloromethane CCl4 (carbon tetrachloride) is quite stable in water. o Nitrogen trichloride NCl3 slowly hydrolyses to give an acid solution. o Chlorine(I) oxide Cl2O hydrolyses to give an acidic solution. o FCl or ClF gives acidic solution ??? o The general trend is for ionic chloride salts to give nearly neutral solutions => covalent chlorides that hydrolyse to give acidic solutions. o -



Reaction of element with water

(Gp 1) 2Li(s) + 2H2O(l) ==> 2LiOH(aq) + H2(g)

(Gp 2) beryllium has no reaction with cold water

(Gp 3) boron has no reaction with water

(Gp 4) C(s) + H2O(g) ==> CO(g) + H2(g) at high temperature

(Gp 5) nitrogen has no reaction with water

(Gp 6) oxygen has no reaction with water

(Gp 7) 2F2(g) + 2H2O(l) ==> 4HF(aq) + O2(g)

(Gp 0) argon has no reaction with water





  

The reaction of the element with water o The reactive metal Li gives the hydroxide and hydrogen. o Beryllium and boron have no reaction with water. o Carbon reacts at high temperature. o Nitrogen and oxygen have no reaction with water. o Fluorine reacts violently forming oxygen and hydrogen fluoride.  Fluorine is such a powerful oxidising agent that it can oxidise water itself! o The 'limited' pattern for period 2 (or any other period), is to have reactive metals on the left forming an alkaline solution of a hydroxide and hydrogen and non-metals on the right forming acid solutions IF they react with water (only fluorine). o The hydrides MHx o For hydrides the difference in electronegativity works both ways! o From left to right across the period you change from an ionic lithium hydride crystal lattice e.g. +  Li H to small non-polar molecule covalent hydrides (methane CH 4)  and then a polar weakly basic covalent hydride molecule (ammonia NH 3)  and finally a quite acidic polar covalent molecule (hydrogen fluoride HF). o The formulae follow a simple period pattern of rising and falling valency combinations.  LiH BeH2 B2H6 (= to BH3) CH4 NH3 H2O HF no neon hydride  valency combing power of the element gives a 1 2 3 4 3 2 1 0 pattern o On reaction with water, the ionic metal hydrides at the start of the period give an alkaline solution  e.g. LiH(s) + H2O(l) ==> LiOH(aq) + H2(g)  MgH2(s) + 2H2O(l) ==> Mg(OH)2(aq/s) + 2H2(g) o In the middle are neutral hydrides like methane which in contact with water do not change the pH. o Then you get weak base ==> moderately strongly acidic hydrides when they dissolve in water e.g. +  weak base: NH3(aq) + H2O(l) NH4 (aq) + OH (aq) +  quite strong acid: HF(aq) + H2O(l) ==> H3O (aq) + F (aq) o So things are a bit complicated with hydrides on period 2 and not the greatest of patterns! o Radii of isoelectronic ions Isoelectronic means species having the same total number of electrons. The table below considers the isoelectronic anions associated with Periods 2 (and the table continues with Period 3 cations).

 isoelectronic Group Group Group Group (Group Group system 4/14 5/15 6/16 7/17 1 0/18) Period

Period 2

[Ne] 10e 2 2 6 1s 2s 2p

C

total nuclear charge

+6

+7

radius in picometre (pm)

260

171

4-

N

3-

Group 3/13

Period 3 -

(Ne)

Na

Mg

+8

+9

(+10)

+11

+12

+13

140

136

(38112*)

95

65

50

O

2-

Group 2

F

+

2+

Al

3+

name of ion carbide nitride oxide fluoride (neon) sodium magnesium aluminium

   

Excluding the noble gases themselves, there is a clear pattern of decreasing ionic radius with increase in nuclear charge (+ atomic/proton number) for the two isoelectronic series tabulated above. From left to right the proton/electron ratio is steadily increasing so that the electrons are experiencing an increasingly greater attractive force of the nucleus for the same number of electrons, hence the steady decrease in radii for an isoelectronic series. * all sorts of values are quoted for noble gas radii e.g. atomic, covalent and ionic, but most don't fit in the pattern above which is quite clear for all the cations and anions listed. More isoelectronic radii are at the end of the Period 3 survey pages or the end of the Period 4 survey pages.

Survey of Period 3: Na across to Ar (8 elements, Z = 11 to 18) 5.1 Survey of the individual elements Na, Mg, Al, Si, P, S, Cl, Ar

Z = 11 Sodium Na in Group 1 Alkali Metals    





 



 

The structure of the element: o Giant lattice metallic structure of immobile positive metal ions surrounded by a 'sea' of freely moving mobile electrons (so-called delocalised electrons). Physical properties: o o o A Soft silvery solid, less dense than water; mpt 98 C; bpt 883 C; good conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 1 o Gp1 Alkali Metal; e.c. 2,8,1 or 1s 2s 2p 3s ; (+1 only) e.g. in NaCl, Na2O etc. Reaction of element with oxygen: + 2o Burns when heated in air to form the ionic white solid oxides, sodium oxide, Na2O or (Na )2O . + 2 4Na(s) + O2(g) ==> 2Na2O(s) sodium oxide (Na )2O . + 2 and Na(s) + O2(g) ==> Na2O2(s) sodium peroxide, (Na )2[O2] Reaction of oxide with water: o Na2O is a very basic oxide, readily dissolving/reacting to form an alkaline solution of sodium hydroxide of pH 14.  Na2O(s) + H2O(l) ==> 2NaOH(aq) Reaction of oxide with acids: o Behaves as a basic oxide dissolving to form the chloride, sulphate and nitrate salt in the relevant dilute acid. o Na2O(s) + 2HCl(aq) ==> 2NaCl(aq) + H2O(l) o Na2O(s) + H2SO4(aq) ==> Na2SO4(aq) + H2O(l) o Na2O(s) + 2HNO3(aq) ==> 2NaNO3(aq) + H2O(l) + + o In all cases the ionic equation is: Na2O(s) + 2H (aq) ==> 2Na (aq) + H2O(l) Reaction of oxide with strong bases/alkalis: o None, sodium oxides are ONLY basic. Reaction of element with chlorine: o Burns when heated in chlorine to form white powder/colourless crystals of ionic sodium chloride, NaCl or + Na Cl .  2Na(s) + Cl2(g) ==> 2NaCl(s) Reaction of chloride with water: o The salt readily dissolves forming a neutral solution of sodium and chloride ions (pH 7). +  NaCl(s) + aq ==> Na (aq) + Cl (aq)  The chloride ion is such a weak base it shows no chemical interaction with water and the sodium ion shows virtually no acidic character, so sodium chloride solution is neutral ~pH 7. Reaction of element with water: o The metal reacts quite quickly forming hydrogen gas and alkaline sodium hydroxide, pH 14.  2Na(s) + 2H2O(l) ==> 2NaOH(aq) + H2(g) Other comments: o -

Z = 12 Magnesium Mg in Group 2 Alkaline Earth Metals    

The structure of the element: o Giant lattice metallic structure of immobile positive metal ions surrounded by a 'sea' of freely moving mobile electrons (so-called delocalised electrons). Physical properties: o o o A moderately hard silvery-white solid; mpt 649 C; bpt 1090 C; good conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 o Gp2 Alkaline Earth Metal; e.c. 2,8,2 or 1s 2s 2p 3s ; (+2 only) e.g. MgCl2, MgO. Reaction of element with oxygen:

2+

o





  





2-

Burns brightly when heated in air to form a white powder of ionic magnesium oxide Mg O when heated strongly in air.  2Mg(s) + O2(g) ==> 2MgO(s) Reaction of oxide with water: o It is slightly soluble in water, and is a basic oxide forming an alkaline solution of magnesium hydroxide 2+ o Mg (OH )2, of about pH12.  MgO(s) + H2O(l) ==> Mg(OH)2(aq) Reaction of oxide with acids: o Behaves as a basic oxide dissolving to form the chloride, sulphate and nitrate salt in the relevant dilute acid. o MgO(s) + 2HCl(aq) ==> MgCl2(aq) + H2O(l) o MgO(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) o MgO(s) + 2HNO3(aq) ==> Mg(NO3)2(aq) + H2O(l) + 2+ o In all cases the ionic equation is: MgO(s) + 2H (aq) ==> Mg (aq) + H2O(l) Reaction of oxide with strong bases/alkalis: o None, sodium oxide is ONLY basic. Reaction of element with chlorine: 2+ o Forms colourless solid ionic magnesium chloride, Mg (Cl )2, when heated in chlorine.  Mg(s) + Cl2(g) ==> MgCl2(s) Reaction of chloride with water: o The salt dissolves in water forming a nearly neutral solution of about pH6. 2+  MgCl2(s) + aq ==> Mg (aq) + 2Cl (aq) o or more correctly ... 2+  MgCl2(s) + 6H2O(l) ==> [Mg(H2O)6] (aq) + 2Cl (aq) o The solution is slightly acidic, because the hexa-aqa magnesium ion can donate a proton to a water + molecule forming the oxonium ion, more simply H (aq). 2+ + +  [Mg(H2O)6] (aq) + H2O(l) [Mg(H2O)5OH] (aq) + H3O (aq) Reaction of element with water: o Very slow reaction with cold water to form hydrogen (bubbles form slowly on the surface) and alkaline magnesium hydroxide.  Mg(s) + 2H2O(l) ==> Mg(OH)2(aq) + H2(g) o Ignited magnesium will continue to burn in steam to form the white powder of magnesium oxide and hydrogen gas and you get the same reaction if steam is passed over heated magnesium.  Mg(s) + H2O(g) ==> MgO(s) + H2(g) Other comments: o -

Z = 13 Aluminium Al in Group 3/13    

 



The structure of the element: o Giant lattice metallic structure of immobile positive metal ions surrounded by a 'sea' of freely moving mobile electrons (so-called delocalised electrons). Physical properties: o o o Moderately hard silvery-white high melting solid; mpt 661 C; bpt 2467 C; good conductor heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 1 o Gp3; e.c. 2,8,3 or 1s 2s 2p 3s 3p ; (+3 only) e.g. Al2O3 and AlCl3. Reaction of element with oxygen: o Reacts when heated strongly in air to form a white powder of aluminium oxide which has a giant ionic 3+ 2structure, (Al )2(O )3.  4Al(s) + 3O2(g) ==> 2Al2O3(s)  The above reaction occurs very rapidly on a freshly cut aluminium surface, but the microscopic oxide layer inhibits any further reaction, giving aluminium a 'lower reactivity' than expected, and its excellent anti-corrosion properties. Reaction of oxide with water: o Insoluble, no reaction but it is an amphoteric oxide and forms salts with both acids and alkali (see below). Reaction of oxide with acids: o It behaves as a basic oxide dissolving to form the chloride, sulphate and nitrate salt in the relevant dilute acid. o Al2O3(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2O(l) o Al2O3(s) + 3H2SO4(aq) ==> Al2(SO4)3(aq) + 3H2O(l) o Al2O3(s) + 6HNO3(aq) ==> 2Al(NO3)3(aq) + 3H2O(l) + 3+ o ionic equation: Al2O3(s) + 6H (aq) ==> 2Al (aq) + 3H2O(l) Reaction of oxide with strong bases/alkalis: o The oxide also behaves as an acidic oxide by dissolving in strong soluble bases to form aluminate(III) salts. o e.g. Al2O3(s) + 2NaOH(aq) + 3H2O(l) ==> 2Na[Al(OH)4](aq)



o forming sodium aluminate(III) with sodium hydroxide. o ionic equation: Al2O3(s) + 2OH (aq) + 3H2O(l) ==> 2[Al(OH)4] (aq) o Therefore aluminium oxide is an amphoteric oxide, because of this dual acid-base behaviour. Reaction of element with chlorine:

o o

*

Burns when heated strongly in chlorine gas to form the white solid aluminium chloride on heating in chlorine gas.  2Al(s) + 3Cl2(g) ==> 2AlCl3(s) *  It is often a faint yellow in colour, due to traces of iron forming iron(III) chloride.

 



 

Aluminium chloride is a curious substance in its behaviour. The solid, AlCl3, consists of an ionic 3+ o lattice of Al ions, each surrounded by six Cl ions, BUT on heating, at about 180 C, the thermal kinetic energy of vibration of the ions in the lattice is sufficient to cause it break down and sublimation takes place (s ==> g). In doing so the co-ordination number of the aluminium changes from six to four to form the readily vapourised covalent dimer molecule, Al2Cl6, shown above. Reaction of chloride with water: o With a little water it rapidly, and exothermically hydrolyses to form aluminium hydroxide and nasty fumes of hydrogen chloride gas.  AlCl3(s) + 3H2O(l) ==> Al(OH)3(s) + 3HCl(g) o However, if a large excess of water is rapidly added, a weakly acidic solution of aluminium chloride is formed, with the minimum of nasty fumes! 3+  AlCl3(s) + aq ==> Al (aq) + 3Cl (aq) 3+  or more correctly: AlCl3(s) + 6H2O(l) ==> [Al(H2O)6] (aq) + 3Cl (aq) o The solution is slightly acidic, because the hexa-aqa aluminium ion can donate a proton to a water molecule forming the oxonium ion. 3+ 2+ +  [Al(H2O)6] (aq) + H2O(l) [Al(H2O)5OH] (aq) + H3O (aq) Reaction of element with water: o None due to protective oxide layer. Reactions of the hexa-aqua aluminium ion: o It gives a gelatinous white precipitate with sodium hydroxide or ammonia solution which displays amphoteric behaviour by dissolving in excess strong alkali (NaOH(aq), NOT NH3(aq)) and acids. 3+  Al (aq) + 3OH (aq) ==> Al(OH)3(s) 3+  or [Al(H2O)6] (aq) + 3OH (aq) ==> [Al(OH)3(H2O)3] + 3H2O(l)  The hydroxide readily dissolves in acids to form salts: + 3+  Al(OH)3(s) + 3H (aq) ==> Al (aq) + 3H2O(l) + 3+  or more elaborately: [Al(OH)3(H2O)3] + 3H3O (aq) [Al(H2O)6] (aq) + 3H2O(l)  Thus showing amphoteric behaviour, since the hydroxide ppt. also dissolves in excess strong alkali (below). 3+ 3 [Al(H2O)6] (aq) + 6OH (aq) ==> [Al(OH)6] (aq) + 6H2O(l) (from original aqueous ion) 3 or [Al(OH)3(H2O)3](s) + 3OH (aq) ==> [Al(OH)6] (aq) + 3H2O(l) (from hydroxide ppt.) 3 or more simply Al(OH)3(s) + 3OH (aq) ==> [Al(OH)6] (aq) (from hydroxide ppt.) o With aqueous sodium carbonate solution, the hydroxide ppt. is formed, and, because of its acidic nature, bubbles of carbon dioxide gas are evolved. 3+ 22+  2[Al(H2O)6] (aq) + CO3 (aq) 2[Al(H2O)5(OH)] (aq) + H2O(l) + CO2(g)  this process of proton donation continues until the gelatinous ppt. [Al(OH)3(H2O)3](s) is formed, but will not dissolve in excess of the weak base/alkali.  See Appendix 1 in Transition Metal Chemistry pages for more examples and equations etc.  No Cr2(CO3)3 is formed because of the acid-base reaction above, due to the acidity of the 3+ 3+ chromium(III) ion. Note the similarly highly charged and small ions Cr and Fe behave in the same way.

Z = 14 Silicon Si in Group 4/14       

   

The structure of the element: o Non-metal existing as a giant covalent lattice, Sin, where n is an extremely large number, held together by tetrahedrally arranged Si-Si bonds. Physical properties: o o o Hard high melting solid; mpt 1410 C; bpt 2355 C; poor conductor of heat/electricity, but with other elements added, conducts better, hence use in microchips. Group, electron configuration (and oxidation states): 2 2 6 2 2 o Gp4; e.c. 2,8,4 or 1s 2s 2p 3s 3p ; (+4 only) e.g. SiO2 and SiCl4 etc. Reaction of element with oxygen: o Reacts when strongly heated in air to form silicon dioxide (silica, silicon(IV) oxide).  Si(s) + O2(g) ==> SiO2(g) Reaction of oxide with water: o None and insoluble. Reaction of oxide with acids: o None, only acidic in nature. Reaction of oxide with bases/alkalis: o It is a weakly acidic oxide dissolving very slowly in hot concentrated sodium hydroxide solution to form sodium silicate. o SiO2(s) + 2NaOH(aq) ==> Na2SiO3(aq) + H2O(l) 2o or simplified ionic equation: SiO2(s) + 2OH (aq) ==> SiO3 (aq) + H2O(l) Reaction of element with chlorine: o On heating in chlorine forms the covalent liquid silicon tetrachloride.  Si(s) + 2Cl2(g) ==> SiCl4(l) Reaction of chloride with water: o Hydrolyses to form gelatinous hydrated silicon oxide and hydrochloric acid.  SiCl4(l) + 2H2O(l) ==> SiO2(s) + 4HCl(aq) Reaction of element with water: o None Other comments: o -

Z = 15 Phosphorus P in Group 5/15    





 

The structure of the element: o Two solid allotropes (red and white) consisting of P4 molecules, also a polymer form. Physical properties: o o o Colourless gas; mpt 44 C; bpt 280 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 3 o Gp5; e.c. 2,8,5 or 1s 2s 2p 3s 3p ; Variety of oxidation states from -3 to +5 e.g. o PH3 (-3), P4O6 (+3), P4O10, PCl5 and H3PO4 (+5). Reaction of element with oxygen: o With limited air/oxygen, on heating the phosphorus, the covalent white solid phosphorus(III) oxide is formed.  P4(s) + 3O2(g) ==> P4O6(s) o With excess air/oxygen, on heating the phosphorus, the covalent white solid phosphorus(V) oxide is formed.  P4(s) + 5O2(g) ==> P4O10(s) Reaction of the oxides with water: Both oxides dissolve in water to form acidic solutions. o Phosphoric(III) oxide forms phosphoric(III) acid.  P4O6(s) + 6H2O(l) ==> 4H3PO3(aq) o Phosphoric(III) oxide forms phosphoric(V) acid.  P4O10(s) + 6H2O(l) ==> 4H3PO4(aq) Reaction of element with chlorine: o With limited chlorine, on heating the phosphorus, the covalent liquid phosphorus(III) chloride is formed.  P4(s) + 3Cl2(g) ==> 4PCl3(l) * o With excess chlorine, on heating the phosphorus, the ionic solid phosphorus(III) chloride is formed.  P4(s) + 5Cl2(g) ==> 4PCl5(s) *  PCl5 is a bit unusual for an 'expected covalent' liquid chloride. +  It is an ionic solid with the structure [PCl4] [PCl6]  Hence its melting point is much greater than the liquid phosphorus(III) chloride, where the molecules are only held together by the inter-molecular forces.  However, gaseous phosphorus(V) chloride consists of PCl5 covalent molecules. Reaction of oxide with acids: o None, only acidic in nature. Reaction of oxide with bases/alkalis:

o o



 

Both oxides dissolve in alkalis to form a whole series of phosphate(III) and phosphate(V) salts. with excess strong bases like sodium hydroxide, the simplified equations are:  P4O6(s) + 12NaOH(aq) ==> 4Na3PO3(aq) + 6H2O(l) sodium phosphate(III) formed from phosphorus(III) oxide 3 ionic equation: P4O6(s) + 12OH (aq) ==> 4PO3 (aq) + 6H2O(l)  P4O10(s) + 12NaOH(aq) ==> 4Na3PO4(aq) + 6H2O(l) sodium phosphate(V) formed from phosphorus(V) oxide 3 ionic equation: P4O10(s) + 12OH (aq) ==> 4PO4 (aq) + 6H2O(l)  If the empirical formulae P2O3 and P2O5 are used, just halve all the balancing numbers.  Other than using excess sodium hydroxide, other salts can be formed.  e.g. P4O10(s) + 4NaOH(aq) + 2H2O(l) ==> 4NaH2PO4(aq) sodium dihydrogen phosphate(V)  or P4O10(s) + 8NaOH(aq) ==> 4Na2HPO4(aq) + 2H2O(l) disodium hydrogen phosphate(V) o Reaction of the chlorides with water: o Phosphorus(III) chloride hydrolyses rapidly and exothermically to form phosphoric(III) acid.  PCl3(l) + 3H2O(l) ==> H3PO3(aq) + 3HCl(aq) o Phosphorus(V) chloride initially hydrolyses to form phosphorus oxychloride and hydrochloric acid.  PCl5(s) + H2O(l) ==> POCl3(aq) + 2HCl(aq)  Then on boiling the aqueous solution, phosphoric(V) acid is formed and more hydrochloric acid.  POCl3(aq) + 3H2O(l) ==> H3PO4(aq) + 3HCl(aq)  overall: PCl5(s) + 4H2O(l) ==> H3PO4(aq) + 5HCl(aq) Reaction of element with water: o None. Other comments: o -

Z = 16 Sulphur S in Group 6/16 

  



The structure of the element: o Three solid allotropes. Two are crystalline lattices based on S8 molecules (rhombic and monoclinic sulfur). A 3rd form is an unstable dark brown-black polymeric form called plastic sulphur, formed when boiling sulphur is poured onto cold water, great fun, but of little use! Physical properties: o o o Colourless gas; mpt 117 C; bpt 445 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 4 o Gp6; e.c. 2,6 or 1s 2s 2p 3s 3p ; ranges from (-2 to +6) e.g. o Na2S (-2), S2Cl2 (+1), SO2 (+4) and H2SO4, SF6, SO3 (all +6). Reaction of element with oxygen: o Burns in air with a pale blue flame to form sulphur dioxide (sulphur(IV) oxide), with a little sulphur trioxide.  S(s) + O2(g) ==> SO2(g) o Sulphur trioxide (sulphur(VI) oxide) has to be made by the industrial Contact Process. o  2SO2(g) + O2(g) == V2O5 catalyst, 450 C ==> 2SO3(g) Reaction of the oxides with water: Both dissolve to form acid solutions. o Sulphur dioxide forms the weak 'fictitious' sulphurous acid.  SO2(g) + H2O(l) ==> H2SO3(aq)  the reaction is better represented ionically as .. +

 





-

 SO2(aq) + H2O(l) H (aq) + HSO3 (aq) o Sulphur trioxide reacts very violently and exothermically to form the oily liquid, strong sulphuric acid.  SO3(g) + H2O(l) ==> H2SO4(l)  In water, the sulphuric acid is almost fully ionised. + 2 H2SO4(aq) + 2H2O(l) ==> 2H3O (aq) + SO4 (aq) Reaction of oxide with acids: o None, only acidic in nature. Reaction of oxide with bases/alkalis: o Sulphur dioxide dissolves in strong bases to form sulphites/sulphate(IV)s o 2NaOH(aq) + SO2(g) ==> Na2SO3(aq) + H2O(l) formation of sodium sulphite/sulphate(IV) 2o ionic equation: 2OH (aq) + SO2(g) ==> SO3 (aq) + H2O(l) o You would NOT attempt to react sulphur trioxide with water, the reaction is very violent and exothermic. o but theoretically: 2NaOH(aq) + SO3(g) ==> Na2SO4(aq) + H2O(l) Reaction of element with chlorine: o When chlorine is passed over molten sulphur a variety of chlorides are formed. o The main product is disulphur dichloride o 2S(s) + Cl2(g) ==> S2Cl2(l) (SiCl2, SiCl4 also possible) Reaction of chloride with water:

o

 

Slowly hydrolyses in water, via a complex reaction, to form an acid solution of several products (not meant to be a balanced equation).  S2Cl2(g) + H2O(l) ==> HCl(aq), S(s), SO2(aq), H2SO3(aq), H2SO4(aq), H2S(aq) - complex redox/hydrolysis reaction! Reaction of element with water: o None. Other comments: o Z = 17 Chlorine Cl in Group 7/17 The Halogens

    

 



The structure of the element: o Non-metal existing as covalent diatomic molecule, Cl2, with a single bond. Physical properties: o o o Pale green gas; mpt -101 C; bpt -34 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 5 o Gp7 Halogens; e.c. 2,8,7 or 1s 2s 2p 3s 3p ; (ranges from -1 to +7) e.g. o HCl and NaCl (-1), NaClO and Cl2O (+1), NaClO2 (+3), KClO3 (+5), Cl2O7 and HClO4 (+7). Reaction of element with oxygen: o None, but there are several oxides made indirectly. Reaction of the oxides with water: o Chlorine(I) oxide forms weak chloric(I) acid.  Cl2O(g) + H2O(l) ==> 2HClO(aq) o Chlorine(V) oxide forms the strong chloric(VII) acid.  Cl2O7(l) + H2O(l) ==> 2HClO4(aq) Reaction of oxide with acids: o None, only acidic in nature. Reaction of oxide with bases/alkalis: o chlorine(I) oxide forms sodium chlorate(I) with sodium hydroxide,  Cl2O(g) + 2NaOH(aq) ==> 2NaClO(aq) + H2O(l)  ionic equation: Cl2O(g) + 2OH (aq) ==> 2ClO (aq) + H2O(l) o and chlorine(VII) oxide will dissolve to form sodium chlorate(VII)  Cl2O7(l) + 2NaOH(aq) ==> 2NaClO4(aq) + H2O(l)  ionic equation: Cl2O7(l) + 2OH (aq) ==> 2ClO4 (aq) + H2O(l) Reaction of element with water: o Slightly reacts to form an acid mixture of chloric(I) acid and hydrochloric acid, but the position of the equilibrium is very much on the left.  



Cl2(g) + H2O(l) HClO(aq) + HCl(aq) or more accurately and ionically ...

 Cl2(g) + 2H2O(l) Other comments: o -

HClO(aq) + H3O

+

-

(aq)

+ Cl (aq)

Z = 18 Argon Ar in Group 0/18 The Noble Gases     

The structure of the element: o Exists as single atoms, sometimes described as 'monatomic molecules'. Physical properties: o o o Colourless gas, more dense than air; mpt -189 C, bpt -186 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 6 o Gp0/8 Noble gas; e.c. 2,8,8 or 1s 2s 2p 3s 3p ; no stable oxidation states (other than 0 for the element itself!) so no compounds! Reaction with anything: o None! Far too stable electron configuration. Other comments: o Last element in the period, as the outer principal quantum level (shell) is full to the maximum number of electrons, conferring extra chemical stability on the atom.

5. Survey of Period 3: Na across to Ar (8 elements, Z = 11 to 18) 5.2 Period 3 trends and explanations of selected physical properties Element

Sodium Magnesium Aluminium

1st ionization energy -1 (kJ mol ) Atomic metallic or covalent radius (pm, /1000 for nm)

496

738

186 160 (metal) (metal)

Silicon

Phosphorus

Sulfur

Chlorine

Argon

577

786

1080

1000

1251

1520

143 (metal)

117 (covalent)

110 (covalent)

104 99 94 (covalent (covalent) (covalent) -theoretical)

Electronegativity (Pauling scale)

0.93

1.31

1.61

1.90

2.19

2.58

3.16

3.20

Melting Point (K)

371

922

934

1683

317

390

172

84

Boiling Point (K)

1156

1363

2740

2628

553

718

239

87

Relative electrical conductivity

0.218

0.224

0.382

0.001

3 mobile delocalised electrons involved in electrical conduction. From Group 4 to 0 the element structure changes to giant covalent lattice or simple molecular structures with no free delocalised electrons within the structure to convey an electric current. Although silicon can be described as a semi-metal (wrongly in my view) it is virtually an insulator unless doped with other elements.

(6) Variation of density across Period 3 The peaks correspond to the metals in the middle of the period with the strongest bonding in the solid. The density increases from sodium to aluminium as the atomic radii decrease and the bonding gets stronger with 1 ==> 3 bonding electrons (delocalised outer valency electrons in the metal lattice). However, they are relatively low densities compared to most metals. Silicon, phosphorus and sulphur have a low densities, typical of non-metallic covalent solids. Chlorine and argon are small covalent molecules and have very low densities being gaseous at room temperature because only weak intermolecular forces act between them.

5. Survey of Period 3: Na across to Ar (8 elements, Z = 11 to 18) 5.3 Period 3 trends in bonding, structure, oxidation state, formulae & reactions +

δ+

-

δ+

M X ionic bond, M -X

polar bond and M-X a relatively non-polar bond (no partial charges shown)

Element

Sodium

Magnesium

Aluminium

Silicon

Phosphorus

Sulfur

Chlorine

Argon

old/latest Group

1

2

3/13

4/14

5/15

6/16

7/17

0/18

ZSymbol

Structure of element

electron configuration

11Na

12Mg

1

[Ne]3s

electronegativity of element

0.93

formula of oxides

Na2O, Na2O2

bonding and structure of oxides

ionic lattice

2.51 electronegativity difference X-O (O is 3.44) nature of Na+ O22bond or O2 NaCl

2

[Ne]3s

max +4

max +5

1.31

1.61

1.90

2.19

MgO

Al2O3

SiO2

ionic lattice

ionic lattice

solid covalent giant structure

2.13

1.83

1.54

Mg

2+

O

2-



Mg

2+

Al

3+

O

2-

AlCl3

1.85

electronegativity difference X-H (H is 2.20) nature of bond

-

Cl

δ+

Si -O

+

-

δ+

Mg -H

Al

3+

1.26 -

Cl

0.59 δ-

4

2

5

δ+

P -O

δ-

δ+

1.25 δ-

δ+

Al -H

δ-

δ+

2

[Ne]3s 3p

2.58

3.16

3.20

SO2, SO3

Cl2O, ClO2, Cl2O6, Cl2O7

-

δ+

S -O

δ-

S2Cl2, ??

δ+

-

0.28

0.58 δ-

gaseous single atoms

6

at Xe can -2, -2, +4, -1, +1, +3, get max of +6 +5, +7 +8 compounds but not max +6 max +7 here!

0.86

PCl3, PCl5

Cl-O Cl2

-

-

0.00 δ-

-

Si -Cl

P -Cl

S -Cl

SiH4

PH3

H2S

HCl

-

small covalent gaseous molecule

small covalent gaseous molecule

small covalent gaseous molecule

-

0.30

0.01

0.38

0.96

Si-H

P-H

H -S

'polymer-like' 'polymer-like' small structure of structure of ionic covalent intermediate intermediate lattice gaseous ionic/covalent ionic/covalent molecule nature nature

Na H

2

[Ne]3s 3p [Ne]3s 3p

solid covalent covalent covalent small small small gaseous gas/liquid molecules molecules molecules 1.25

δ-

18Ar

ionic lattice, covalent vaporises to covalent liquid covalent small covalent small covalent small diatomic dimer liquid small liquid gaseous molecules molecules molecules molecules molecule Al2Cl6

AlH3

0.89

3

P4O6, P4O10

SiCl4

1.55

MgH2

1.27

2

[Ne]3s 3p

max +3

2.23

NaH

2

max +2

MgCl2

-

2

[Ne]3s 3p

+3, +5

electronegativity difference X-Cl (Cl is 3.16) nature of bond

+

1

17Cl

solid gaseous solid small small small covalent covalent covalent molecules molecules molecules P4 S8 Cl2

+4

ionic lattice

Na H

2

[Ne]3s 3p

16S

+3

ionic lattice

bonding and structure of hydride

15P

+2

bonding in chlorides

Formula of hydride

14Si

solid metallic solid solid metallic solid metallic lattice giant 2+ 3+ lattice of Al + lattice of Mg of Na covalent and free e and free e and lattice Sin free e

common +1 oxidation states e.g. in oxides, chlorides, hydrides max +1

formula of chlorides

13Al

δ+

δ-

Cl-Cl

δ+

δ-

H -Cl

The structure and physical properties of the elements o The trend is metal lattice ==> giant covalent structure ==> small covalent molecules

-

o





Sodium Na, magnesium Mg and aluminium Al are silvery solids, with a metal lattice structure, high boiling points and are good conductors of heat/electricity due to the delocalised free electrons moving between the immobile metal ions.  The melting/boiling points increase from Na ==> Mg ==> Al due to 1 ==> 2 ==> 3 potential number of delocalised electrons that may contribute to bonding. o Si has a non-metallic giant covalent structure based on a tetrahedral arrangement of S-Si bonds and is a poor conductor of heat/electricity.  The strong 3D bonding gives silicon a high melting/boiling point and great hardness. o Phosphorus P4, sulfur S8 and chlorine Cl2 are simple-small covalent molecules and Ar consists of single atoms. The molecules are only held together by the weakest of the intermolecular forces, namely the instantaneous dipole - induced dipole forces, and consequently have very low melting/boiling points.  From left to right the elements become less metallic and more non-metallic. o Electron configuration and oxidation states 2 2 6 1 2 2 6 2 6 o Electron configurations of 2,8,1 or 1s 2s 2p 3s to 2,8,8 or 1s 2s 2p 3s 3p  Filling the s orbital (max 2 e-'s) gives the metallic s-block elements of Groups 1-2,  filling the p orbitals gives the predominantly non-metallic p block elements of Group 3-7, 0 (Gps 1318) bar aluminium for Period 3. o Oxidation states in compounds (numerically = valency) are: sodium Na (+1 only), magnesium Mg (+2 only), aluminium Al (+3 only), Si (+4, -4 with electropositive metals), P (usually -3, +3 or +5), S (-2, +4 and +4), Cl (-1, +1, +3, +5 and +7), Ar has no stable compounds due to the full outer quantum level (shell) being full, conferring extra electronic stability on the atom.  From Na to Cl the maximum oxidation state is equal to the 'old' group number and the 'highest' oxide formulae can be predicted up to chlorine and the chloride formula up to P (there is no stable SCl6 but there is a ClF7.  So in the 'highest' oxides you can go from +1 to +7 for groups 1 to 7/17  (at Xe on Period 5 you can reach +8, but not for Ar)  Na2O, MgO, Al2O3, SiO2, P4O10 (= P2O5), SO3, Cl2O7 (at Xe you can have XeO4)  i.e. using all available 1-7 outer 3s and 3p electrons (valence electrons) are all used in the bonding of the highest possible oxide.  and similarly in the 'highest' chlorides (upto Group 5), and fluorides (for Groups 6 and 7) you also go from maximum oxidation state of +1 to +7 in the halide compounds irrespective of bond character.  NaCl, MgCl2, AlCl3, SiCl4, PCl5, SF6, ClF7 (at Xe you can have XeF8) o Reaction of element with oxygen and the structure of the oxide

(Gp 1) 4Na(s) + O2(g) ==> 2Na2O(s) and Na2O2 on heating the metal in air

(Gp 2) 2Mg(s) + O2(g) ==> 2MgO(s) on heating metal in air

(Gp 3) 4Al(s) + 3O2(g) ==> 2Al2O3(s) needs high temperature

(Gp 4) Si(s) + O2(g) ==> SiO2(g) needs high temperature

(Gp 5) P4(s) + 5O2(g) ==> P4O10(s) on heating in air

(Gp 6) S(s) + O2(g) ==> SO2(g) and a little SO3 on heating in air

(Gp 7) Chlorine - no reaction

(Gp 0) Argon - no reaction



Reaction with oxygen and oxide structure o The metals Na, Mg and Al burn to form a giant ionic oxide lattices  sodium oxide/peroxide, magnesium oxide and aluminium oxide ... + 2+ 22+ 23+ 2 (Na )2O and (Na )2O2 , Mg O and (Al )2(O )3 respectively. o Silicon Si forms a giant covalent lattice of (SiO2)n where n is very larger number o Phosphorus P forms two simple molecular covalent solid oxides P4O6 and P4O10. o Sulphur/sulfur S can form two simple molecular covalent gas molecules SO 2 and SO3 o Chlorine Cl forms oxide molecules of Cl2O, Cl2O7 (and others). o Argon has no reaction. o The overall pattern, from left to right is  giant ionic lattice => giant covalent lattice ==> small covalent molecules. o The change in bonding character from ionic to covalent in the oxide, follows the decreasing difference in electronegativity between that of the period 3 element and oxygen. o -



Reaction of the oxides with water, acids and alkalis

(Gp 1) Na2O(s) + H2O(l) ==> 2NaOH(aq) pH 13-14 strong base

(Gp 2) MgO(s) + H2O(l) ==> Mg(OH)2(aq/s) ~pH 11-12 weak base

or Na2O2(s) + 2H2O(l) ==> 2NaOH(aq) + H2O2(aq) (Gp 3) Al2O3, insoluble, no reaction with water, but

(Gp 4) SiO2, insoluble, no reaction with water, but weakly

amphoteric with respect to acids and strong alkalis (Gp 5) P4O6(s) + 6H2O(l) ==> 4H3PO3(aq) ~pH 2 weak acid P4O10(s) + 6H2O(l) ==> 4H3PO4(aq) pH 1 strong acid

acidic and will dissolve a little in strong alkali e.g. conc. NaOH(aq) (Gp 6) SO2(aq) + H2O(l) H acid

+

-

(aq)

+ HSO3 (aq) pH 2-3 weak

SO3(g) + H2O(l) ==> H2SO4(aq) pH 0-1 strong acid

(Gp 7) Cl2O(g) + H2O(l) ==> 2HClO(aq) ~pH 3? weak acid (Gp 0) argon has no oxide Cl2O7(l) + H2O(l) ==> 2HClO4(aq) pH 1 strong acid 



The chemical character of the oxides - reaction of the Period 3 oxides with water, acids or alkalis. o Sodium oxide/peroxide Na2O/Na2O2 and magnesium oxide MgO are basic and form an alkali in water and salts with acids.  MgO(s) + 2HCl(aq) ==> MgCl2(aq) + H2O(l) o Aluminium oxide Al2O3 has no reaction, insoluble, but is amphoteric and forms salts with acids and alkalis.  Al2O3(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2O(l)  Al2O3(s) + 2NaOH(aq) + 3H2O(l) ==> 2Na[Al(OH)4](aq) o Silicon(IV) oxide (silicon dioxide) SiO2 has no reaction but is weakly acidic forming salts with alkalis. o Phosphorus(III) oxide P4O6 and phosphorus(V) oxide P4O10 are moderately-strong acidic oxides forming phosphoric(III) acid H3PO3 and phosphoric(V) acid H3PO4 on reaction with water.  Generally speaking, in a series of oxides for the same element, the higher the oxidation state of X in a 'XxOy' series, the more acidic is the oxide, so H 3PO4 is a stronger acid than H3PO3.  The oxides or acids are readily neutralised to give phosphate salts e.g.  H3PO4 (aq) + NaOH(aq) ==> NaH2PO4(aq) + H2O(l)  Two further reactions are possible with the sodium hydroxide to give Na2HPO4 and Na3PO4. o Chlorine(I) oxide Cl2O and chlorine(VII) oxide Cl2O7 are moderate to strong acidic in water. o The overall patterns, from left to right across Period 3 is ... o giant ionic lattice ==> small covalent molecules  The change in bonding character from ionic to covalent in the oxide follows the decreasing difference in electronegativity between that of the element and oxygen. o In terms of overall chemical character ...  metal basic oxides ==> amphoteric oxides ==> non-metal oxides o This is chemically characteristic of metallic ==> non-metallic element character. o Reaction of element with chlorine and the structure of the chloride

(Gp 1) 2Na(s) + Cl2(g) ==> 2NaCl(s)

(Gp 2) Mg(s) + Cl2(g) ==> MgCl2(s)

(Gp 3) 2Al(s) + 3Cl2(g) ==> 2AlCl3(s)

(Gp 4) Si(s) + 2Cl2(g) ==> SiCl4(l)

(Gp 5) P4(s) + 3Cl2(g) ==> 4PCl3(l) (Gp 6) 2S(s) + Cl2(g) ==> S2Cl2(l) also unstable SiCl2, SiCl4 P4(s) + 5Cl2(g) ==> 4PCl5(s) (Gp 7) chlorine itself 

(Gp 0) no reaction with argon

Reaction with chlorine and chloride structure o All of Na to S will combine directly on heating in chlorine to give the chloride. o Sodium, magnesium and aluminium give giant ionic lattices + 2+ 3+  sodium chloride Na Cl , magnesium chloride Mg (Cl )2 and aluminium chloride Al (Cl )3 respectively. o  Note that aluminium chloride on heating sublimes above 180 C to form small Al2Cl6 covalent dimer molecules. o The non-metal elements give covalent chlorides. o Silicon forms the molecular covalent liquid silicon(IV) chloride SiCl4 (silicon tetrachloride) o Phosphorus forms phosphorus(III) chloride PCl3 (phosphorus trichloride) with limited chlorine  and phosphorus(V) chloride PCl5 (phosphorus pentachloride) with excess chlorine. o Sulphur gives disulfur dichloride S2Cl2. by direct combination (and unstable SCl2 and SCl4 can also be formed). o There is no stable argon chloride. o The overall pattern, from left to right across period 3 is ...  giant ionic lattice => polymeric covalent lattice ==> small covalent molecules.  This is chemically characteristic of metallic ==> non-metallic element character.  The change in bonding character from ionic to covalent in the chloride, follows the decreasing difference in electronegativity between that of the element and oxygen, as in the case of oxides.  The formulae largely follow a pattern of rising formulae based on the use of all outer electrons in bonding (1-5) and then a decline in valency (oxidation state of the Period 3 element).

 



NaCl (+1), MgCl2 (+2), AlCl3 (+3), SiCl4 (+4), PCl5 (+5), S2Cl2 (+1), Cl2, Ar no chloride So the number of atoms of chlorine combined with the Period 3 element (the valency) follows the pattern  1 2 3 4 5 1 1 0

o Reaction of the chlorides with water

(Gp 1) NaCl(s) + aq ==> Na 7

+

-

(aq)

+ Cl (aq) just dissolves, ~pH

(Gp 3) AlCl3(s) + 3H2O(l) ==> Al(OH)3(s) + 3HCl(g) with limited water you get hydrolysis to give acid fumes 3+

-

AlCl3(s) + aq ==> Al (aq) + 3Cl (aq) excess water, weakly + acidic solution due to the acidity of [Al(H2O)6]3

(Gp 2) MgCl2(s) + aq ==> Mg ~pH 7

2+

-

(aq)

+ 2Cl (aq) just dissolves,

(Gp 4) SiCl4(l) + 2H2O(l) ==> SiO2(s) + 4HCl(aq) hydrolysis to give strongly acid solution

(Gp 5) PCl3(l) + 3H2O(l) ==> H3PO3(aq) + 3HCl(aq) hydrolysis to give weakly acid solution PCl5(s) + 4H2O(l) ==> H3PO4(aq) + 5HCl(aq) strongly acid solution

(Gp 6) S2Cl2(g) + H2O(l) ==> HCl(aq), S(s), SO2(aq), H2SO3(aq), H2SO4(aq), H2S(aq) - complex redox/hydrolysis hydrolysis to give reaction but final solution is quite acidic

(Gp 7) chlorine itself 



Gp 0 argon has no chloride

Reaction of the chloride with water o The ionic sodium chloride NaCl and magnesium chloride MgCl2 dissolve in water to form a nearly neutral solution of hydrated ions. o The ionic AlCl3 and the covalent all hydrolyse to form acid solutions.  aluminium chloride Al2Cl6 ==> hydrochloric acid or weakly acidic aluminium ion  silicon(IV) chloride SiCl4, (silicon tetrachloride) ==> hydrated silicon dioxide + hydrochloric acid  phosphorus(III) chloride PCl3 (phosphorus trichloride) ==> phosphoric(III) acid + hydrochloric acid  with phosphorus(V) PCl5 (phosphorus pentachloride) ==> phosphoric(V) acid + hydrochloric acid  Phosphorus(III) chloride hydrolyses rapidly and exothermically to form phosphoric(III) acid.  PCl3(l) + 3H2O(l) ==> H3PO3(aq) + 3HCl(aq)  Phosphorus(V) chloride initially hydrolyses to form phosphorus oxychloride and hydrochloric acid.  (i) PCl5(s) + H2O(l) ==> POCl3(aq) + 2HCl(aq)  If the aqueous solution is boiled, phosphoric(V) acid is formed and more hydrochloric acid.  (ii) POCl3(aq) + 3H2O(l) ==> H3PO4(aq) + 3HCl(aq)  overall (i) + (ii): PCl5(s) + 4H2O(l) ==> H3PO4(aq) + 5HCl(aq)  and disulphur dichloride S2Cl2 ==> a variety products including acidic sulphur dioxide and hydrochloric acid. o The general trend is for ionic metal chloride salts to give nearly neutral solutions => metal/non-metal covalent chlorides that hydrolyse to give acidic solutions. o Reaction of element with water

(Gp 1) 2Na(s) + 2H2O(l) ==> 2NaOH(aq) + H2(g)

(Gp 2) Mg(s) + 2H2O(l) ==> Mg(OH)2(aq) + H2(g)

(Gp 3) aluminium has no reaction with water

(Gp 4) silicon has no reaction with water

(Gp 5) phosphorus has no reaction with water

(Gp 6) sulfur has no reaction with water

(Gp 7) Cl2(g) + H2O(l) HClO(aq) + HCl(aq)

(Gp 0) argon has no reaction with water



Reaction of element with water o The reactive metal sodium Na rapidly gives the alkaline sodium hydroxide and hydrogen, o as does magnesium Mg BUT much more slowly. o Aluminium Al, silicon Si, phosphorus P and sulfur S have no reaction with water. o Chlorine, Cl2 forms a weakly acidic solution in water. o Argon has no reaction. o The 'limited' pattern for period 3 (or any other period), is to have reactive metals on the left forming an alkaline solution and a reactive non-metal on the right forming an acid solutions IF they react with water. o -



The hydrides MHx o For hydrides the difference in electronegativity works both ways!

o

   

From left to right across the period you change from an +  ionic sodium hydride crystal lattice Na H  to small non-polar molecule covalent hydrides (silane SiH4 and phosphine PH3)  and then a weakly acidic polar covalent hydride molecule (hydrogen sulfide H 2S)  and finally a strongly acidic polar covalent molecule (hydrogen chloride HCl). o The formulae follow a simple period pattern of rising and falling valency for the Period 3 elements.  NaH MgH2 AlH3 SiH4 PH3 H2S HCl (Ar)  the element valency pattern being 1 2 3 4 3 2 1 0 o On reaction with water, the ionic metal hydrides at the start of the period give an alkaline solution  e.g. NaH(s) + H2O(l) ==> NaOH(aq) + H2(g) o In the middle are neutral hydrides like phosphine which in contact with water do not change the pH. o Then you get weakly acidic ==> strongly acidic hydrides when they dissolve in water e.g. +  weak acid: H2S(aq) + H2O(l) H3O (aq) + HS (aq) +  strong acid: HCl(aq) + H2O(l) ==> H3O (aq) + Cl (aq) o So things are a bit complicated with hydrides on period 3 due to the left and right sided differences in electronegativity! + δ+ δ δ δ+  You go from X H ==> X -H ==> X-H ==> X -H o Radii of isoelectronic ions Isoelectronic means species having the same total number of electrons. The table below considers the isoelectronic cations and anions associated with Periods 2, 3 and 4.

isoelectronic system

Group 4/14

Group 5/15

Period [Ne] 10e 2 2 6 1s 2s 2p

C

total nuclear charge

+6

+7

+8

radius in picometre (pm)

260

171

140

name of ion

carbide

nitride



Group 1

4-

N

3-

O

2-

Group 2

Group 3/13

Period 3 F

-

+

2+

Na

Mg

+9

(+10)

+11

+12

+13

136

(38112*)

95

65

50

oxide fluoride (neon)

sodium

Period 3

Al

3+

(Ne)

magnesium aluminium Period 4

[Ar] 18e 2 2 6 2 6 1s 2s 2p 3s 3p

Si

P

S

Cl

(Ar)

K

nuclear charge

+14

+15

+16

+17

(+18)

+19

+20

+21

radius in picometre (pm)

271

212

184

181

(71154*)

133

99

81

calcium

scandium

name of ion



Group (Group 7/17 0/18)

Period 2

Period



Group 6/16

4-

3-

2-

-

+

silicide phosphide sulfide chloride (argon) potassium

Ca

2+

3+

Sc

Excluding the noble gases themselves where this is frankly, something of a data problem!, o there is a clear pattern of decreasing ionic radius with increase in total nuclear charge (+ atomic/proton number) for the two isoelectronic series tabulated above,  based on the electron configurations of neon and argon. From left to right the proton/electron ratio is steadily increasing so that the electrons are experiencing an increasingly greater attractive force of the nucleus, hence the steady decrease in radii for an isoelectronic series. * all sorts of values are quoted for noble gas radii e.g. atomic, covalent and ionic, but most don't fit in the pattern above which is quite clear for all the cations and anions listed.

6.1 Survey of Period 4 elements K to Kr (18 elements Z = 19 to 36)

Z = 19 Potassium in Group 1 Alkali Metals 

Potassium's chemistry is almost identical to sodium, a typical Alkali Metal

Z = 20 Calcium in Group 2 Alkaline Earth Metals 

Calcium's chemistry is almost identical to magnesium, a typical Alkaline Earth Metal

Z = 21 Scandium in 3d-block 

The first 3d block element, a typical hard high melting metal whose chemistry is dominated by the formation of the 3+ scandium(III) ion, Sc .

Z = 22 Titanium in 3d-block - Transition Element  

The first of the true transition metals with principal oxidation states of +2, +3 and +4. White titanium(IV) oxide is an important pigment in the paints industry.

Z = 23 Vanadium in 3d-block - Transition Element 

Vanadium is a very typical transition metal, high density and high melting point, various oxidation states (+2 to +5) and its compounds have catalytic properties.

Z = 24 Chromium in 3d-block - Transition Element 

Chromium is a very typical transition metal, high density and high melting point, various oxidation states (+2, +3 and +6) and its many complex ions and compounds have a variety of colours.

Z = 25 Manganese in 3d-block - Transition Element 

Manganese is encountered as its oxide, manganese(IV) oxide which catalyses the decomposition of hydrogen peroxide and the purple ion manganate(VII). Its principal oxidation states are +2 and +7, but+4, +5 and +6 states are also known.

Z = 26 Iron in 3d-block - Transition Element 

Iron is a very typical transition metal, high density and high melting point, several important oxidation states (+2 and +3) and its use in numerous steel alloys.

Z = 27 Cobalt in 3d-block - Transition Element 

Cobalt is a very typical transition metal, high density and high melting point, several important oxidation states (+2 and +3) and its use in alloys and forms a variety of complex ions.

Z = 28 Nickel in 3d-block - Transition Element 

Nickel is also very typical transition metal, high density and high melting point, the most important oxidation state is +2 but +3 complexes are also known and is used in numerous alloys.

Z = 29 Copper in 3d-block - Transition Element 

Copper is one of the most well known transition metals with two principal oxidation states (+1 and +2) and is characterised by the familiar and characteristic blue and green colour of copper(II) complex ions or compounds.

Z = 30 Zinc in 3d-block element



Zinc is not a true transition metal but is the last member of the 3d block series. Its chemistry is dominated by the formation of zinc(II) compounds, its only oxidation state is +2 in compounds.

Z = 31 Gallium in Group 3/13 

Gallium is a soft bluish-white metal whose chemistry is a bit like that of aluminium

Z = 32 Germanium in Group 4/14 

An important semi-metallic element in the semi-conductor industry for manufacturing computer chips and mobile phone technology.

Z = 33 Arsenic in Group 5/15 

A semi-metallic element whose compounds are somewhat poisonous!

Z = 34 Selenium in Group 6/16 

Selenium is another semi-metal.

Z = 35 Bromine in Group 7/17 The Halogens 

Bromine's chemistry is very similar to that of chlorine another of the Halogen elements.

Z = 36 Krypton in Group 0/18 The Noble Gases 

Krypton is very similar to Argon, another of the Noble Gases and has virtually no meaningful chemistry. o Krypton(II) fluoride and krypton(IV) fluoride have been prepared.

6.2 Period 4 Trends in physical properties The data for Period 4 is NOT shown from left to right, due to the number of columns required!

Z symbol

Name

o

o

mpt/ C mpt/K bpt/ C bpt/K

Pauling electronegativity

1st ionisation Atomic energy radius kJ/mol pm

Relative electrical conductivity

Density 3 g/cm

19

K

potassium

64

337

774

1047

0.82

419

231

0.143

0.8

20

Ca

calcium

839

1112

1484

1757

1.00

590

197

0.218

1.6

21

Sc

scandium

1541

1814

2836

3109

1.36

631

161

0.015

3.0

22

Ti

titanium

1668

1941

3287

3650

1.54

658

145

0.024

4.5

23

V

vanadium

1910

2183

3380

3653

1.63

650

132

0.040

6.1

24

Cr

chromium

1857

2130

2672

2945

1.66

653

125

0.078

7.2

25

Mn

manganese 1246

1519

1962

2235

1.55

717

124

0.054

7.4

26

Fe

iron

1538

1811

2861

3134

1.83

759

124

0.100

7.9

27

Co

cobalt

1495

1768

2870

3143

1.88

760

125

0.160

8.9

28

Ni

nickel

1455

1728

2730

3003

1.91

737

125

0.145

8.9

29

Cu

copper

1083

1356

2567

2840

1.90

745

128

0.593

9.0

30

Zn

zinc

420

693

907

1180

1.65

906

133

0.167

7.1

31

Ga

gallium

30

303

2400

2673

1.81

579

122

0.058

5.9

32

Ge

germanium

937

1210

2830

3103

2.01

762

122

amphoteric ==> acidic  BUT within the 3d block the patterns are quite complicated and the oxidation state of the metal very much determines the structural and chemical character of the compound. Oxidation states o From potassium (+1) to manganese (+7) the maximum oxidation state is determined by

the maximum number of outer valency electrons. After Mn, there is a tendency to fall to a stable +2 state e.g. cobalt, nickel, copper and zinc. Beyond zinc, the last element in the 3d block, the maximum oxidation state is governed by the maximum number of s + p electrons beyond the full 3d sub-shell i.e. from gallium to bromine the maximum oxidation state rises from +3 to +7. Oxides - formulae, bonding and chemical character o Ionic lattice ==> covalent character of the oxides and chlorides o and in chemical character the oxide changes from basic ==> amphoteric ==> acidic. o The oxide of maximum oxidation state for potassium to manganese is determined by the maximum number of outer valency electrons (from 1 to 5). o Again the patterns within the 3d block are complicated. Chlorides - formulae, bonding and chemical character o the ionic ==> covalent character of the oxides and chlorides as the electronegativity difference in M-X decreases (M to the left of X) Radii of isoelectronic ions Isoelectronic means species having the same total number of electrons. The table below considers the isoelectronic ions associated with Periods 2, 3 and 4. o o



    

isoelectronic system

Group 4/14

Group 5/15

Period [Ne] 10e 2 2 6 1s 2s 2p

C

total nuclear charge

+6

+7

+8

radius in picometre (pm)

260

171

140

name of ion

carbide

nitride



Group 1

4-

N

3-

O

2-

Group 2

Group 3/13

Period 3 F

-

+

2+

Na

Mg

+9

(+10)

+11

+12

+13

136

(38112*)

95

65

50

oxide fluoride (neon)

sodium

Period 3

Al

3+

(Ne)

magnesium aluminium Period 4

[Ar] 18e 2 2 6 2 6 1s 2s 2p 3s 3p

Si

P

S

Cl

(Ar)

K

nuclear charge

+14

+15

+16

+17

(+18)

+19

+20

+21

radius in picometre (pm)

271

212

184

181

(71154*)

133

99

81

calcium

scandium

name of ion



Group (Group 7/17 0/18)

Period 2

Period



Group 6/16

4-

3-

2-

-

+

silicide phosphide sulfide chloride (argon) potassium

Ca

2+

3+

Sc

Excluding the noble gases themselves, there is a clear pattern of decreasing ionic radius with increase in nuclear charge (+ atomic/proton number) for the two isoelectronic series tabulated above. From left to right the proton/electron ratio is steadily increasing so that the electrons are experiencing an increasingly greater attractive force of the nucleus, hence the steady decrease in radii for an isoelectronic series. * all sorts of values are quoted for noble gas radii e.g. atomic, covalent and ionic, but most don't fit in the pattern above which is quite clear for all the cations and anions listed.

6.4 A summary of important physical and chemical trends down a Group

GROUP DATA TABLES - data aligned down the group z = at. no., cond'y = conductivity, config. = configuration, oxidation states are numerically equal to valencies M = metal, SM = semi-metal (metalloid), NM = non-metallic element

Group 1 Alkali Metals

Electron configuratio n

ZElement

3Li lithium

1

M

[He]2s

11Na M sodium

[Ne]3s

19K potassiu M m

[Ar]4s

37Rb

rubidium

55Cs

caesium

87Fr

francium

     

Boiling Point

o

o

C

K

C

K

Atomi Pauling c Relative chlorid 1st ionisati Densit ox. oxide electro- metalli electric e on energy y stat formula negativit c al formula 3 kJ/mol g/cm e e y radius cond'y e pm

18 45 134 162 1 4 7 0

0.98

152

513

0.5

0.108

+1

Li2O

LiCl

1

98

37 115 883 1 6

0.93

186

496

1.0

0.218

+1

Na2O, Na2O2

NaCl

1

64

33 104 774 7 7

0.82

231

419

0.8

0.143

+1

K2O, K2O2, KO2

KCl

1

39

31 688 961 2

0.82

244

403

1.5

0.080

+1

Rb2O, Rb2O2, RbO2

RbCl

1

29

30 679 952 2

0.79

265

376

1.9

0.053

+1

Cs2O, Cs2O2, CsO2

CsCl

+1

assume d Fr2O, assume Fr2O2, d FrCl FrO2

M

[Kr]5s

M

[Xe]6s

M

Meltin g Point

[Rn]7s

1

30 27 677 950 0

0.70

270

400

2.0

0.014

One of the most consistent sets of data for any group of the periodic table, helped by the fact they are all metals. The melting/boiling points, electronegativity, 1st ionisation energy and electrical conductivity decrease down the group. The atomic radius and density increase down the group. The formulae are perfectly consistent with what you might expect from the principles of the Periodic Table e.g. the chlorides, the metal having the only possible oxidation state of +1 (valency 1). Down the group the peroxide (M2O2) and the 'superoxide' (MO2) become more stable with respect to the expected oxide formula (M2O). 7 Detailed advanced level chemistry notes s-block Group 1 Alkali Metals

Group 2 Alkaline Earth Metals

Electron configurati on

ZElement

4Be M beryllium

2

[He]2s

Melting Point

Boiling Point

o

o

C

K

1.57

111

900

1.8

0.250

+2

BeO

BeCl2

109 136 0 3

1.31

160

738

1.7

0.224

+2

MgO, MgO2

MgCl2

2

649 922

2

839

111 148 175 2 4 7

1.00

197

590

1.6

0.218

+2

CaO, CaO2

CaCl2

2

769

104 138 165 2 4 7

0.95

215

550

2.5

0.043

+2

SrO, SrO2

SrCl2

2

729

100 163 191 2 7 0

0.89

215

503

3.6

0.016

+2

BaO, BaO2

BaCl2

+2

assum ed RaO, RaO2

assum ed RaCl2

[Ne]3s

20Ca M calcium

[Ar]4s

38Sr M strontium

[Kr]5s

56Ba

barium

88Ra

radium

    

M

M

K

127 155 297 324 8 1 0 3

12Mg

magnesiu M m

C

Atomi Pauling c Relative chlorid 1st ionisati Densit ox. oxide electro- metalli electrical e on energy y stat formula negativi c conductivi formula 3 kJ/mol g/cm e e ty radius ty e pm

[Xe]6s

[Rn]7s

2

700 973

114 141 0 3

0.89

223

509

5.0

0.010

A reasonably consistent sets of data for a group of the periodic table, helped by the fact they are all metals.. The melting/boiling points, electronegativity, 1st ionisation energy and electrical conductivity decrease down the group. The atomic radius and density increase down the group. There is total consistency in the formulae for the expected oxidation state of +2 (valency 2) for the chloride and 'normal oxide'. From magnesium onwards they all form a peroxide (MO 2) as well as the 'normal' expected oxide (MO). Group 3/13 Melting Point

ZElemen

Electron configuration

t

o

5B boron

N M

[He]2s 2p

13Al alumini M um

[Ne]3s 3p

C

K

Boiling Atomi Point Pauling c Relative chlorid 1st ionisat Densi ox. oxide electro- metall electrical e ion energy ty state formul negativi ic conductiv formul 3 kJ/mol g/cm s ae o ty radius ity ae C K pm

230 257 365 393 0 3 9 2

2.04

90 (cov)

801

2.3

M2SO4(aq) + H2O(l) to give the soluble sulphate salt o M2O(s) + 2CH3COOH(aq) ==> 2CH3COOM(aq) + H2O(l) to give the soluble ethanoate salt The reaction of Group 2 metals with oxygen (a redox reaction) Group 2 metals: 2M(s) + O2(g) ==> 2MO(s) (M = Be, Mg, Ca, Sr, Ba) o shows the formation of the oxide expected from their position in the periodic table when the element is heated or burned in air. Oxidation state changes: M from 0 to +2, and oxygen from 0 to –2. o The oxide, apart from beryllium, is slightly soluble in water forming the alkaline hydroxide, which increases in strength of basic character down the group.  MO(s) + H2O(l) ==> M(OH)2(s/aq) (not a redox change, M = Be, Mg, Ca, Sr, Ba) 2+ 2–  ionically: M O (s) + H2O(l) ==> M(OH)2(s/aq) 2+ 2– 2+ –  if the hydroxide is soluble: M O (s) + H2O(l) ==> M (aq) + 2OH (aq)  Bronsted–Lowry acid–base reaction, the oxide base accepts proton from the water.  The mixture of magnesium hydroxide and water is sometimes called milk of magnesia.  The formation of calcium hydroxide (slaked lime) when water is added to calcium oxide (quicklime) is very exothermic!  The pH of the resulting solution ranges from ~pH 10 to ~pH 13 for Mg(OH)2 to Ba(OH)2  All the oxides are basic and readily neutralised by acids (not a redox change).  MO(s) + 2HCl(aq) ==> MCl2(aq) + H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to give the soluble chloride salt 2+ 2– + 2+  ionically: M O (s) + 2H (aq) ==> M (aq) + H2O(l)  This applies to all four acid reactions examples in this section, acid proton donation to the oxide ion base. – – 2–  In each case the chloride Cl , nitrate NO3 and sulphate SO4 are spectator ions.  MO(s) + 2HNO3(aq) ==> M(NO3)2(aq) + H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to give the soluble nitrate salt  MO(s) + H2SO4(aq) ==> MSO4(aq/s) + H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to form the sulphate salt (soluble => insoluble)  but reaction increasingly slower for calcium oxide ==> barium oxide as the sulphate becomes less insoluble.

 

MO(s) + 2CH3COOH(aq) ==> (CH3COO)2M(aq) + H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to give the ethanoate salt Beryllium oxide BeO is amphoteric (another Be Gp 2 anomaly) and dissolves in strong bases like sodium hydroxide.  The equation below shows the formation of a hydroxo beryllate complex ion (not a redox change).  BeO(s) + 2NaOH(aq) + H2O(l) ==> Na2[Be(OH)4](aq) (beryllate salt) 2+ 2– – 2–  ionically: Be O (s) + 2OH (aq) + H2O(l) ==> [Be(OH)4] (aq)



7.6. Reaction of s–block metals and water & their hydroxide (OH ) chemistry The oxides and hydroxides are usually white ionic solids.  

The reaction of group 1 metals with water (a redox reaction) Group 1 metal hydroxide formation + – o 2M(s) + 2H2O(l) ==> 2M OH (aq) + H2(g) (M = Li, Na, K, Rb, Cs) o shows the formation of the alkaline metal hydroxide and hydrogen.  Oxidation state changes: M from 0 to +1, one H per water remains unchanged in oxidation number and one changes from +1 to 0 in H2. o M = Li (slow at first), Na (fast), K (faster – may ignite hydrogen to give a lilac coloured flame* from hot potassium atoms), Rb, Cs, Fr (very explosive) i.e. the reactivity increases down the group.  The reactivity trend is explained in section 7.4  * 2H2(g) + O2(g) ==> 2H2O(l) i.e. the chemistry of the lit splint pop! o The hydroxides, MOH, are white ionic solids, all very soluble (except LiOH), strong bases, getting stronger down the group.  All Group 1 hydroxides are soluble in water giving strongly alkaline solutions,  and their aqueous solutions readily neutralised by acids (not a redox change) e.g.  MOH(aq) + HCl(aq) ==> MCl(aq) + H2O(l) (M = Li, Na, K, Rb, Cs)  to give the soluble chloride salt* – +  ionically: OH (aq) + H (aq) ==> H2O(l)  an acid–base reaction, same for all four examples in this section  MOH(aq) + HNO3(aq) ==> MNO3(aq) + H2O(l) (M = Li, Na, K, Rb, Cs)  to give the soluble nitrate salt  2MOH(aq) + H2SO4(aq) ==> M2SO4(aq) + 2H2O(l) (M = Li, Na, K, Rb, Cs)  to give the soluble sulphate salt  MOH(aq) + CH3COOH(aq) ==> CH3COOM(aq) + H2O(l) (M = Li, Na, K, Rb, Cs) – +  to give the soluble ethanoate salt CH3COO M  * The hydroxide solutions are readily titrated with standardised hydrochloric acid (burette) using phenolphthalein indicator, the colour change is from pink to colourless.

 

The reaction of group 1 metals with water (a redox reaction) Group 2 metal hydroxide formation o M(s) + 2H2O(l) ==> M(OH)2(aq/s) + H2(g) (M = Mg, Ca, Sr, Ba) o shows the formation of the hydroxide and hydrogen with cold water. 2+ – o ionically: M(s) + 2H2O(l) ==> M (aq) + 2OH (aq) + H2(g) o oxidation number changes, M is 0 to +2, for one H per water it changes from +1 to 0 in H 2. o M = Be (no reaction, anomalous), Mg (very slow reaction), Ca, Sr, Ba (fast to very fast).  i.e. the reactivity increases down the group.  The reactivity trend for Group 2, and its explanation, are similar to that above for the Group 1 Alkali Metals.  The reactivity trend for s–block metals is explained below  Magnesium hydroxide and calcium hydroxide (limewater) are sparingly soluble, but the solubility increases down the group, so barium hydroxide is moderately soluble.  As previously mentioned, a mixture of magnesium oxide/hydroxide and water is sometimes called milk of magnesia and the saturated aqueous solution of calcium hydroxide is called limewater. If the metal is heated in steam the oxide is formed: o e.g. Mg(s) + H2O(g) ==> MgO(s) + H2(g)  NOT an experiment you would do with Alkali Metals! but beryllium gives little reaction. o The oxide is formed because the hydroxide is thermally unstable at higher temperatures













 M(OH)2(s) ==> MO(s) + H2O(g) (M = Be, Mg, Ca, Sr, Ba) REACTIVITY TREND THEORY: The Group 1/2 metal gets more reactive down the group because ... o When an alkali metal atom reacts, it loses an electron to form a singly positively charged ion. + –  e.g. Na ==> Na + e  in terms of electrons 2.8.1 ==> 2.8 and so forming a stable ion with a noble gas electron arrangement. o As you go down the group from one element down to the next the atomic radius gets bigger due to an extra filled electron shell as you go down from one period to the next one. o This means the outer electron is further and further from the nucleus. o This also means the outer electron is also shielded by the extra full electron shell of negative charge. o Due to this shielding the effective nuclear charge on the external electron is ~ +1 (~ proton number – number of noble gas inner core electrons). o Further more, the effective nuclear charge of ~+1 is acting over a larger 'surface area' as the atomic radius increases. o Therefore both of these factors combine to make the outer electron less and less strongly held by the positive nucleus as the atomic number increases (down the group). + o So, the outer electron is more easily lost, and the M ion more easily formed, and so the element is more reactive as you go down the group – best seen in the laboratory with their reaction with water. o The reactivity argument mainly comes down to increasingly lower ionisation energy down the group (i.e. ease of ion formation) and a similar argument applies to the Group 2 metals, but two electrons are removed to form the cation. o The enthalpy change in forming the hydrated cation from the solid metal does not appear to be as important here.  At a more advanced and detailed level, this change can be theoretically split into the  enthalpies of (i) atomisation, (ii) ionisation, (iii) hydration of gaseous ion ... (BUT not here!). θ θ o The reactivity trend is also paralleled by the increasingly negative half–cell potential (E M/M+ and E M/M2+) down groups, 1 and 2 i.e. increasing potential to acts as a reducing agent – an electron donor. o As with water, the reaction of a group 1/2 metal with oxygen or halogens gets more vigorous as you descend the group. All the hydroxides are basic with increasing strength down the group and readily neutralised by acids (not redox reactions). Magnesium hydroxide is sparingly soluble in water but the solubility increases down the group. o M(OH)2(aq/s) + 2HCl(aq) ==> MCl2(aq) + 2H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to give the soluble chloride salt* – +  all base (OH ) ... acid (H ) reactions – +  ionically if soluble the reaction is: OH (aq) + H (aq) ==> H2O(l) 2+ – + 2+  ionically if insoluble: M (OH )2(s) + 2H (aq) ==> M (aq) + 2H2O(l) o M(OH)2(aq/s) + 2HNO3(aq) ==> M(NO3)2(aq) + 2H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to give the soluble nitrate salt o M(OH)2(aq/s) + H2SO4(aq) ==> M2SO4(aq/s) + 2H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to give the sulphate salt o M(OH)2(aq/s) + 2CH3COOH(aq) ==> (CH3COO)2M(aq) + 2H2O(l) (M = Be, Mg, Ca, Sr, Ba)  to give the ethanoate salt o * Saturated calcium hydroxide solution (limewater) can be titrated with standardised hydrochloric acid (burette, low molarity) to determine its solubility. You normally use phenolphthalein indicator and the end–point colour change is from pink to colourless. The Group 2 hydroxides, M(OH)2, get more soluble down the group: o If the hydroxide more or less insoluble (e.g. for Be and Mg), they can be made by adding excess sodium/potassium hydroxide solution to a solution of a soluble salt of a Group 2 metal e.g. three 'double decompositions' are shown below ...  (i) calcium chloride + sodium hydroxide ==> sodium chloride + calcium hydroxide  CaCl2(aq) + 2NaOH(aq) ==> 2NaCl(aq) + Ca(OH)2(s)  (ii) magnesium sulphate + potassium hydroxide ==> potassium sulphate + magnesium hydroxide  MgSO4(aq) + 2KOH(aq) ==> K2SO4(aq) + Mg(OH)2(s)  (iii) beryllium nitrate + sodium hydroxide ==> sodium nitrate + beryllium hydroxide  Be(NO3)2(aq) + 2NaOH(aq) ==> 2NaNO3(aq) + Be(OH)2(s) 2+ –  or ionically: M (aq) + 2OH (aq) ==> M(OH)2(s) for any Group 2 metal M  All the hydroxides are white powders or white gelatinous precipitates. Beryllium hydroxide is amphoteric (an anomaly in the group), because apart from the reactions above, it dissolves in strong alkalis like sodium hydroxide to form a hydroxo–complex ion salts called 'beryllates' e.g. o Be(OH)2(s) + 2NaOH(aq) ==> Na2[Be(OH)4](aq) (not a redox change) – 2– o ionically: Be(OH)2(s) + 2OH (aq) ==> [Be(OH)4] (aq) showing formation of a complex ion For the reaction of Group 1 and 2 hydroxides with carbon dioxide to form the carbonates and hydrogen carbonates, see section 7.9

7.7. The reaction of s–block metals with acids  





Group 1 metals are far too reactive to contemplate adding them to acids in a school laboratory! Group 2 metals, apart from beryllium (another anomaly), readily react with acids, with increasing vigour down the group (explanation in section 7.4). A redox reaction to form the soluble chloride salt. o M(s) + 2HCl(aq) ==> MCl2(aq) + H2(g) (M = Mg, Ca, Sr, Ba) + 2+  ionically for all four examples: M(s) + 2H (aq) ==> M (aq) + H2(g)  oxidation state changes: one M at (0) and two H's at (+1) ==> one M (+2) and two H's at (0)  the metal is oxidised, electron loss, increase in oxidation state  hydrogen ions are reduced, electron gain, decrease in oxidation state o M(s) + 2HNO3(aq) ==> M(NO3)2(aq) + H2(g)  to form the soluble nitrate salt  Looks ok in principle, and does this with Mg and very dilute nitric acid, but rarely this simple, the nitrate(V) ion can get reduced to nasty brown nitrogen(IV) oxide gas (nitrogen dioxide, NO 2) and – other products, NO gas?, NO2 ion? o M(s) + H2SO4(aq) ==> MSO4(aq/ s) + H2(g)  to form soluble ==> insoluble sulphate salt  The reaction from magnesium to barium becomes increasingly slower as the sulphate becomes less soluble, it coats the metal, inhibiting the reaction. o M(s) + 2CH3COOH(aq) ==> (CH3COO)2M(aq) + H2(g) to form soluble ethanoate salt  This reaction is much slower than the previous three because ethanoic acid is a weak acid (about 2% ionised, so the fizzing appears a lot less vigorous than the other three acids using solutions of similar molarity). In aqueous solutions the metal cations formed are hydrated to aqa–complex ions. n+ o not quite the simple isolated ions M (aq) which we use in most equations for brevity. n+ o e.g. [M(H2O)6] (aq) where n=1 for Gp 1 and n=2 for group 2.  There may be several layers of water molecules around the ion, so the six is not the whole story, but is typical for the number of 'nearest neighbours', albeit weakly dative covalently bonded water molecules in this case.  The six is called the co–ordination number and each water molecule (or anything else attached to the central metal ion) is called a ligand.  The shape of such an ion is 'octahedral' and its simplified structure is shown above on the right. The middle 'blob' is the metal ion and the six outer 'blobs' are the water molecules.  (I will replace with proper diagrams later)  However lithium and beryllium are anomalous (M = Li n = 1, or Be n = 2), because of electronic quantum level restrictions, they can have a maximum co–ordination number of four, so their aqueous cations should be written as n+ [M(H2O)4] (aq) which has a tetrahedral shape (shown on the left). As described above, The soluble groups 1/2 salt solutions contain the hydrated cations derived from the metal: + 2+ o tetra–aqua cations [Li(H2O)4] (aq) and [Be(H2O)4] (aq) + o or the hexa–aqua ions [M(H2O)6] (aq) M = Na, K etc. for Group 1 2+ o and [M(H2O)6] (aq) where M = Mg, Ca etc. for Group 2 o The tetraaqua beryllium ion and the hexaaqua magnesium ions generate a slight acidity in their salt 2+ 2+ solutions due to the significant polarising power of the ions (Be very small and double charged, Mg double charged) e.g. 2+

o

for beryllium: [Be(H2O)4] (aq) + H2O(l)

o

or magnesium: [Mg(H2O)6] (aq) + H2O(l)

2+

+

+

[Be(H2O)3(OH)] (aq) + H3O (aq) +

+

[Mg(H2O)5(OH)] (aq) + H3O (aq)



7.8. The reaction of s–block metals with chlorine & halide (X ) salts The salts are white or colourless crystalline solids 

Group 1 metals readily react with halogens (a redox reaction) o e.g. heating the metal in chlorine will cause it to burn forming the chloride o 2M(s) + Cl2(g) ==> 2MCl(s) (redox reaction, M = Li, Na, K, Rb, Cs)  Oxidation state changes: M from 0 to +1, X = F, Cl, Br & I from 0 to –1 + –  The salt products, M X , are white–colourless crystalline ionic solids that dissolve in water to give neutral solutions of about pH 7. The crystalline solids have high melting and boiling points.





The solids do not conduct electricity (no mobile ions or electrons) but will conduct and undergo electrolysis when molten or dissolved in water when ions are free to move to electrodes. – o The halogen is in the –1 oxidation state in the halide ion X o The halides of groups 1–2 are important raw materials e.g.  sodium chloride ==> sodium hydroxide from rock salt by electrolysis of aqueous solution  potassium bromide/iodide ==> elemental bromine/iodine from seawater by oxidation  calcium chloride ==> calcium metal by electrolysis of molten chloride Group 2 metals (except Be) readily react on heating with halogens (a redox reaction) o e.g. heating in chlorine the chloride is formed o M(s) + Cl2(g) ==> MCl2(s) (M = Mg, Ca, Sr, Ba)  Oxidation state changes: M from 0 to +2, X = F, Cl, Br & I from 0 to –1 2+ – + –  The salt products, M (X )2, are similar in properties to the Group 1 M X compounds.  However, beryllium chloride has a polymeric covalent structure, due to the high polarising influence of beryllium in its +2 oxidation state and the smaller difference in electronegativity between Be–Cl compared to chlorine and the other group 1 and 2 metals.

2–



7.9. The properties and chemistry of the carbonates (CO 3 ) & hydrogencarbonates (HCO3 ) The carbonates and hydrogencarbonates are white ionic solids 





Group 1 carbonates M 2CO3: Formed on bubbling carbon dioxide into excess hydroxide solution o 2MOH(aq) + CO2(g) ==> M2CO3(aq) + H2O(l) (M = Li, Na, K, Rb, Cs) – 2– o ionic equation: 2OH (aq) + CO2(g) ==> CO3 (aq) + H2O(l) o The carbonates are white solids, quite soluble in water, and, apart from lithium, thermally stable to red–heat. (see section 7.11) o Hydrated sodium carbonate, Na2CO3.10H2O, is known as washing soda and is used to soften water by precipitating magnesium and calcium salts as their less soluble carbonates (see section 7.10). Group 1 hydrogencarbonates MHCO3: Formed on bubbling excess carbon dioxide into the hydroxide solution o The reaction above happens first and then:  M2CO3(aq) + H2O(l) + CO2(g) ==> 2MHCO3(aq) (M = Li, Na, K, Rb, Cs) 2– –  ionic equation: CO3 (aq) + H2O(l) + CO2(g) ==> 2HCO3 (aq)  They are white solids, slightly soluble in water, weakly alkaline and readily decompose on heating to form the carbonate and carbon dioxide gas. o  e.g. at 270 C: 2NaHCO3(s) ==> Na2CO3(s) + H2O(l) + CO2(g)  An alternative to yeast in baking is to use sodium hydrogencarbonate ('sodium bicarbonate' or 'baking soda'). The rising action is also due to carbon dioxide gas formation from reacting with an acid (e.g. an organic acid like tartaric acid) and nothing to do with enzymes:  self–raising baking powder = carbonate base + a solid organic acid, giving  sodium hydrogencarbonate + acid ==> sodium salt of acid + water + carbon dioxide Group 1 carbonates and hydrogencarbonates are readily neutralised by acids: 2– – + o these are base(proton accepting CO3 or HCO3 )–acid(H from HCl etc.) reactions giving a salt, water and carbon dioxide o In all cases M = Li, Na, K, Rb, Cs e.g. o M2CO3(aq) + 2HCl(aq) ==> 2MCl(aq) + H2O(l) + CO2(g) to give the chloride salt* 2– +  ionically for any soluble carbonates: CO3 (aq) + 2H (aq) ==> H2O(l) + CO2(g) o M2CO3(aq) + 2HNO3(aq) ==> 2MNO3(aq) + H2O(l) + CO2(g) to give the soluble nitrate salt o M2CO3(aq) + H2SO4(aq) ==> M2SO4(aq) + H2O(l) + CO2(g) to give the soluble sulphate salt o M2CO3(aq) + 2CH3COOH(aq) ==> CH3COOM(aq) + H2O(l) + CO2(g) to give the soluble ethanoate salt o MHCO3(aq) + HCl(aq) ==> MCl(aq) + H2O(l) + CO2(g) to give the soluble chloride salt – +  ionically for any soluble hydrogencarbonates: HCO3 (aq) + H (aq) ==> H2O(l) + CO2(g) o MHCO3(aq) + HNO3(aq) ==> MNO3(aq) + H2O(l) + CO2(g) to give the nitrate salt o 2MHCO3(aq) + H2SO4(aq) ==> M2SO4(aq) + 2H2O(l) + CO2(g) to give the sulphate salt o MHCO3(aq) + CH3COOH(aq) ==> CH3COOM(aq) + H2O(l) + CO2(g) to give the ethanoate salt o * The group 1 carbonates e.g. sodium carbonate can be titrated with standardised hydrochloric acid using methyl orange indicator (red in acid, yellow in alkali, the end point is a sort of 'pinky orange').









Group 2 carbonates MCO3: formed on bubbling carbon dioxide into the hydroxide solution or 'slurry', but beryllium carbonate is not stable (another anomaly). Non of them are very soluble. o M(OH)2(aq) + CO2(g) ==> MCO3(s) + H2O(l) (M = Mg, Ca, Sr, Ba) o When M = Ca, this the reaction of limewater when positively testing for carbon dioxide gas. o They can also be prepared by a double decomposition precipitation reaction (see section 7.10). o The carbonates decompose on heating to give the oxide and carbon dioxide and exhibit a clear thermal stability trend (see section 7.11). Group 2 hydrogencarbonates M(HCO3)2: formed when excess carbon dioxide is bubbled through a slurry of the carbonate and they are only stable in solution and their reaction with acids is not important: o MCO3(s) + H2O(l) + CO2(g) M(HCO3)2(aq) (M = Mg, Ca, Sr, Ba)  This the reaction of 'temporary hard water' formation from calcium and magnesium carbonate minerals like limestone and dolomite. Boiling the solution reverses the reaction, so removing the metal cations from solution, and referred to as removing 'temporary hardness'. Group 2 carbonates MCO3 readily neutralised by acids to form salt, water and carbon dioxide: 2– + o These are Bronsted–Lowry base (proton accepting CO3 )–acid (H from HCl etc.) reactions giving a salt, water and carbon dioxide e.g. for M = Mg, Ca, Sr, Ba o MCO3(s) + 2HCl(aq) ==> MCl2(aq) + H2O(l) + CO2(g) to give the chloride salt (M = Mg, Ca, Sr, Ba) 2+ 2– + 2+  ionically for all four examples: M CO3 (s) + 2H (aq) ==> M (aq) + H2O(l) + CO2(g) o MCO3(s) + 2HNO3(aq) ==> M(NO3)2(aq) + H2O(l) + CO2(g) to give the nitrate salt o MCO3(s) + H2SO4(aq) ==> M2SO4(aq) + H2O(l) + CO2(g) to give the sulphate salt o MCO3(s) + 2CH3COOH(aq) ==> (CH3COO)2M(aq) + H2O(l) + CO2(g) to give the ethanoate salt The thermal decomposition of carbonates and nitrates is covered in detail in section 7.11

7.10. Solubility Trends of Group 2 compounds – linked to preparations  





All the nitrates, M(NO3)2, are soluble in water. (M = Be, Mg, Ca, Sr, Ba) The hydroxides M(OH)2, get more soluble down the group: (M = Be, Mg, Ca, Sr, Ba) o If more or less insoluble, they can be made by adding sodium hydroxide solution to a solution of a soluble salt of M e.g.  magnesium chloride + sodium hydroxide ==> sodium chloride + magnesium hydroxide  MgCl2(aq) + 2NaOH(aq) ==> 2NaCl(aq) + Mg(OH)2(s) 2+ –  or ionically: Mg (aq) + 2OH (aq) ==> Mg(OH)2(s)  Magnesium hydroxide is almost insoluble in water i.e. sparingly soluble.  Calcium hydroxide is slightly soluble in water, so–called 'limewater' used in the simple test for carbon dioxide gas.  Barium hydroxide is moderately soluble in water. The sulphates, MSO4, get less soluble down the group. (M = Be?, Mg, Ca, Sr, Ba) o Magnesium sulphate is very soluble in water, in fact it was first crystallised from spring water e.g. the chalk downs of Southern England, hence known as Epsom Salts.  the heptahydrate salt MgSO4.7H2O o Calcium sulphate is slightly soluble in water. o If more or less insoluble, they can be made by adding dilute sulphuric acid or sodium sulphate solution to a solution of a soluble salt of M. o This reaction is used as a test for a sulphate by adding an acidified barium chloride/dil. hydrochloric acid or barium nitrate/dil. nitric acid solution to a solution of the suspected sulphate. A dense white precipitate of barium sulphate forms in a positive result and also illustrates the preparation too e.g. o barium chloride + sodium sulphate ==> sodium chloride + barium sulphate o BaCl2(aq) + Na2SO4(aq) ==> 2NaCl(aq) + BaSO4(s) o or o barium nitrate + magnesium sulphate ==> magnesium nitrate + barium sulphate o Ba(NO3)2(aq) + MgSO4(aq) ==> Mg(NO3)2(aq) + BaSO4(s) 2+ 2–  or ionically in each case Ba (aq) + SO4 (aq) ==> BaSO4(s)  Why is the acidification necessary? The addition of dilute hydrochloric acid is to prevent the precipitation of other insoluble salts like barium sulphite which would be confusing and make the test less specific. The carbonates, MCO3, get less soluble down the group. (M = Mg, Ca, Sr, Ba) o If insoluble, they can be made by adding sodium carbonate solution to a solution of a soluble salt of M e.g. the 'double decomposition' ... o magnesium chloride + sodium carbonate ==> sodium chloride + magnesium carbonate o MgCl2(aq) + Na2CO3(aq) ==> 2NaCl(aq) + MgCO3(s)

   







2+

2–

or ionically: Mg (aq) + CO3 (aq) ==> MgCO3(s) You can also use the nitrate and in the case of magnesium, its sulphate too. + The spectator ions are Na and the chloride or sulphate etc. anion from the original group II salt. Hydrated sodium carbonate, Na2CO3.10H2O, is known as washing soda and is used to soften water by using the above reaction.  e.g. calcium sulphate (gypsum) + sodium carbonate ==> sodium sulphate (soluble) + calcium carbonate (insoluble)  CaSO4(aq) + Na2CO3(aq) ==> Na2SO4(aq) + CaCO3(s) (ionic equation similar to above) Explanation of solubility trends (usually dealt with later in course e.g. in UK A2 advanced level) o The simplest approach is to consider the two enthalpy change trends.  The process of dissolving can be analysed in terms of two theoretical stages e.g. for simple cation–anion ionic compound. n+ n–  In the arguments outlined below M could be Gp1 or Gp2 metal cation etc., X could be halide, oxide, hydroxide, sulphate, carbonate anion etc., and n is the charge on ion – the n's may be different or the same): n+ n– n+ n–  (1) M aX b(s) ==> aM (g) + bX (g) (breaking the lattice apart into its constituent ions)  This process is always endothermic, and is called the lattice enthalpy. Its usually defined in the opposite direction by saying it is 'the energy released when 1 mole of an ionic lattice is formed from its constituent gaseous ions' (at 298K, 1 atmos./101kPa pressure).  * The lattice enthalpy decreases down the group as the cation radius increases (anion radius constant for a particular series e.g. sulphates). Therefore, energetically, the solvation in terms of lattice energy is increasingly favoured down the group. n+ n+ n– n–  (2) M (g) + aq ==> M (aq) and X (g) + aq ==> X (aq)  Representing the solvation–hydration of ions.  The equations above represent to the two 'hydration enthalpies', the heat released when an isolated gaseous ion becomes solvated by water to form an aqueous solution (1M, 298K, 1 atmos./101kPa pressure)  * The hydration enthalpy for the cation decreases down the group as the radius gets larger. Therefore, energetically, the solvation is less favoured down the group as the cation radius increases. o * In both cases the numerical enthalpy value increases the smaller the radii as charges closer, and the greater the ionic charge (constant for a series), both factors increase the electrical attraction of either cation–anion in the crystal or ion–water in aqueous solution. We therefore have two competing trends! o So, one approach is to say which 'energy change' trend outweighs the other to explain the solubility trend ... o e.g. for Group 2 hydroxides, energetically, the decrease in lattice enthalpy more than compensates for the 2+ decrease in the hydration enthalpy of the M cation as it gets larger down the group so leading to greater solubility. Unfortunately the above is hardly an explanation of a correct prediction! and neither is entropy taken into consideration. o The explanations offered are argued after the fact and unsatisfactory! o There is no simple explanation possible and ultimately the solubility is dependent on the entropy changes, a notoriously difficult concept area. o If there was an appropriate AS–A2 answer, it would be in the textbooks by now! o See below on Jim Clarks website for an intelligent discussion on the matter. Jim's Group 2 pages solubility descriptions and trends and discussions and theory of solubility

7.11. Thermal decomposition & stability trends of Group 1 and Group 2 compounds 

One theory of the thermal instability trend o The lower down the metal in the group the more thermally stable is its hydroxide, nitrate, carbonate or sulphate etc. o This is because the polarising power of the cation increases up the group with the smaller ionic radius,  AND, in most cases discussed here, the smaller the cation the greater the lattice enthalpy of the oxide formed on decomposition (meaning the oxide is more thermodynamically stable up the group). + 2+  Particularly for the tiny Li and Be ions, the polarising effect considerably reduces the stability of their compounds (e.g. BeCO3 is quite unstable and Li2CO3 decomposes on gentle heating) o The 'polarising power' of a cation is a measure of its electric field effect to attract and distort electron charge on a neighbouring anion:





The cation polarising power increases with increase in charge on the ion or decreasing the radius of the ion, both of which increase the intensity of the electric field effect. o The 'polarisability of an anion' is how easily the electron charge clouds are 'distorted' by a neighbouring cation.  The anion is more easily distorted the larger the anion radius and the higher its charge. The general 'polarising effect' is shown in the diagram below. n+ n– o Think of the M cation as the Gp1 or Gp2 cation and the XO3 anion as the nitrate ion or the carbonate ion (and the think the same way for a hydroxide ion or a sulphate – in fact any 'oxyanion').

o o o













The electrical field of the cation distorts or polarises the anion, and at the decomposition temperature, a 'residual' oxide ion is attracted to the cation and the rest of the original larger anion is released as a gas or gases. Notes:  (i) The residual oxide ion is smaller and less polarisable.  (ii) These reactions eventually become favourable at higher temperature because of the large increase in the 'systems' entropy when gases formed.  (iii) the smaller oxide ion means the resulting oxide has a higher lattice enthalpy than the carbonate or nitrate etc. and increases up the group making the decomposition more favourable.  Point (iii) forms the basis of a purely thermodynamic argument to explain the thermal stability trend and even predict decomposition temperatures (see the end of this section).

Trends in thermal stability: o In all cases, for a particular set of e.g. Gp1 or Gp2 compounds, the thermal stability increases down the group as the ionic radius of the cation increases, and its polarising power decreases. o Group 1 compounds tend to be more thermally stable than group 2 compounds because the cation has a smaller charge and a larger ionic radius, and so a lower polarising power, particularly when adjacent metals on the same period are compared. Group 1 Carbonates: o lithium carbonate readily decomposes: Li2CO3(s) ==> Li2O(s) + CO2(g) o but the others are quite stable to red heat, so again lithium is anomalous by its comparative carbonate instability. Group 2 Carbonates: o The carbonates thermally decompose into the metal oxide and carbon dioxide gas. o MCO3 ==> MO(s) + CO2(g) (M = Be, Mg, Ca, Sr, Ba) o Thermal Tdecomp order BaCO3 > SrCO3 > CaCO3 > MgCO3 > BeCO3 o This is the reaction that converts calcium carbonate (limestone) into calcium oxide (quicklime) in a limekiln at o about 900 C. beryllium BeCO3 is unstable at room temperature, magnesium carbonate MgCO3 decomposes o o o at about 400 C, strontium carbonate SrCO3 at 1280 C and barium carbonate BaCO3 at 1360 C. Group 1 Nitrates: o lithium nitrate is the least stable and decomposes readily on heating to form lithium oxide, nitrogen dioxide and oxygen. o 4LiNO3(s) ==> 2Li2O(s) + 4NO2(g) + O2(g) – – o The other group 1 Alkali Metal nitrates [NO3 , nitrate(V)] decompose to form the nitrite [NO2 , nitrate(III)] salt + and oxygen gas. Lithium is anomalous due to the particularly high polarising power of the Li ion. o 2MNO3(s) ==> 2MNO2(s) + O2(g) (M = Na, K, Rb, Cs) o The nitrites, or nitrate(III)'s, are very thermally stable white solids, soluble in water giving neutral solutions in water. Group 2 Nitrates: o For M = Mg, Ca, Sr, Ba the nitrate decompose to form the metal oxide, nasty brown nitrogen dioxide [nitrogen(IV) oxide] gas and oxygen gas when strongly heated. o 2M(NO3)2(s) ==> 2MO(s) + 4NO2(g) + O2(g) (M = Be?, Mg, Ca, Sr, Ba) o Thermal Tdecomp order Ba(NO3)2 > Sr(NO3)2 > Ca(NO3)2 > Mg(NO3)2 > BeNO3)2 Group 1 and Group 2 hydroxides o The general thermal decomposition equations are ...  Group 1: 2MOH(s) ==> M2O(s) + H2O(g) (M = Li, Na, K, Rb, Cs)  Group 2: M(OH)2(s) ==> MO(s) + H2O(g) (M = Be?, Mg, Ca, Sr, Ba)  The thermal stability trend is just the same as for carbonates, nitrates (and even sulphates) i.e. they become more stable down the group with increasing atomic number of the metal M.  So for group 1 the Tdecomp sequence is CsOH > RbOH > KOH > NaOH > LiOH

 

and for group 2 the Tdecomp sequence is Ba(OH)2 > Sr(OH)2 > Ca(OH)2 > Mg(OH)2 > Be(OH)2

o – Comparing the stabilities of Group 1 and Group 2 compounds: o Group 1 compounds are more stable than group 2 compounds, particularly when group adjacent elements on the same period. + o The reason being the polarising effect of the group I cation M , is much less than the polarising power of the 2+ smaller and more highly charged group II M ion, particularly when comparing adjacent metals on the same period.  The group 2 cation is both smaller and more highly charged than the corresponding group 1 cation. A thermodynamic discussion on the thermal stability of s–block compounds I'm referring to oxyanion compounds like carbonates, sulfates, hydroxides and sulfates

 





What ensues is a completely alternative explanation of the thermal stability trends of the oxyanion compounds of alkali metals and alkaline earth metals without any reference to the relative polarising effect of the cation. It may seem curious at first sight, but large cations can stabilise large anions in a crystal lattice (and vice versa). o The decomposition temperatures of thermally unstable compounds containing large anions e.g. carbonates, increases with cation radius. o The stabilizing influence of a cation can be explained in terms of trends in lattice enthalpies (lattice energies). o The arguments given below are purely in terms of the thermodynamics and explain the Group 2 carbonate stability trend without reference to the polarising power of the cation (which is the argument required by most UK GCE A level syllabuses). An initial discussion of the Group II carbonate stability trend illustrates the points made above. o A study of the thermal decomposition temperatures of the Group 2 Alkaline Earth carbonates proves most instructive. o The general decomposition equation for group II carbonates to give the group II oxide is  MCO3(s) ==> MO(s) + CO2(g)  M = Be, Mg, Ca, Sr and Ba  Beryllium carbonate is not very stable (can be stabilised in an atmosphere of CO 2) and radium carbonate would be rather too radioactive to study!  MCO3 and MO are sometimes referred to as the alkaline–earth carbonates and alkaline–earth oxides. Most of the data tabulated below was obtained from 'Inorganic Chemistry' 2nd edition, by Shriver, Atkins and Langford and the Nuffield Science Book of Data (revised edition 1988) plus internet research for research papers quoting as upto date values as I could find. Thermodynamic and decomposition temperature data for Group II carbonates of the periodic table. the basis for the purely thermodynamic argument for the Group II carbonate thermal stability trend

Decomposition data Ø

–1

Ø

–1

Be

Mg

Ca

Sr

Ba

?

+48.3

+130.4

+183.8

+218.1

?

+100.6

+178.3

+234.6

+269.3

?

+175.0

+160.6

+171.0

+172.1

?

302, 575

837, 1110

1099, 1372

1292, 1565

~100 C

400 C

900 C

1280 C

1360 C

?

3180

2987

2720

2615

4443

3960

3489

3248

3011

ΔG (kJ mol ) ΔH (kJ mol ) Ø

ΔS (J K

–1

–1

mol )

o

Tdecomp ( C, K) theoretical Typical quoted decomposition temperatures Ø MCO3

LE

Ø MO

LE

o

–1

(kJ mol ) –1

(kJ mol )

o

o

o

o

ΔLE(MO–MCO3) M

2+

radius in nm

? 2+

Be

= 0.034

other radii: oxide ion O        



Ø

2–

780 Mg

2+

= 0.078

502 2+

Ca

= 0.100

528 2+

Sr

= 0.127

396 2+

Ba

= 0.143

2–

= 0.140 nm, carbonate ion CO3 = 0.176 nm

ΔG = standard Gibbs free energy change for the thermal decomposition of the carbonate MCO 3 (at 298K, 1 atm) Ø ΔH = standard enthalpy change for the thermal decomposition of the carbonate MCO 3 (at 298K, 1 atm) Ø ΔS = standard entropy change for the thermal decomposition of MCO 3 (at 298K, 1 atm) Ø Ø Ø o Note ΔS = ΔS system and is NOT ΔS surroundings ... o ... see theory of the method of calculating the decomposition temperature below. Tdecomp = decomposition temperature in Celsius and Kelvin when the equilibrium pressure pCO 2 = 1 atm (101kPa) Ø LE MCO3 = lattice enthalpy/lattice energy of the group 2 carbonate MCO3 (at 298K, 1 atm. pressure) Ø LE MO = lattice enthalpy/lattice energy of the group 2 oxide MO (at 298K, 1 atm. pressure) ΔLE(MO–MCO3) is the difference between the lattice enthalpies of the group 2 oxide and the corresponding group 2 carbonate The decomposition temperature was calculated as follows ... o (i) From the Gibbs free energy equation Ø Ø Ø  ΔG = ΔH – TΔS (terms defined above)  The criteria for equilibrium is when ΔG = 0 Ø Ø  therefore at equilibrium: ΔH – TΔS = 0, and rearranging terms and signs gives Ø Ø Ø Ø  TΔS = ΔH , therefore T = ΔH / ΔS  = decomposition temperature (K) to give an equilibrium pressure of 1 atm of carbon dioxide gas o (ii) From the total entropy change equation  This is if your course doesn't involve free energy, or just an alternative method depending on what data is given or available. Ø Ø Ø  ΔS total = ΔS system + ΔS surroundings Ø  The criteria for equilibrium is that ΔS total = 0 Ø Ø  therefore at equilibrium: 0 = ΔS system + ΔS surroundings, rearranging so ... Ø Ø  at equilibrium: –ΔS surroundings = ΔS system Ø Ø  since ΔS surroundings = –ΔH / T i.e. minus the enthalpy change divided by the absolute temperature Ø Ø  then: –(–ΔH / T) = ΔS system Ø Ø  ΔH / T = ΔS system, rearranging ... Ø Ø  gives: Tdecomposition = ΔH / ΔS system  i.e. the identical expression derived from the Gibbs free energy expression. o (iii) All you have to do now is substitute in the numerical values from the data table ...  ... and note that ΔG and ΔH are usually in kJ BUT S or ΔS values are usually in J so don't forget to multiply the ΔG and ΔH values by 1000, therefore ...  Tdecomp(MgCO3) = 100600 / 175.0 = 575 K  Tdecomp(CaCO3) = 178300 / 160.6 = 1110 K  Tdecomp(SrCO3) = 234600 / 171.0 = 1372 K  Tdecomp(BaCO3) = 269300 / 172.1 = 1565 K o Assumptions and comments  The calculations have been based on the enthalpy, free energy and entropy value changes at 298K.  Enthalpy values (H) do vary with temperature and entropy values (S) increase with temperatures.  These factors have been ignored in the calculation BUT the values seem to be roughly born out by experiment as far as I can gather from the values quoted in the literature.  Note that the entropy change is almost constant because the increase in entropy is primarily due to the formation of 1 mole of carbon dioxide gas in each case.  Remember Sgas >> Sliquid > Ssolid  Giving a large increase in entropy, the formation of a gas is a powerful driving force to facilitate the decomposition of these essentially stable compounds BUT this factor applies almost equally to all the group 2 carbonates, therefore entropy cannot be used to explain the stability trend.  To explain the thermal stability trend thermodynamically, we must look at the enthalpy changes and the lattice enthalpies of the carbonate and the oxide residue. The thermodynamic argument to explain the thermal stability trend of Group 2 carbonates. o To follow the argument you need to x–reference with the numerical values in the data table above. o Irrespective of the validity of the theoretical values calculated for the group 2 carbonate decomposition temperatures, what is clearly predicted is that they become more thermally stable down the group i.e. with increase in atomic number of the metal. o This increase in stability trend matches the experimental values which in turn are of the same order as those calculated theoretically despite the decrease in lattice enthalpy of the carbonate down the group!

o o



The first point to be made is that the endothermic enthalpy of reaction increases down the group. This is itself a clear indication that the decomposition is becoming much less energetically favourable down the group and remember the entropy change is almost constant down the group. o The pivotal point in the argument–explanation rests on the differences between the lattice enthalpies (LEs) of the reacting carbonate and the oxide product as you descend the group. o The difference in their LEs is primarily the reason for the rise in the endothermic enthalpy of reaction leading to the rising decomposition temperature. o Down the group the lattice enthalpy of both the carbonate and the oxide decrease because the cation radius increases.  Lattice enthalpy is a function of two factors (other than the spatial positions of the ions)  (i) the charge on the positive and negative ions attracting each other, both constant in this case and ...  (ii) the radius of the ions. Here the carbonate ion and the oxide ion radii are constant, but the cation radius is increasing with atomic number of the group 2 metal. This increase the nuclear (+) ... (–) ion distance, reducing the force of attraction and hence reducing the lattice enthalpy. o However, generally speaking (3/4 values!), as you go down the group the lattice enthalpy of the oxide decreases more rapidly than the lattice enthalpy of the carbonate. o This means the difference between the two enthalpies becomes less and less down the group making the enthalpy more and more positive/endothermic and resulting in an increasingly higher temperature to effect the thermal decomposition (to the extent of producing an equilibrium partial pressure of 1 atmosphere of carbon dioxide gas). 2+ 2– 2+ 2–  Breaking up the M CO3 lattice is endothermic, but the formation of the M O lattice is exothermic and numerically greater than for MCO3 and particularly the lower the atomic number of the metal (i.e.) higher up the group). o Note that, although I do not have the comparable data for beryllium, the quoted lattice energy for beryllium 2+ oxide (BeO) is very high due to the very small beryllium cation Be , and therefore extrapolating up the group, you would expect beryllium carbonate to have a much lower decomposition temperature. o  This is usually quoted as ~100 C and completely fits in with both the theoretical thermal stability trend and the experimental thermal decomposition values quoted in the literature–textbooks. Further extension of the ideas – looking at other thermal stability trends o The anhydrous Group 2 sulphates show a similar thermal stability trend to the carbonates...  i.e. for the reaction: MSO4(s) ==> MO(s) + SO3(g)  the Tdecomp is in the order BaSO4 > SrSO4 > CaSO4 > MgSO4  – o The effect of the cation radius also shows up when comparing the thermal stability of Group 1 carbonates and Group 2 carbonates.  Comparing the two thermal decomposition reactions ... 2+ +  Because of the greater charge on the Group 2 cation (M ) compared to the Group 1 cation (M ) the lattice enthalpy of the Group 2 oxide is much greater than for the Group 1 oxide.  So, for the s–block metals on the same period, for Tdecomp the trend is M2CO3 > MCO3  The lattice enthalpies are ... –1  2478 kJ mol for Na2O (1239 kJ per mol Na) –1  and 3960 kJ mol for MgO (3960 kJ per mol Mg)  The very high lattice enthalpy of MgO compared to that the LE for Na2O contributes to a much less endothermic enthalpy of decomposition for the Group 2 carbonate compared to the Group 1 carbonate and hence a lower decomposition temperature for the MCO 3.  – o The stability trend for Group 1 alkali metal carbonates is similar to that of the Group 2 carbonates ...  i.e. for Tdecomp the trend is K2CO3 > Na2CO3 > Li2CO3 etc. ...  ... for exactly the same reasons argued above for MCO3 stability. o Comparing the thermal stability of Group 1 nitrates [nitrate(V)] and Group 2 nitrates [nitrate(V)].  In group 1, only lithium nitrate readily decomposes to the oxide ...  4LiNO3(s) ==> 2Li2O(s) + 2NO2(g) + O2(g)  whereas all the other nitrates initially give the thermally stable nitrite [nitrate(III)] ...  2MNO3(s) ==> 2MNO2(s) + O2(g) (M = Na, K, Rb, Cs) +  The relatively much smaller size of the lithium cation (Li ) produces a particularly high lattice enthalpy for lithium oxide compared to the other group 1 oxides, hence the direct formation of the oxide.  However, due to the much higher MO lattice enthalpies, the oxide is formed directly in each case for the group 2 nitrates ...  2M(NO3)2(s) ==> 2MO(s) + 4NO2(g) + O2(g) (M = Mg, Ca, Sr and Ba)  and the thermal stability trend will be Ba(NO3)2 > Sr(NO3)2 > Ca(NO3)2 > Mg(NO3)2  as in the case of the group 2 carbonates and sulfates etc. o –

7.12. some examples of the uses of Group 1 and 2 Metals and their Compounds.               

MCl & MCl2 The Group 1 and Group 2 chlorides are used as sources of metal extraction by electrolysis. Na & Mg Sodium and magnesium are then used to extract titanium from its chloride by displacement. Na Sodium vapour is used in the yellow–orange street lamps. NaCl Sodium chloride 'common salt' is used as a food flavouring and preservative, source of chlorine, hydrogen, sodium metal and sodium hydroxide via electrolytic processes. NaHCO3 is used in baking powders – heat or a weak organic acid (e.g. citric acid) is used in baking powders to form carbon dioxide gas to produce the 'rising' action in baking. Na2CO3 Sodium carbonate is used in the manufacture of glass and the treatment of hard water. NaOH Sodium hydroxide, an important strong alkali, is used in the manufacture of sodium salts, soaps, detergents, bleaches, rayon. KNO3 Potassium nitrate is used in NPK fertilisers. Mg Magnesium metal is used in the manufacture of alloys, particularly those of aluminium. Mg(OH)2 Magnesium hydroxide is used in antacid indigestion powders to neutralise excess stomach (hydrochloric) acid. When magnesium hydroxide mixed with water it is known commercially as 'milk of magnesia' as an antacid remedy avoiding the use indigestion tablets. CaCO3 Calcium carbonate (limestone) and calcium oxide (quicklime, from thermal decomposition of limestone in kiln) are both used in agriculture to reduce the acidity of soil to improve its fertility. CaCO3 Limestone is used directly as building and road foundation material. CaCO3 Limestone is heated with clay (aluminium silicates) to make cement. BaSO4 Barium sulphate is used in medicine for X–ray colonoscopy of the bowel ('barium meal'), the dense white solid shows up clearly as a white or dark shadow and hence the physical topography of the intestines. and there are lots of other examples if you dig around.

Part 8. The p-block elements: 8.1 Group 3/13 Boron and Aluminium in particular

Group 3/13 Introduction

down group 3/13 ===> property\Zsymbol, name

5B

Boron

13Al

Aluminium

31Ga

Gallium

49In

Indium

81Tl

Thallium

Period

2

3

4

5

6

Appearance (RTP)

brown solid

silvery solid

silvery solid

silvery solid

silvery solid

o

2300

661

30

156

304

o

3659

2467

2400

2080

1457

2.3

2.7

5.9

7.3

11.9

2B2O3(s) Reaction of oxide with water: o Insoluble, no reaction but it is a weakly acidic oxide. Reaction of oxide with acids: o None, only acidic in acid-base behaviour. Reaction of oxide with strong bases/alkalis: o Presumably dissolves to give a solution of sodium borate.

 



Reaction of element with chlorine: o Forms covalent liquid boron trichloride on heating in chlorine gas.  2B(s) + 3Cl2(g) ==> 2BCl3(l) Reaction of chloride with water: o It hydrolyses to form boric acid and hydrochloric acid. *  BCl3(l) + 3H2O(l) ==> B(OH)3(aq) + 3HCl(aq) *  can also be, but less accurately, written as H3BO3 Reaction of element with water: o None.

Some molecule shapes and bond angles Three bond pairs of electrons gives TRIGONAL PLANAR shape. The Q-X-Q bond angle is o exactly 120 e.g. for gaseous boron hydride BH3 (X = B, Q = H).

Three bond pairs of electrons gives TRIGONAL PLANAR shape. The Q-X-Q bond o angle is exactly 120 e.g. for gaseous boron trifluoride BF3 (Q = F, Cl and X = B)

H3N:=>BF3

Boron trifluoride (3 bonding pairs, 6 outer electrons) acts as a lone pair acceptor (Lewis acid) and ammonia (3 bond pairs) and lone pair which enables it to act as a Lewis base - a an electron pair donor. It donates the lone pair to the 4th 'vacant' boron orbital to form a sort of 'adduct' compound. Its shape is essentially the same as ethane, a o sort of double tetrahedral with H-N-H, N-B-F and F-B-F bond angles of ~109 .

-

Boron compound reducing agents in organic chemistry    

 

Derivatives of boron hydride are useful reducing agents in organic chemistry. o All the reduction reactions are shown as simplified equations. Sodium tetrahydrioborate(III), NaBH4 (sodium borohydride) reduces aldehydes to primary alcohols and ketones to secondary alcohols. These reactions are essentially the reduction of the carbony1 group >C=O to >CHOH. The reaction can be carried out in water. The reduction mechanism is very complicated, but can be considered in a simplistic way as involving the donation of a hydride ion to the aldehyde/ketone. o aldehyde: RCHO + 2[H] ==> RCH2OH (R = H, alkyl or aryl) o ketone: R2C=O + 2[H] ==> R2CHOH (R = alkyl or aryl) NaBH4, is not a powerful enough reducing agent to reduce carboxylic acids to a primary aliphatic alcohol. NaBH4, is not a powerful enough reducing agent to reduce nitro-aromatic compounds to primary aromatic amines.

ALUMINIUM - brief summary of a few points    

 





The structure of the element: o Giant lattice metallic structure of immobile positive metal ions surrounded by a 'sea' of freely moving mobile electrons (so-called delocalised electrons). Physical properties: o o o Moderately hard high melting solid; mpt 661 C; bpt 2467 C; good conductor heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 1 o Gp3; e.c. 2,8,3 or 1s 2s 2p 3s 3p ; (+3 only) e.g. Al2O3 and AlCl3. Reaction of element with oxygen: o Reacts when heated strongly in air to form a white powder of aluminium oxide which has a giant ionic 3+ 2structure, (Al )2(O )3.  4Al(s) + 3O2(g) ==> 2Al2O3(s)  The above reaction occurs very rapidly on a freshly cut aluminium surface, but the microscopic oxide layer inhibits any further reaction, giving aluminium a 'lower reactivity' than expected, and its excellent anti-corrosion properties. Reaction of oxide with water: o Insoluble, no reaction but it is an amphoteric oxide and forms salts with both acids and alkali (see below). Reaction of oxide with acids: o It behaves as a basic oxide dissolving to form the chloride, sulphate and nitrate salt in the relevant dilute acid. o Al2O3(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2O(l) o Al2O3(s) + 3H2SO4(aq) ==> Al2(SO4)3(aq) + 3H2O(l) o Al2O3(s) + 6HNO3(aq) ==> 2Al(NO3)3(aq) + 3H2O(l) + 3+ o ionic equation: Al2O3(s) + 6H (aq) ==> 2Al (aq) + 3H2O(l) Reaction of oxide with strong bases/alkalis: o The oxide also behaves as an acidic oxide by dissolving in strong soluble bases to form aluminate(III) salts. o e.g. Al2O3(s) + 2NaOH(aq) + 3H2O(l) ==> 2Na[Al(OH)4](aq) o forming sodium aluminate(III) with sodium hydroxide. o ionic equation: Al2O3(s) + 2OH (aq) + 3H2O(l) ==> 2[Al(OH)4] (aq) o Therefore aluminium oxide is an amphoteric oxide, because of this dual acid-base behaviour. Reaction of element with chlorine: * o Burns when heated strongly in chlorine gas to form the white solid aluminium chloride on heating in chlorine gas.  2Al(s) + 3Cl2(g) ==> 2AlCl3(s) *  It is often a faint yellow in colour, due to traces of iron forming iron(III) chloride.

 

 

Aluminium chloride is a curious substance in its behaviour. The solid, AlCl3, consists of an ionic 3+ o lattice of Al ions, each surrounded by six Cl ions, BUT on heating, at about 180 C, the thermal kinetic energy of vibration of the ions in the lattice is sufficient to cause it break down and sublimation takes place (s ==> g). In doing so the co-ordination number of the aluminium changes from six to four to form the readily vapourised covalent dimer molecule, Al2Cl6, shown above. Reaction of element with water: o None due to protective oxide layer. Reactions of the hexa-aqua aluminium ion: o It gives a gelatinous white precipitate with sodium hydroxide or ammonia solution which displays amphoteric behaviour by dissolving in excess strong alkali (NaOH(aq), NOT NH3(aq)) and acids. 3+  Al (aq) + 3OH (aq) ==> Al(OH)3(s) 3+  or [Al(H2O)6] (aq) + 3OH (aq) ==> [Al(OH)3(H2O)3] + 3H2O(l)  The hydroxide readily dissolves in acids to form salts: + 3+  Al(OH)3(s) + 3H (aq) ==> Al (aq) + 3H2O(l) + 3+  or more elaborately: [Al(OH)3(H2O)3] + 3H3O (aq) [Al(H2O)6] (aq) + 3H2O(l)  Thus showing amphoteric behaviour, since the hydroxide ppt. also dissolves in excess strong alkali (below). 3+ 3 [Al(H2O)6] (aq) + 6OH (aq) ==> [Al(OH)6] (aq) + 6H2O(l) (from original aqueous ion) 3 or [Al(OH)3(H2O)3](s) + 3OH (aq) ==> [Al(OH)6] (aq) + 3H2O(l) (from hydroxide ppt.) 3 or more simply Al(OH)3(s) + 3OH (aq) ==> [Al(OH)6] (aq) (from hydroxide ppt.) o With aqueous sodium carbonate solution, the hydroxide ppt. is formed, and, because of its acidic nature, bubbles of carbon dioxide gas are evolved. 3+ 22+  2[Al(H2O)6] (aq) + CO3 (aq) 2[Al(H2O)5(OH)] (aq) + H2O(l) + CO2(g)

 

this process of proton donation continues until the gelatinous ppt. [Al(OH)3(H2O)3](s) is formed, but will not dissolve in excess of the weak base/alkali. Sodium carbonate is not a strong enough base-alkali to dissolve the aluminium hydroxide precipitate.

The extraction of aluminium Aluminium is obtained from mining the mineral bauxite. The purified bauxite ore of aluminium oxide is continuously fed in. Cryolite is added to lower the melting point and dissolve the ore. Ions must be free to move to the electrode connections called the 3+ cathode (-, negative), attracting positive ions e.g. Al , and the anode (+, 2positive) which attracts negative ions e.g. O . When the d.c. current is passed through aluminium forms at the negative cathode (metal*) and sinks to the bottom of the tank. At the positive anode, oxygen gas is formed (non-metal*). This is quite a problem. At the high temperature of the electrolysis cell it burns and oxidises away the carbon electrodes to form toxic carbon monoxide or carbon dioxide. So the electrode is regularly replaced and the waste gases dealt with! It is a costly process (6x more than Fe!) due to the large quantities of expensive electrical energy needed for the process. * Two general rules:  

Metals and hydrogen (from positive ions), form at the negative cathode electrode. Non-metals (from negative ions), form at the positive anode electrode.

Raw materials for the electrolysis process:     

3+

2-

Bauxite ore of impure aluminium oxide [Al2O3 made up of Al and O ions] Carbon (graphite) for the electrodes. Cryolite reduces the melting point of the ore and saves energy, because the ions must be free to move to carry the current Electrolysis means using d.c. electrical energy to bring about chemical changes e.g. decomposition of a compound to form metal deposits or release gases. The electrical energy splits the compound! At the electrolyte connections called the anode electrode (+, attracts - ions) and the cathode electrode (-, attracts + ions). An electrolyte is a conducting melt or solution of freely moving ions which carry the charge of the electric current.

The redox details of the electrode processes:  

 

At the negative (-) cathode, reduction occurs (electron gain) when the positive aluminium ions are attracted to it. They gain three electrons to change to neutral Al atoms. 3+ o Al + 3e ==> Al At the positive (+) anode, oxidation takes place (electron loss) when the negative oxide ions are attracted to it. They lose two electrons forming neutral oxygen molecules. 2o 2O ==> O2 + 4e 2o or 2O - 4e ==> O2 Note: Reduction and Oxidation always go together! The overall electrolytic decomposition is ... o aluminium oxide => aluminium + oxygen o 2Al2O3 ==> 4Al + 3O2 o and is a very endothermic process, lots of electrical energy input!

   



GENERAL NOTE ON ELECTROLYSIS: Any molten or dissolved material in which the liquid contains free moving ions is called the electrolyte. + Ions are charged particles e.g. Na sodium ion, or Cl chloride ion, and their movement or flow constitutes an electric current, because a current is moving charged particles. What does the complete electrical circuit consist of? o There are two ion currents in the electrolyte flowing in opposite directions: 3+  positive cations e.g. Al attracted to the negative cathode electrode, 2 and negative anions e.g. O attracted to the positive anode electrode,  BUT remember no electrons flow in the electrolyte, only in the graphite or metal wiring! o The circuit of 'charge flow' is completed by the electrons moving around the external circuit e.g. copper wire or graphite electrode, from the positive to the negative electrode o This e flow from +ve to -ve electrode perhaps doesn't make sense until you look at the electrode reactions, electrons released at the +ve anode move round the external circuit to produce the electron rich negative cathode electrode. Electron balancing: In the above process it takes the removal of four electrons from two oxide ions to form one oxygen molecule and the gain of three electrons by each aluminium ion to form one aluminium atom. Therefore for every 12 electrons you get 3 oxygen molecules and 4 aluminium atoms formed.

The properties and uses of aluminium 

Aluminium can be made more resistant to corrosion by a process called anodising. Iron can be made more useful by mixing it with other substances to make various types of steel. Many metals can be given a coating of a different metal to protect them or to improve their appearance. o Aluminium is a reactive metal but it is resistant to corrosion. This is because aluminium reacts in air to form a layer of aluminium oxide which then protects the aluminium from further attack.  This is why it appears to be less reactive than its position in the reactivity series of metals would predict. o For some uses of aluminium it is desirable to increase artificially the thickness of the protective oxide layer in a process is called anodising.  This involves removing the oxide layer by treating the aluminium sheet with sodium hydroxide solution.  The aluminium is then placed in dilute sulphuric acid and is made the positive electrode (anode) used in the electrolysis of the acid.  Oxygen forms on the surface of the aluminium and reacts with the aluminium metal to form a thicker protective oxide layer.

o



Aluminium can be alloyed to make 'Duralumin' by adding copper (and smaller amounts of magnesium, silicon and iron), to make a stronger alloy used in aircraft components (low density = 'lighter'!), greenhouse and window frames (good anti-corrosion properties), overhead power lines (quite a good conductor and 'light'), but steel strands are included to make the 'line' stronger and poorly electrical conducting ceramic materials are used to insulate the wires from the pylons and the ground.  There is a note about structure of metal alloys on the metallic bonding page. Reactions of aluminium o Reaction with chlorine

o o

If dry chlorine gas Cl2 is passed over heated iron or aluminium, the chloride is produced. The experiment (shown above) should be done very carefully by the teacher in a fume cupboard.  2Al(s) + 3Cl2(g) ==> 2AlCl3(s)



o

o

o

o

o

The aluminium can burn intensely with a violet flame, white fumes of aluminium chloride sublime from the hot reacted aluminium and the white solid forms on the cold surface of the flask (its often discoloured yellow from the trace chlorides of copper or iron that may be formed).  2Fe(s) + 3Cl2(g) ==> 2FeCl3(s)  Aluminium chloride reacts exothermically as it is hydrolysed by water to give the metal hydroxide and fumes of hydrogen chloride, and so dry conditions are needed.  Aluminium chloride cannot be made in an anhydrous form from aqueous solution neutralisation. This is because on evaporation the compounds contain 'water of crystallisation'. On heating the hydrated salt it hydrolyses and decomposes into water, the oxide or hydroxide and fumes of hydrogen chloride.  Reaction of chloride with water:  With a little water it rapidly, and exothermically hydrolyses to form aluminium hydroxide and nasty fumes of hydrogen chloride gas.  AlCl3(s) + 3H2O(l) ==> Al(OH)3(s) + 3HCl(g)  However, if a large excess of water is rapidly added, a weakly acidic solution of aluminium chloride is formed, with the minimum of nasty fumes! 3+  AlCl3(s) + aq ==> Al (aq) + 3Cl (aq) 3+  or more correctly: AlCl3(s) + 6H2O(l) ==> [Al(H2O)6] (aq) + 3Cl (aq)  The solution is slightly acidic, because the hexa-aqa aluminium ion can donate a proton to a water molecule forming the oxonium ion. 3+ 2+ +  [Al(H2O)6] (aq) + H2O(l) [Al(H2O)5OH] (aq) + H3O (aq) The surface of aluminium goes white when strongly heated in air/oxygen to form white solid aluminium oxide. Theoretically its quite a reactive metal but an oxide layer is readily formed even at room temperature and this has quite an inhibiting effect on its reactivity. Even when scratched, the oxide layer rapidly reforms, which is why it appears to be less reactive than its position in the reactivity series of metals would predict but the oxide layer is so thin it is transparent, so aluminium surfaces look metallic and not a white matt surface.  aluminium + oxygen ==> aluminium oxide  4Al(s) + 3O2(g) ==> 2Al2O3(s)  Under 'normal circumstances' in the school laboratory aluminium has virtually no reaction with water, not even when heated in steam due to a protective aluminium oxide layer of Al2O3. (see above) The metal chromium behaves chemically in the same way, forming a protective layer of chromium(III) oxide, Cr2O3, and hence its anti-corrosion properties when used in stainless steels and chromium plating. Although this again illustrates the 'under-reactivity' of aluminium, the Thermit Reaction shows its rightful place in the reactivity series of metals. The Thermit reaction: However the true reactivity of aluminium can be spectacularly seen when its grey powder is mixed with brown iron(III) oxide powder. When the mixture is ignited with a magnesium fuse (needed because of the very high activation energy!), it burns very exothermically in a shower of sparks to leave a red hot blob of molten=>solid iron and white aluminium oxide powder. Note the high temperature of the magnesium fuse flame is so high, the oxide layer (to the delight of all pupils) fails to inhibit the displacement reaction! yippee!  aluminium + iron(III) oxide ==> iron + aluminium oxide  aluminium + iron(III) oxide ==> aluminium oxide + iron  2Al(s) + Fe2O3(s) ==> Al2O3(s) + 2Fe(s)  This is a typical displacement reaction by a more reactive metal displacing a less reactive metal from one of its compounds. Slow reaction with dilute hydrochloric acid to form the colourless soluble salt aluminium chloride and hydrogen gas.  aluminium + hydrochloric acid ==> aluminium chloride + hydrogen  2Al(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2(g) The reaction with dilute sulphuric acid is very slow to form colourless aluminium sulphate and hydrogen.  aluminium + sulphuric acid ==> aluminium sulphate + hydrogen  2Al(s) + 3H2SO4(aq) ==> Al2(SO4)3(aq) + 3H2(g) If the surface of aluminium is treated with less reactive metal salt, it is still possible to get a displacement reaction. Check this out by leaving a piece of aluminium foil in copper(II) sulphate solution and a patchy pink colour of copper metal slowly appears over many hours/days?  aluminium + copper(II) sulphate ==> aluminium sulphate + copper  2Al(s) + 3CuSO4(aq) ==> Al2(SO4)3(aq) + 3Cu(s) 2+ 3+  ionic redox equation: 2Al(s) + 3Cu (aq) ==> 2Al (aq) + 3Cu(s)

Amphoteric nature of aluminium hydroxide and acidity of the hexaaquaaluminium ion 





The addition of limited amounts of the bases sodium hydroxide or ammonia solution to an aluminium salt solution. 3+ o [Al(H2O)6] (aq) + 3OH (aq) ==> [Al(H2O)3(OH)3](s) + 3H2O(aq) o A white gelatinous precipitate of aluminium hydroxide is formed. 3+  Simplified equation: Al (aq) + 3OH (aq) ==> Al(OH)3(s) The further addition of excess sodium hydroxide or ammonia solution. o With excess ammonia there is no effect, but with excess sodium hydroxide the aluminium hydroxide dissolves to form a soluble aluminate complex anion - therefore exhibiting amphoteric behaviour. since the hydroxide will also dissolve in acids (paragraph below NaOH equation). 3o [Al(H2O)3(OH)3](s) + 3OH (aq) ==> *[Al(OH)6] (aq) + 3H2O(aq) 

Simplified equation: Al(OH)3(s) + 3OH (aq) ==> *[Al(OH)6]



*The products will be an equilibrium mixture including [Al(H2O)2(OH)4]-(aq) and [Al(H2O)(OH)5]2-(aq)

-

(aq)

too. You could write the equation in terms of forming these species too and any of the three possibilities should get you the marks. o To complete the 'amphoteric' picture of aluminium hydroxide we consider it dissolving in mineral acids to form typical salts e.g. aluminium chloride, aluminium nitrate and aluminium sulphate.  Al(OH)3(s) + 3HCl(aq) ==> AlCl3(aq) + 3H2O(l)  Al(OH)3(s) + 3HNO3(aq) ==> Al(NO3)3(aq) + 3H2O(l)  2Al(OH)3(s) + 3H2SO4(aq) ==> Al2(SO4)3(aq) + 6H2O(l) The addition of sodium carbonate solution to an aluminium salt solution. o Bubbles of carbon dioxide and a white gelatinous precipitate of aluminium hydroxide are formed. 3+ 2 2[Al(H2O)6] (aq) + 3CO3 (aq) ==> 2[Al(H2O)3(OH)3](s) + 3CO2(g) + 3H2O(aq)  There several equation 'permutations' to represent this quite complicated reaction, so I've just composed one that shows the formation of both observed products. Since sodium carbonate solution is alkaline you can legitimately write a hydroxide ppt. equation as for sodium hydroxide above but it doesn't show the formation of carbon dioxide.  You can write an equation to show the formation of carbon dioxide leaving a soluble cationic complex of aluminium in solution and this equation fits in well with the acid-base nature of this reaction. 3+ 2+  [Al(H2O)6] (aq) + CO3 (aq) ==> 2[Al(H2O)4(OH)2] (aq) + CO2(g) + 3H2O(aq)  This equation shows the hexaaquaaluminium ion acting as a Bronsted-Lowry acid donating two protons to the carbonate ion (B-L base) to form carbon dioxide and water.  This reaction shows why 'aluminium carbonate' 'Al2(CO3)3' cannot exist. The hydrated highly charged central metal ion is too acidic to co-exist with a carbonate ion. The same situation applies to 3+ 3+ the chromium(III) Cr and iron(III) Fe ions i.e. no chromium(III) carbonate or iron(III) carbonate exists. However with a lesser charged, lesser acidic ion, carbonates can exist, so there is an iron(II) carbonate FeCO3. o Aluminium salt solutions are slightly acidic for the same reasons as the carbonate reaction - namely the acidity of the hexaaquaaluminium ion i.e. a acting as a proton donor. 3+



3-

2+

+

o [Al(H2O)6] (aq) + H2O(l) [Al(H2O)5(OH)] (aq) + H3O (aq) The addition of excess sodium carbonate solution has no further effect. Sodium carbonate is too weak a base to effect the amphoteric nature of aluminium hydroxide and dissolve the aluminium hydroxide precipitate. o For strong alkalis like sodium hydroxide the whole sequence of each theoretical step of aluminium hydroxide precipitation and its subsequent dissolving in strong base-alkali is shown the series of diagrams below. o All are, for simplicity, treated as octahedral complexes of 6 ligands - either water H2O or hydroxide ion OH-. 3+ 2+ + o [Al(H2O)6] => [Al(OH)(H2O)5] => [Al(OH)2(H2O)4] => [Al(OH)3(H2O)3](s) precipitate 23o dissolving => [Al(OH)4(H2O)3] => [Al(OH)5(H2O)] => [Al(OH)6]

The sequence of aluminium hydroxide precipitate formation and its subsequent dissolving in excess strong alkali. Each step is essentially one of proton removal from each complex (from 3+ to 3-). -

1 2 3 4

From 1 to 7 happen as you add more alkali, increasing pH and the OH concentration, removing protons from the aluminium complex.

5 6 7 *

From 7 back to1 represents what happens when you add acid, decreasing pH, increasing H /H3O concentration and protonating the aluminium complex.

+

+

Aluminium compound reducing agents in organic chemistry  

    

Lithium tetrahydridoaluminate(III), LiAlH 4 (lithium tetrahydride) reduces aldehydes to primary alcohols and ketones to secondary alcohols. LiAlH4 is a more powerful reducing agent than NaBH4 and reacts violently with water, so the reaction must be carried out in an inert solvent like ethoxyethane ('ether'). The initial product is hydrolysed by dil. sulphuric acid. o aldehyde: RCHO + 2[H] ==> RCH2OH (R = H, alkyl or aryl) o ketone: R2C=O + 2[H] ==> R2CHOH (R = alkyl or aryl) LiAlH4 is a more powerful reducing agent than NaBH4, and in ether solvent, readily reduces carboxylic acids to primary alcohols. The reaction can be summarised as: o RCOOH + 4[H] ==> RCH2OH + H2O (R = H, alkyl or aryl) LiAlH4 is a more powerful reducing agent than NaBH4 and in ether solvent will reduce nitriles to primary aliphatic amines. o RC N + 4[H] ==> RCH2NH2 (R = H, alkyl or aryl) LiAlH4 is a more powerful reducing agent than NaBH4 and in ether solvent readily reduces nitro-aromatics to primary aromatic amines. o C6H5NO2 + 6[H] ==> C6H5NH2 + 2H2O methylnitrobenzenes would be reduced to methylphenylamine primary amines, i.e. o CH3C6H4NO2 + 6[H] ==> CH3C6H4NH2 + 2H2O as will any aromatic compound with a nitro group (-NO2) directly attached to a benzene ring.

Part 8. The p-block elements: 8.2 Group 4/14 carbon & silicon in particular

Group 4/14 Introduction

down group 4/14 ===> property\Zsymbol, name

6C

carbon

14Si

Silicon

32Ge Germanium

50Sn

Tin

82Pb

Lead

Period

2

3

4

5

6

Appearance (RTP)

soft black solid (graphite) / hard clear light coloured crystal (diamond)

black amorphous powder or bluegrey metalloid pure for semiconductors

silvery white brittle metal

soft silvery metal

soft grey dull-silvery metal

o

3547 sub

1410

937

232

328

o

4827 sub

2355

2830

2270

1740

2.3 (graphite)/3.5 (diamond)

2.3

5.3

5.8

11.4

-1

1086

786

762

709

716

-1

2350

1580

1540

1410

1450

-1

4610

3230

3300

2940

3080

4th IE/kJmol

-1

6220

4360

4390

2930

4080

5th IE/kJmol

-1

37800

16000

8950

7780

6700

atomic covalent or metallic radius/pm

77 (cov)

117 (cov)

139 (met)

158 (met)

175 (met)

Van der Waals radius/pm

170

210

na

190

200

melting pt./ C boiling pt./ C density/ gcm

-3

1st IE/ kJmol

2nd IE/kJmol 3rd IE/kJmol

M

2+

radius/pm

na

na

90

93

132

M

4+

radius/pm

na

na

na

74

84

2+

na

na

-0.25V

-0.14V

-0.13V

na

na

0.00V

+0.15V

+1.69V

electronegativity

2.55

1.90

2.01

1.96

2.33

simple electron config.

2,4

2,8,4

2,8,18,4

2,8,18,18,4

2,8,18,32,18,3

electron configuration

[He]2s 2p

principal oxidation states

e.g. -4 CH4, +2 CO, +4 CO2

El'de p'l M(s)/M El'de p'l M

2+

(aq)/M

(aq) 4+ (aq)

property\Zsymbol, name

  

2

6C

2

carbon

2

[Ne]3s 3p

2

+4, -4

14Si

Silicon

10

2

[Ar]3d 4s 4p +2, +4 32Ge Germanium

2

10

2

[Kr]4d 5s 5p

2

14

+2, +4

50Sn

Tin

10

2

[Xe]4f 5d 6s 6p +2, +4

82Pb

Generally speaking down a p block group the element becomes more metallic in chemical character. Carbon and silicon are essentially non-metals, germanium is a metalloid. Tin is basically metallic with a little non-metallic chemical character, lead is a metal.

Lead

2

CARBON - brief summary of a few points 

  



 





 

The structure of the element: o Non-metal existing as three allotropes covalently bonded. Diamond (tetrahedral bond network) and graphite (layers of connected hexagonal rings) have giant covalent structures Cn where n is an extremely large number, and a series of large molecules (3rd allotrope) called fullerenes e.g. C60. o Bonding details and diagrams of the allotropes of carbon. Physical properties: o o o mpt 3547 C; bpt 4827 C; very hard colourless/light coloured diamond (poor conductor) or dark softish/slippery crystals of graphite (moderate conductor of heat/electricity). Group, electron configuration (and oxidation states): 2 2 2 o Gp4; e.c. 2,4 or 1s 2s 2p ; (can be +2, but usually +4) e.g. o (+2) CO, (+4) CO2 and CCl4 etc. Reaction of element with oxygen: o Burns when heated in air to form carbon dioxide gas.  C(s) + O2(g) ==> CO2(g)  In limited air/oxygen, carbon monoxide would be formed too.  2C(s) + O2(g) ==> 2CO(g) Reaction of carbon dioxide with water: o Quite soluble to form a weakly acid solution of pH 4-5. So called carbonic acid, H2CO3, does not really exist, but the dissolved carbon dioxide reacts with water to form hydrogen/oxonium ions and hydrogencarbonate ions. The equilibrium is very much on the left - hence the fizz in 'fizzy drinks'! +  CO2(aq) + 2H2O(l) H3O (aq) + HCO3 (aq) Reaction of oxide with acids: o None, only acidic in acid-base behaviour. Reaction of oxide with bases/alkalis: o It is a weakly acidic oxide dissolving sodium hydroxide solution to form sodium carbonate. o CO2(g) + 2NaOH(aq) ==> Na2CO3(aq) + H2O(l) 2o ionic equation: CO2(g) + 2OH (aq) ==> CO3 (aq) + H2O(l) o With excess of carbon dioxide, sodium hydrogencarbonate is formed. o CO2(g) + Na2CO3(aq) + H2O(l) ==> 2NaHCO3(aq) 2o ionic equation: CO2(g) + CO3 (aq) + H2O(l) ==> 2HCO3 (aq) Reaction of element with chlorine: o None directly. o Tetrachloromethane (carbon tetrachloride) is made by fully chlorinating methane in a multi-stage reaction.  CH4(g) + 4Cl2(g) ==> CCl4(l) + 4HCl(g) Reaction of chloride with water: * o None. CCl4(l) cannot readily act as a Lewis acid and accept a lone pair from a water molecule at the polar CCl bond to start the hydrolysis process. *  In the case of SiCl4, 3d orbitals can be used to accept a lone pair from water, so providing a mechanistic route for hydrolysis to occur. (compare with silicon). Reaction of element with water: o No reaction with cold water but red hot carbon reacts with steam to form carbon monoxide and hydrogen.  C(s) + H2O(g) ==> CO(g) + H2(g) Other comments: o All of organic chemistry is based on the compounds of carbon except for the oxides and carbonates.

The structure of the elements carbon and silicon and their oxides The structure of the three allotropes of carbon (diamond, graphite and fullerenes), silicon and silicon dioxide (silica)   

It is possible for many atoms to link up to form a giant covalent structure or lattice. The atoms are usually non-metals. This produces a very strong 3-dimensional covalent bond network or lattice. This gives them significantly different properties from the small simple covalent molecules mentioned above.

DIAGRAMS Carbon-DIAMOND and silicon

  

 



      

  

 

In carbon dioxide, the smaller C atom can form a double bond with oxygen, but in silicon dioxide (silicon(IV) oxide, silica, quartz) the larger Si atom can only form single Si-O bonds. The result is SiO2 has a giant 3D covalent lattice structure or network in which each silicon atom forms 4 bonds to an oxygen in a 'tetrahedral' spatial arrangement. The structure is therefore held together by strong covalent bonds (not weak intermolecular forces) and so it is far more thermally stable giving a high melting point, and insoluble in any solvent, because solvation energies are much lower than covalent bond energies. Carbon and silicon are two elements which form giant covalent structures i.e. they are high melting and insoluble solids. Carbon (diamond) and silicon form networks based on tetrahedral arrangements of C-C or Si-Si bonds around each atom. Both are very hard substances because of the strong bonding, diamond is harder because the smaller C atoms give shorter stronger bonds. Both are poor conductors of electricity because the outer electrons are strongly held and localised between the two atoms of any bond. However, carbon in the form of graphite, forms hexagonal ring layers in which the three C-C single bonds are supplemented by delocalised electron bonding from the 4th out electron of carbon. This makes graphite a moderately good electrical conductor as the electrons can move freely through a layer. The layers are held together by weak inter-molecular forces and easily slip over each other making graphite a 'slippery' brittle solid. But as a giant covalent structure it is still high melting and insoluble. Relatively recently (and another case of serendipity!) a 3rd form of carbon has been discovered in the form of the 'ball shaped' fullerenes. TYPICAL PROPERTIES of GIANT COVALENT STRUCTURES This type of giant covalent structure is thermally very stable and has a very high melting and boiling points because of the strong covalent bond network (3D or 2D in the case of graphite below). A relatively large amount of energy is needed to melt or boil giant covalent structures. Energy changes for the physical changes of state of melting and boiling for a range of differently bonded substances are compared in a section of the Energetics Notes. They are usually poor conductors of electricity because the electrons are not usually free to move as they can in metallic structures. Also because of the strength of the bonding in all directions in the structure, they are often very hard, strong and will not dissolve in solvents like water. The bonding network is too strong to allow the atoms to become surrounded by solvent molecules Silicon dioxide [silicon(IV) oxide, silica, SiO2] has a similar 3D structure and properties to carbon (diamond) e.g. very hard, very high melting point and virtually insoluble in anything! o This contrasts sharply with the structure and properties of the gas carbon dioxide which is a small covalent molecule. o With only weak intermolecular forces between the O=C=O molecules it o consequently has a very low melting/boiling point (actually it sublimes at -78 C). Carbon dioxide readily dissolves in solvents such as water and organic polar solvents. δ+ do Carbon dioxide has two polar bonds, C =O , but because of the linearity of the molecule the two permanent dipoles cancel out to give overall a non-polar molecule o Note that carbon + oxygen, instead of forming a 3D network of O-C-O single bonds, with the smaller carbon atom, it is energetically more favourable to form C=O double bonds and thus forming a small triatomic molecule. The hardness of diamond enables it to be used as the 'leading edge' on cutting tools. Energy changes for the physical changes of state of melting and boiling for a range of differently bonded substances is given in a section of the Energetics Notes. Many naturally occurring minerals are based on -O-X-O- linked 3D structures where X is often silicon (Si) and aluminium (Al), three of the most abundant elements in the earth's crust. o Silicon dioxide is found as quartz in granite (igneous rock) and is the main component in sandstone - which is a sedimentary rock formed the compressed erosion products of igneous rocks. o Many some minerals that are hard wearing, rare and attractive when polished, hold great value as gemstones. Carbon also occurs in the form of graphite. The carbon atoms form joined hexagonal rings forming layers 1 atom thick. There are three strong covalent bonds per carbon (3 C-C bonds in a planar

SILICA silicon dioxide

GRAPHITE

   



   



arrangement from 3 of its 4 outer electrons), BUT, the fourth outer electron is 'delocalised' or shared between the carbon atoms to form the equivalent of a 4th bond per carbon atom (this situation requires advanced level concepts to fully explain, and this bonding situation also occurs in fullerenes described below, and in aromatic compounds you deal with at advanced level). The layers are only held together by weak intermolecular forces shown by the dotted lines NOT by strong covalent bonds. Like diamond and silica (above) the large molecules of the layer ensure graphite has typically very high melting point because of the strong 2D bonding network (note: NOT 3D network).. Graphite will not dissolve in solvents because of the strong bonding BUT there are two crucial differences compared to diamond ... o Electrons, from the 'shared bond', can move freely through each layer, so graphite is a conductor like a metal (diamond is an electrical insulator and a poor heat conductor). Graphite is used in electrical contacts e.g. electrodes in electrolysis. o The weak forces enable the layers to slip over each other so where as diamond is hard material graphite is a 'soft' crystal, it feels slippery. Graphite is used as a lubricant. These two different characteristics described above are put to a common use with the electrical contacts in electric motors and dynamos. These contacts (called brushes) are made of graphite sprung onto the spinning brass contacts of the armature. The graphite brushes provide good electrical contact and are self-lubricating as the carbon layers slide over each other. A 3rd form of carbon are fullerenes or 'bucky balls'! It consists of hexagonal rings like graphite and alternating pentagonal rings to allow curvature of the surface. Buckminster Fullerene C60 is shown and the bonds form a pattern like a soccer ball. Others are oval shaped like a rugby ball. It is a black solid insoluble in water. They are NOT considered giant covalent structures and are classed as simple molecules. They do dissolve in organic solvents giving coloured solutions (e.g. deep red in petrol hydrocarbons, and although solid, their melting points are not that high. They are mentioned here to illustrate the different forms of carbon AND they can be made into continuous tubes to form very strong fibres of 'pipe like' molecules called 'nanotubes'. These 'molecular size' particles behave quite differently to a bulk carbon material like graphite. Uses of Nanotubes: o They can be used as semiconductors in electrical circuits. o They act as a component of industrial catalysts for certain reactions whose economic efficiency is of great importance (time = money in business!).  The catalyst can be attached to the nanotubes which have a huge surface are per mass of catalyst 'bed'.  They large surface combined with the catalyst ensure two rates of reaction factors work in harmony to increase the speed of the industrial reaction. o Nanotube fibres are very strong and so they are used in 'composite materials' e.g. reinforcing graphite in carbon fibre tennis rackets. o Nanotubes can 'cage' other molecules and can be used as a means of delivering drugs in controlled way to the body.

FULLERENES

-

SILICON - brief summary of a few points  

The structure of the element: o Non-metal existing as a giant covalent lattice, Sin, , where n is an extremely large number, held together by tetrahedrally arranged Si-Si bonds. Physical properties:

o

o

    

  

o

Hard high melting solid; mpt 1410 C; bpt 2355 C; poor conductor of heat/electricity, but with other elements added, conducts better, hence use in microchips. Group, electron configuration (and oxidation states): 2 2 6 2 2 o Gp4; e.c. 2,8,4 or 1s 2s 2p 3s 3p ; (+4 only) e.g. SiO2 and SiCl4 etc. Reaction of element with oxygen: o Reacts when strongly heated in air to form silicon dioxide (silica, silicon(IV) oxide).  Si(s) + O2(g) ==> SiO2(g) Reaction of oxide with water: o None and insoluble. Reaction of oxide with acids: o None, only acidic in nature. Reaction of oxide with bases/alkalis: o It is a weakly acidic oxide dissolving very slowly in hot concentrated sodium hydroxide solution to form sodium silicate. o SiO2(s) + 2NaOH(aq) ==> Na2SiO3(aq) + H2O(l) 2o or simplified ionic equation: SiO2(s) + 2OH (aq) ==> SiO3 (aq) + H2O(l) Reaction of element with chlorine: o On heating in chlorine forms the covalent liquid silicon tetrachloride.  Si(s) + 2Cl2(g) ==> SiCl4(l) Reaction of chloride with water: o Hydrolyses to form gelatinous hydrated silicon oxide and hydrochloric acid.  SiCl4(l) + 2H2O(l) ==> SiO2(s) + 4HCl(aq) Reaction of element with water: o None

Shapes and bond angles of some molecules and ions of carbon and silicon

With 4 bond pairs of bonding electrons and no lone pairs you get a TETRAHEDRAL shape: e.g. methane CH 4, o + silicon hydride SiH4 with H-X-H bond angle of 109 and similarly ions like the ammonium ion NH4 . Note: No lone pair, no extra repulsion, no reduction in angle, therefore perfect tetrahedral angle (Q = H, X = C, Si, Ge etc. in group 4)

Similarly with 4 bond pairs, again a TETRAHEDRAL shape: e.g. tetrachloromethane CCl4 or SiCl4 with exact Cl-C-Cl o and Cl-Si-Cl bond angles of 109

2-

o

Carbonate ion, CO3 is trigonal planar in shape with a O-C-O bond angle of 120 because of three groups of bonding electrons and no lone pairs of electrons.

The shape is deduced below using dot and cross diagrams and VSEPR theory and illustrated below.

valence bond dot and cross diagram

The chemistry of carbonates Carbonates and hydrogencarbonates of Groups 1-2 are dealt with in s-block notes sections 7.9 to 7.12 and Notes on limestone - calcium carbonate

Semi-metals or 'metalloids'

Gp Gp Gp Gp 3/13 4/14 5/15 6/16

B

Al

C

Si

Ga Ge

N

P

As

BASIC IDEA: A narrow diagonal band of elements can show both metallic and nonmetallic physical or chemical properties and are referred to as 'semi-metals' or 'metalloids'. Although most tend to be nearer being a metal or a non-metal, they do represent the point elements change from metal to non-metal as you move from left to right across the Periodic Table BUT please read the notes below carefully!

O

To me boron, B, is clearly a non-metal, showing no real metallic character and I'm not sure why it is sometimes shown as a semi-metal on some periodic tables? and is very different in character to metallic aluminium below it in the same group. Boron's oxide is acidic only, and the solid element consists of a non-conducting giant covalent structure, both classic non-metallic properties. Carbon, C, is also clearly a non-metal, its oxide is acidic and in the form of diamond, it is a non-electrical conducting 3D giant covalent structure. However, in the form of graphite, it has a layered 2D giant covalent structure that does allow electricity to conduct through the layers.

S

Physically and chemically aluminium, Al, is very much a metal, but the oxide/hydroxide reacts with both acids (metallic) and alkalis (acidic) to form salts showing dual character. Silicon is mainly non-metallic character e.g. the oxide is acidic but, although the solid element has a giant covalent structure, it shows slight electrical conducting properties (semi-conductor), especially when doped with other elements and used in computer chip technology. To me, neither are true semi-metals.

Se

Germanium, Ge, is considered as a true semi-metal (metalloid). Like silicon, germanium is a semi-conductor and used in electronic technology. Its oxide/hydroxide react with both acids/alkalis showing dual metal/non-metal character. Arsenic, As, is also a true metalloid with oxides/hydroxides that react both with acids/ and alkalis to form salts and the element

exists in two allotropic* crystalline forms. One form is less dense, non-conducting and covalent in structure (non-metal) and the other is more dense and weakly electrical conducting (metallic) and used in transistors. Selenium, Se, is also a semi-conductor with metallic and non-metallic properties and is used in photo-electric cells (solar cells) and xerography (photocopying). (*Allotropes are different physical forms of the same element in the same physical state.)

In

Sn

Sb

Te

Arsenic, As, (like antimony in the same group), is also a true semi-metal (metalloid) with oxides/hydroxides that react both with acids/ and alkalis to form salts and the element exists in two allotropic* crystalline forms (non-metallic and metallic). Tellurium, Te, is also a semiconductor with metallic and non-metallic properties. Both As and Te are used in electronic devices.

Part 8. The p-block elements: 8.3 Group 5/15 nitrogen in particular

Group 5/15 Introduction

down group 5/15 ===> property\Zsymbol, name

7N

Nitrogen

15P

Phosphorus

33As

Arsenic

51Sb

Antimony

83Bi

Bismuth

Period

2

3

4

5

6

Appearance (RTP)

colourless gas

white/red solid allotropes

grey solid (also yellow/black allotropes)

grey metalloid solid (also yellow allotrope)

silver-white brittle metal

o

-210

44

sublimes

631

272

boiling pt./ C

o

-1.96

280

616?

1635

1560

-3

2NO(g) o and the nitrogen(II) oxide rapidly reacts in air to form nitrogen(IV) oxide (nitrogen dioxide).  2NO(g) + O2(g) ==> 2NO2(g) o The theoretical highest oxide is N2O5 nitrogen(V) oxide (nitrogen pentoxide) and does exist. Reaction of nitrogen oxides with water: o Nitrogen(IV) oxide dissolves to form an acidic solution of weak nitrous acid and strong nitric acid.  2NO2(g) + H2O(l) ==> HNO2(aq) + HNO3(aq) +  or 2NO2(g) + 2H2O(l) ==> HNO2(aq) + H3O (aq) + NO3 (aq) o NO and N2O are neutral oxides but nitrogen(V) oxide is strongly acidic and dissolves to form nitric acid.  N2O5(s) + H2O(l) ==> 2HNO3(aq) +  or N2O5(s) + 3H2O(l) ==> 2H3O (aq) + 2NO3 (aq) Reaction of nitrogen oxides with acids: o None, only acidic (N2O3 (very unstable), NO2 and N2O5) or neutral (N2O and NO), in nature. Reaction of nitrogen oxides with bases/alkalis: o Nitrogen(IV) oxide or nitrogen dioxide forms sodium nitrite and sodium nitrate with sodium hydroxide solution. o 2NO2(g) + 2NaOH(aq) ==> NaNO2(aq) + NaNO3(aq) + H2O(l) o ionic equation: 2NO2(g) + 2OH (aq) ==> NO2 (aq) + NO3 (aq) + H2O(l) o As well as being a neutralisation reaction, it is also a redox reaction, the oxidation states of oxygen (-2) and hydrogen (+1) do not change BUT the oxidation state of nitrogen changes from two at (+4) to one at (+3) and one at (+5). The simultaneous change of an element into an lower and upper oxidation sate is sometimes called disproportionation. Reaction of nitrogen with chlorine: o None, but the unstable yellow oily liquid chloride can be made indirectly. Reaction of chloride with water: o Slowly hydrolyses to form weak nitrous acid and strong hydrochloric acid.  NCl3(l) + 2H2O(l) ==> HNO2(aq) + 3HCl(aq) +  or NCl3(l) + 2H2O(l) ==> HNO2(aq) + 3H (aq) + 3Cl (aq) +  or NCl3(l) + 5H2O(l) ==> HNO2(aq) + 3H3O (aq) + 3Cl (aq) Reaction of nitrogen with water: o Slightly soluble but NO reaction. Other comments: o An essential element for plants, hence need for nitrogen compounds in compost and artificial fertilisers (NPK bags!).

PHOSPHORUS - brief summary of a few points

   





 





The structure phosphorus: o Two solid allotropes (red and white) consisting of P4 molecules, also a polymer form. Physical properties of phosphorus: o o o Colourless gas; mpt 44 C; bpt 280 C; poor conductor of heat/electricity. Group, electron configuration (and oxidation states): 2 2 6 2 3 o Gp5; e.c. 2,8,5 or 1s 2s 2p 3s 3p ; Variety of oxidation states from -3 to +5 e.g. o PH3 (-3), P4O6 (+3), P4O10, PCl5 and H3PO4 (+5). Reaction of phosphorus with oxygen: o With limited air/oxygen, on heating the phosphorus, the covalent white solid phosphorus(III) oxide is formed.  P4(s) + 3O2(g) ==> P4O6(s) o With excess air/oxygen, on heating the phosphorus, the covalent white solid phosphorus(V) oxide is formed.  P4(s) + 5O2(g) ==> P4O10(s) Reaction of phosphorus oxides with water: Both oxides dissolve in water to form acidic solutions. o Phosphoric(III) oxide forms phosphoric(III) acid.  P4O6(s) + 6H2O(l) ==> 4H3PO3(aq) o Phosphoric(III) oxide forms phosphoric(V) acid.  P4O10(s) + 6H2O(l) ==> 4H3PO4(aq) Reaction of phosphorus with chlorine: o With limited chlorine, on heating the phosphorus, the covalent liquid phosphorus(III) chloride is formed.  P4(s) + 3Cl2(g) ==> 4PCl3(l) * o With excess chlorine, on heating the phosphorus, the ionic solid phosphorus(V) chloride is formed.  P4(s) + 5Cl2(g) ==> 4PCl5(s) *  PCl5 is a bit unusual for an 'expected covalent' liquid chloride. +  It is an ionic solid with the structure [PCl4] [PCl6]  Hence its melting point is much greater than the liquid phosphorus(III) chloride, where the molecules are only held together by the inter-molecular forces.  However, gaseous phosphorus(V) chloride consists of PCl5 covalent molecules. Reaction of phosphorus oxides with acids: o None, only acidic in nature. Reaction of phosphorus oxides with bases/alkalis: o Both oxides dissolve in alkalis to form a whole series of phosphate(III) and phosphate(V) salts. o with excess strong bases like sodium hydroxide, the simplified equations are:  P4O6(s) + 12NaOH(aq) ==> 4Na3PO3(aq) + 6H2O(l) sodium phosphate(III) formed from phosphorus(III) oxide 3 ionic equation: P4O6(s) + 12OH (aq) ==> 4PO3 (aq) + 6H2O(l)  P4O10(s) + 12NaOH(aq) ==> 4Na3PO4(aq) + 6H2O(l) sodium phosphate(V) formed from phosphorus(V) oxide 3 ionic equation: P4O10(s) + 12OH (aq) ==> 4PO4 (aq) + 6H2O(l)  If the empirical formulae P2O3 and P2O5 are used, just halve all the balancing numbers.  Other than using excess sodium hydroxide, other salts can be formed.  e.g. P4O10(s) + 4NaOH(aq) + 2H2O(l) ==> 4NaH2PO4(aq) sodium dihydrogen phosphate(V)  or P4O10(s) + 8NaOH(aq) ==> 4Na2HPO4(aq) + 2H2O(l) disodium hydrogen phosphate(V) o Reaction of phosphorus chlorides with water: o Phosphorus(III) chloride hydrolyses rapidly and exothermically to form phosphoric(III) acid.  PCl3(l) + 3H2O(l) ==> H3PO3(aq) + 3HCl(aq) o Phosphorus(V) chloride initially hydrolyses to form phosphorus oxychloride? and hydrochloric acid.  PCl5(s) + H2O(l) ==> POCl3(aq) + 2HCl(aq)  Then on boiling the aqueous solution, phosphoric(V) acid is formed and more hydrochloric acid.  POCl3(aq) + 3H2O(l) ==> H3PO4(aq) + 3HCl(aq)  overall: PCl5(s) + 4H2O(l) ==> H3PO4(aq) + 5HCl(aq) Reaction of element with water: o None.

The shapes of some molecules and ions of nitrogen and phosphorus

electrons: three bond pairs and one lone pair, PYRAMIDAL or TRIGONAL PYRAMID shape: e.g. ammonia NH3 with o o bond angle of approximately 109 . Note: the exact H-N-H angle is 107 due to the extra repulsion of one lone pair (for H-X-H angles: NH3 > H2O and < CH4).

electrons: three bond pairs and one lone pair, PYRAMIDAL or TRIGONAL PYRAMID shape. e.g. nitrogen trifluoride/trichloride, NCl3, or phosphorus(III) fluoride/chloride (phosphorus trifluoride/trichloride), PF3/PCl3, with bond o + angles Q-X-Q of approximately 109 and similarly with ions like the oxonium ion H3O (Q = F, Cl etc. X = N, P etc.)

electrons: 5 bond pairs, TRIGONAL BIPYRAMID shape: e.g. phosphorus(V) fluoride (phosphorus pentafluoride) PF5, o o o gaseous phosphorus(V) chloride, PCl5, with bond angles 90 and 180 based on the vertical Q-X-Q bond and 120 based on the central trigonal planar arrangement. Note that solid PCl5 has an ionic structure and is not a trigonal + bipyramid molecule - a tetrahedral [PCl4] ion and an octahedral [PCl6] ion.

H3N:=>BF3 Boron trifluoride (3 bonding pairs, 6 outer electrons) acts as a lone pair acceptor (Lewis acid) and ammonia (3 bond pairs) and lone pair which enables it to act as a Lewis base - a an electron pair donor. It donates the lone pair to the 4th 'vacant' boron orbital to form a sort of 'adduct' compound. Its shape is essentially the same as ethane, a o sort of double tetrahedral with H-N-H, N-B-F and F-B-F bond angles of ~109 .

Nitrogen(IV) oxide, NO2 (nitrogen dioxide) is bent o shaped (angular), O-N-O bond angle ~120 because of two bonding groups of bonding electrons and a single lone electron in the same plane as the bonding pairs of electrons. -

The nitrate(III) ion, NO2 (nitrite ion) is bent shaped o (angular), O-N-O bond angle ~120 due to two groups of bonding electrons and one lone pair of electrons. -

The nitrate(V) ion, NO3 (nitrate ion) is trigonal planar, o O-N-O bond angle 120 due to three bonding groups of electrons and no lone pairs of electrons. +

The nitronium ion, NO2 , is linear, O-N-O bond angle o of 180 because there are two groups of bonding electrons and no lone pairs of electrons (you easily see this from the NO2 neutral molecule diagram below). The shapes are deduced below using dot and cross diagrams and VSEPR theory and illustrated in the valence bond dot and cross diagrams below.

Ammonia can act as an electron pair donor ligand in transition metal ion complexes e.g. tetraamminedichlorochromium(III) complex ion - cis/trans isomers (Z/E isomers)

With excess aqueous ammonia a pale blue hexa-ammine complex is formed with hexaaquanickel(II) ions [Ni(H2O)6]

2+ (aq)

+ 6NH3(aq)

[Ni(NH3)6]

2+ (aq)

+ 6H2O(l)

and also excess aqueous ammonia a pale blue hexa-ammine complex is formed with aqueous copper(II) ions [Cu(H2O)6]

2+ (aq)

or [Cu(H2O)4]

+ 4NH3(aq)

2+ (aq)

+ 4NH3(aq)

[Cu(NH3)4(H2O)2] [Cu(NH3)4]

2+

2+ (aq)

(aq)

+ 4H2O(l)

+ 4H2O(l)

The Synthesis of ammonia - The Haber Process 

Ammonia gas is synthesised in the chemical industry by reacting nitrogen gas with hydrogen gas in what is known as the Haber-Bosch Process, named after two highly inventive and subsequently famous chemists.

 

    

 

   

The nitrogen is obtained from liquified air (80% N2). Air is cooled and compressed under high pressure to form liquid air (liquefaction). The liquid air is fractionally distilled at low temperature to separate oxygen (used in welding, hospitals etc.), nitrogen (for making ammonia), Noble Gases e.g. argon for light bulbs, helium for balloons). The hydrogen is made by reacting methane (natural gas) and water or from cracking hydrocarbons (both reactions are done at high temperature with a catalyst). o CH4 + H2O ==> 3H2 + CO o e.g. C8H18 ==> C8H16 + H2 The synthesis equation for this reversible reaction is ... o N2(g) + 3H2(g) 2NH3(g) .. which means an equilibrium will form, so there is no chance of 100% yield even if you use, as you actually do, the theoretical reactant ratio of nitrogen : hydrogen of 1 : 3 In forming ammonia 92kJ of heat energy is given out (i.e. exothermic, 46kJ of heat released per mole of ammonia formed). Four moles of 'reactant' gas form two moles of 'product' gas, so there is net decrease in gas molecules on forming ammonia. So applying the Le Chatelier equilibrium rules, the formation of ammonia is favoured by ... o (a) Using high pressure because you are going from 4 to 2 gas molecules (the high pressure also speeds up the reaction because it effectively increases the concentration of the gas molecules), but higher pressure means more dangerous and more costly engineering. o (b) Carrying out the reaction at a low temperature, because it is an exothermic reaction favoured by low temperature, but this may produce too slow a rate of reaction, o So, the idea is to use a set of optimum conditions to get the most efficient yield of ammonia and this involves getting a low % yield (e.g. 8% conversion) but fast. Described below are the conditions to give the most economic production of ammonia. o these arguments make the point that the yield* of an equilibrium reaction depends on the conditions used.  * The word 'yield' means how much product you get compared to the theoretical maximum possible if the reaction goes 100%.  For more on chemical economics see Extra Industrial Chemistry page. In industry pressures of 200 - 300 times normal atmospheric pressure are used in line with the theory. Theoretically a low temperature would give a high yield of ammonia BUT ... o Nitrogen is very stable molecule and not very reactive i.e. chemically inert, so the rate of reaction is too slow at low temperatures. o o To speed up the reaction an iron catalyst is used as well as a higher temperature (e.g. 400-450 C). o The higher temperature is an economic compromise, i.e. it is more economic to get a low yield fast, than a high yield slowly! o Note: a catalyst does NOT affect the yield of a reaction, i.e. the equilibrium position BUT you do get there faster! At the end of the process, when the gases emerge from the iron catalyst reaction chamber, the gas mixture is cooled under high pressure, when only the ammonia liquefies and is so can be removed and stored in cylinders. Any unreacted nitrogen or hydrogen (NOT liquified), is recycled back through the reactor chamber, nothing o o o is wasted! [nitrogen (-196 C) and hydrogen (-252 C) have much lower boiling points than ammonia (-33 C). Boiling points increase with pressure, but these normal atmospheric pressure values offer a fair comparison]. To sum up: A low % yield of ammonia is produced quickly at moderate temperatures and pressure, and is more economic than getting a higher % equilibrium yield of ammonia at a more costly high pressure and a slower lower temperature reaction. AND there are some more general notes on Chemical Economics on the Industrial Chemistry page.

The Uses of Ammonia and derived compounds Ammonia is used to manufacture nitric acid      

Ammonia is oxidised with oxygen from air using a hot platinum catalyst to form nitrogen monoxide and water. 4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(g) The gas is cooled and reacted with more oxygen to form nitrogen dioxide. 2NO(g) + O2(g) ==> 2NO2(g) This is reacted with more oxygen and water to form nitric acid. 4NO2(g)+ O2(g) + 2H2O(l) ==> 4HNO3(aq)

 

Nitric acid is used to make nitro-aromatic compounds from which dyes are made. It is also used in the manufacture of artificial nitrogenous fertilisers (like ammonium nitrate, see below). Ammonia is used to manufacture 'artificial' nitrogenous fertilisers

    









Ammonia is a pungent smelling alkaline gas that is very soluble in water. The gas or solution turns litmus or universal indicator blue because it is a soluble weak base or weak alkali and is neutralised by acids to form salts. Ammonium salts are used as 'artificial' or 'synthesised' fertilisers i.e. nitrogenous fertilisers 'man-made' in a chemical works, and used as an alternative to natural manure or compost etc. The fertiliser salts are made by neutralising ammonia solution with the appropriate acid. The resulting solution is heated, evaporating the water to crystallise the salt e.g. o ammonia + sulphuric acid ==> ammonium sulphate o 2NH3(aq) + H2SO4(aq) ==> (NH4)2SO4(aq) o AND o ammonia + nitric acid ==> ammonium nitrate o NH3(aq) + HNO3(aq) ==> NH4NO3(aq) The salt Ammonium chloride is used in zinc-carbon dry cell batteries. The slightly acid paste, made from the salt, slowly reacts with the zinc to provide the electrical energy from the chemical reaction. o ammonia + hydrochloric acid ==> ammonium chloride o NH3(aq) + HCl(aq) ==> NH4Cl(aq)  or NH4OH(aq) + HCl(aq) ==> NH4Cl(aq) + H2O(l) If ammonium salts are mixed with sodium hydroxide solution, free ammonia is formed (detected by smell and damp red litmus turning blue). o e.g. ammonium chloride + sodium hydroxide ==> sodium chloride + water + ammonia o NH4Cl + NaOH ==> NaCl + H2O + NH3 Ammonium sulphate or nitrate salts are widely used as 'artificial or synthetic fertilisers (preparation reactions above). There are several advantages to using artificial fertilisers in the absence of sufficient manure-silage etc. e.g. relatively cheap mass production, easily used to make poor soils fertile or quickly enrich multi-cropped fields. Artificial fertilisers are important to agriculture and used on fields to increase crop yields but they should be applied in a balanced manner. o Fertilisers usually contain compounds of three essential elements for healthy and productive plant growth to increase crop yield. They replace nutrient minerals used by a previous crop or enriches poor soil and more nitrogen gets converted into plant protein.  Nitrogen (N) e.g. from ammonium or nitrate salts like ammonium sulphate, ammonium sulphate or ammonium phosphate or urea (e.g. look for the N in the formula of ammonium salts)  Phosphorus (P) e.g. from potassium phosphate or ammonium phosphate  Potassium (K) e.g. from potassium phosphate, potassium sulphate.  The fertiliser is marked with an 'NPK' value, i.e. the nitrogen : phosphorus : potassium ratio o Fertilisers must be soluble in water to be taken in by plant roots. Problems with using 'artificial' fertilisers

   





Overuse of ammonia fertilisers on fields can cause major environmental problems as well as being uneconomic. Ammonium salts are water soluble and get washed into the groundwater, rivers and streams by rain contaminating them with ammonium ions and nitrate ions. This contamination causes several problems. Excess fertilisers in streams and rivers cause eutrophication. o Overuse of fertilisers results in appreciable amounts of them dissolving in rain water. o This increases levels of nitrate or phosphate in rivers and lakes. o This causes 'algal bloom' i.e. too much rapid growth of water plants on the surface where the sunlight is the strongest. o This prevents light from reaching plants lower in the water. o These lower plants decay and the active aerobic bacteria use up any dissolved oxygen. o This means any microorganisms or higher life forms relying on oxygen cannot respire. o All the eco-cycles are affected and fish and other respiring aquatic animals die. o The river or stream becomes 'dead' below the surface as all the food webs are disrupted. Nitrates are potentially carcinogenic (cancer or tumour forming). o The presence in drinking water is a health hazard. o Rivers and lakes can be used as initial sources for domestic water supply. o You cannot easily remove the nitrate from the water, it costs too much! o So levels of nitrate are carefully monitored in our water supply. Cost - The hydrogen for the Haber Process for manufacturing ammonia is usually obtained from hydrocarbon sources e.g. methane gas. Therefore, as oil becomes more scarce, the cost of producing 'artificial' fertilisers will increase.

The Nitrogen Cycle for the gaseous element N2(g)  





Nitrogen is an extremely important element for all plant or animal life! It is found in important molecules such as amino acids, which are combined to form proteins. Protein is used everywhere in living organisms from muscle structure in animals to enzymes in plants/animals. Nitrogen from the atmosphere: o Nitrifying bacteria, e.g. in the root nodules of certain plants like peas/beans (the legumes), can directly convert atmospheric nitrogen into nitrogen compounds in plants e.g. nitrogen => ammonia => nitrates which plants can absorb.  However, most plants can't do this conversion from nitrogen => ammonia, though they can all absorb nitrates, so the 'conversion' or 'fixing' ability might be introduced into other plant species by genetic engineering. o The nitrogen from air is converted into ammonia in the chemical industry, and from this artificial fertilisers are manufactured to add to nutrient deficient soils. However, some of the fertiliser is washed out of the soil and can cause pollution. o The energy of lightning causes nitrogen and oxygen to combine and form nitrogen oxides which dissolve in rain that falls on the soil adding to its nitrogen content. 1. N2(g) + O2(g) ==> 2NO(g), then 2. then 2NO(g) + O2(g) ==> 2NO2(g) 3. NO2(g) + water ==> nitrates(aq) in rain/soil 4. Incidentally, reactions 1. and 2. can also happen in a car engine, and NO2 is acidic and adds to the polluting acidity of rain as well as providing nutrients for plants! Nitrogen recycling apart from the atmosphere: o Nitrogen compounds, e.g. protein formed in plants or animals, are consumed by animals higher up the food chain and then bacterial and fungal decomposers break down animal waste and dead plants/animals to release nitrogen nutrient compounds into the soil (e.g. in manure/compost) which can then be re-taken up by plants. Nitrogen returned to the atmosphere: o However, denitrifying bacteria will break down proteins completely and release nitrogen gas into the atmosphere.

Some nitrogen oxides chemistry  

 





The equilibria between oxygen O2, nitrogen (II) oxide NO, nitrogen(IV) oxide NO2 and its dimer N2O4. NO2 can be made from the irreversible thermal decomposition of lead(II) nitrate in a pyrex boiling tube connected 3 to a 100 cm gas syringe in a fume cupboard. o lead(II) nitrate ==> lead(II) oxide + nitrogen(IV) oxide + oxygen o 2Pb(NO3)2(s) ==> 2PbO(s) + 4NO2(g) + O2(g) -1 (a) 2NO2(g, brown) 2NO(g, colourless) + O2(g, colourless) (ΔH = +113 kJ mol ) o o (a) The temperature effect can be observed by strongly heating the gases in the pyrex tube above 400 C. -1 (b) 2NO2(g, brown) N2O4(g, colourless) (ΔH = -58 kJ mol ) o o (b) The temperature effect can be observed by cooling and warming below 100 C. o (b) The pressure effect can be observed by sealing the cool gases in the gas syringe and compressing and decompressing it. Temperature and energy change (ΔH) o (a) Increases in temperature favours the endothermic decomposition of NO 2 to NO and O2, so at high temperatures the brown colour fades. o (b) Decrease in temperature favours the exothermic formation of the dimer N 2O4 from NO2, so the brown colour fades on cooling the gas mixture. Gas pressure change (ΔV) o (a) Increase in pressure favours the LHS, more NO2, because 2 mol gas 1 mol gas, so the mixture would get lighter in colour.  (b) This can be demonstrated by compressing/decompressing the gas mixture in the syringe to see the brown colour intensity increase/decrease.  In fact you can even see the dynamic equilibrium 'kinetics' in operation here. There is a time lag of about 1-2 seconds before the new equilibrium position is established as the 'imposed' colour intensity change becomes constant. Concentration change o (a) Theoretically an increase in O2 would lead to decrease in NO and increase in NO 2, so the mixture would get darker. o (b) Increase in NO2 would increase N2O4, but overall the colour would still be darker because not all of the 'extra' NO2 can be converted to maintain the equilibrium. The formation of nitrogen(II) oxide at high temperature e.g. in a car engine -1 o N2(g) + O2(g) 2NO(g) (ΔH = +181 kJ mol ) o Temperature and energy change (ΔH)  Increase in temperature favours the endothermic formation of NO.  This reaction does not happen at room temperature but is formed at the high temperatures in car engines.  Unfortunately when released through the car exhaust, it cools to normal temperatures when NO irreversibly reacts with oxygen in air to form nitrogen(IV) oxide, NO 2,  2NO(g) + O2(g) ==> 2NO2(g)  Nitrogen dioxide is acidic, a lung irritant and a reactive free radical molecule involved in the chemistry of photochemical smog not good! o Its concentration in car exhaust gases can be reduced, along with that of carbon monoxide, by using a catalytic converter. o Using platinum, and other transition metal, based catalysts, the following reaction can be made to take place producing harmless nitrogen and carbon dioxide.  2NO(g) + 2CO(g) ==> N2(g) + 2CO2(g)

Some acid-base chemistry of ammonia 

Definition and examples of WEAK BASES o A weak base is only weakly or partially ionised in water e.g. o A good example is ammonia solution, which is only about 2% ionised : +  NH3(aq) + H2O(l) NH4 (aq) + OH (aq) +  Ammonia is the base and the ammonium ion NH4 is its conjugate acid.  Water is the acid and the hydroxide ion is its conjugate base.  This equilibrium is sometimes referred to as a base hydrolysis.  The low % of ionisation gives a less alkaline solution of lower pH than for strong soluble bases (alkalis), but pH is still > 7.  The concentration of water is considered constant and to solve simple problems, the base ionisation equilibrium expression is written as:  +

Kb =

 

-

[NH4 (aq)] [OH (aq)] ----------------------------[NH3(aq)] -3

Kb is the base ionisation/dissociation constant (mol dm ) for any base.

Salts of weak bases and strong acids give acidic solutions. o e.g. ammonium chloride. The chloride ion is such a weak base that there is no acid-base reaction with water, but the ammonium ion is an effective proton donor. As a general rule, the conjugate acid of a weak base is quite strong. The result here is that ammonium salt solutions have a pH of 3-4. + + o NH4 (aq) + H2O(l) NH3(aq) + H3O (aq) o In zinc-carbon batteries an acidic ammonium chloride paste dissolves the zinc in the cell reaction, though an oxidising agent must be added (MnO2) to oxidise the hydrogen formed into water, or batteries might regularly explode!

o



If you place a piece of magnesium ribbon or a zinc granule in ammonium chloride or ammonium sulphate solution you will see fizzing as hydrogen gas is formed. + 2+  2H3O (aq) + M(s) ==> M (aq) + H2O(l) + H2(g)  M = zinc or magnesium Buffering action ammonium salts o A mixture of a weak base and the salt of the weak base with a strong acid. + o e.g. ammonia NH3 and ammonium chloride NH4 Cl + o NH4 and NH3 constitute a conjugate acid-base pair. o In solution most of the ammonia is NOT ionised (and even suppressed by the ammonium ions from the salt).  It is the weak base that 'removes' most of any added hydrogen ions. +

o

+

 NH3(aq) + H (aq) NH4 (aq) The salt is fully ionised in solution giving relatively high concentrations of the ammonium ion.  It is the ammonium ion that removes most of any added hydroxide ions. 

+

NH4

-

(aq)

+ OH (aq)

NH3(aq) + H2O(l)

The thermal decomposition of ammonium chloride  



o

On heating strongly above 340 C, the white solid ammonium chloride, thermally decomposes into a mixture of two colourless gases ammonia and hydrogen chloride. On cooling the reaction is reversed and solid ammonium chloride reforms. o This is an example of sublimation but here it involves both physical and chemical changes. o When a substance sublimes it changes directly from a solid into a gas without melting and on cooling reforms the solid without condensing to form a liquid. o

Ammonium chloride + heat

o o

NH4Cl(s) NH3(g) + HCl(g) so the thermal decomposition of ammonium chloride is the forward reaction, and the formation of ammonium chloride is the backward reaction.

Note: o o o o o

ammonia + hydrogen chloride

Reversing the reaction conditions reverses the direction of chemical change, typical of a reversible reaction. Thermal decomposition means using 'heat' to 'break down' a molecule into smaller ones. The decomposition is endothermic (heat absorbed or heat taken in) and the formation of ammonium chloride is exothermic (heat released or heat given out). This means if the direction of chemical change is reversed, the energy change is also reversed. o o o Ammonium fluoride (>? C), ammonium bromide (>450 C) and ammonium iodide (>550 C), with a similar formula, all sublime in a similar physical-chemical way when heated, so the equations will be similar i.e. just swap F, Br or I for the Cl. o  Similarly, ammonium sulphate also sublimes when heated above 235 C and thermally decomposes into ammonia gas and sulphuric acid vapour. 

 (NH4)2SO4(s) NH3(g) + H2SO4(g) Ammonium nitrate does not undergo a reversible sublimation reaction, it melts and then decomposes into nitrogen(I) oxide gas (dinitrogen oxide) and water vapour.

 

o

NH4NO3(s) N2O(g) + 2H2O(g) This is very different reaction, in fact its an irreversible redox reaction. The nitrate ion, NO3 , or any nitric acid formed, HNO3, act as an oxidising agent and oxidise the ammonium ion. If the products are cooled, ammonium nitrate is NOT reformed. For more on sublimation, see the States of Matter webpage.

REDOX analysis of selected reactions

The oxidation of ammonia with molecular oxygen    

The concept of oxidation state can now be fully applied to reactions which do not involve ions e.g. The oxidation of ammonia via a Pt catalyst at high temperature which is part of the chemistry of nitric acid manufacture. 4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(g) The oxidation number analysis is: o 4N at (-3) each in NH3 and 10O all at (0) in O2 change to ... o 4N at (+2) each in NH3, 4O at (-2) each and 6 O at (-2) each in H2O. o H is +1 throughout i.e. does not undergo an ox. state change. o Oxygen is reduced from ox. state (0) to (-2). o Nitrogen is oxidised from ox. state (-3) to (+2). o The total increase in ox. state change of 4 x (-3 to +2) for nitrogen is balanced by the total decrease in ox. state change of 10 x (0 to -2) for oxygen i.e. 20 e or ox. state units change in each case. o Oxygen is the oxidising agent (gain/accept e s, lowered ox. state) and ammonia is the reducing agent (loses e s, inc. ox. state of N). The reaction between ammonium and nitrate(III) (nitrite) ions

      

+

-

NH4 (aq) + NO2 (aq) ==> 2H2O(l) + N2(g) Here its the opposite of disproportionation where two species of an element in different oxidation states react to produce one species of a single intermediate oxidation state. Ox. state changes: Nitrogen in a (-3) and a (+3) state both end up in the (0) state. Oxygen at (-2) and hydrogen (+1) remain unchanged in oxidation state. The nitrite ion acts as the oxidising agent and gets reduced (N +3 to 0, 3e's gained, decrease of 3 ox. state units) and the ammonium ion acts as the reducing agent and gets oxidised (N -3 to 0, 3 e's lost, inc. ox. state 3 units). The nitrite ion acts as the oxidising agent (gains/accepts e s, lowered ox. state of N) and the ammonium ion acts as the reducing agent (loses/donates e s, inc. ox. state of N).

Part 8. The p-block elements: 8.4 Group 6/16 oxygen and sulfur in particular

Group 6/16 Introduction

down group 6/16 ===> property\Zsymbol, name Period

8O

Oxygen 2

Appearance (RTP)

16S

Sulphur 3

34Se

Selenium 4

yellow solid silver metalloid or colourless red powder (monoclinic/rhombic gas allotropes allotropes)

52Te

Tellurium

84Po

Polonium

(radioactive)

5

6

silver white metalloid

radioactive silvery solid

o

-218

117

217

450

254

boiling pt./ C

o

-183

445

685

990

962

-3

ammonium sulphate (a fertiliser salt) o 2NH3(aq) + H2SO4(aq) ==> (NH4)2SO4(aq) => evaporation to get crystals o Its acid action make it good for cleaning metal surfaces in industry. Sulphuric acid is manufactured from the raw materials sulphur, air and water and involves the production of sulphur trioxide in the Contact Process. (1) Sulfur is burned in air to form sulphur dioxide (exothermic). o In the reaction the sulphur is oxidised (O gain) (1a) S(s) + O2(g) ==> SO2(g) o Sulfur dioxide can also be indirectly obtained from the process of extracting copper from copper sulphide ores e.g. in a copper smelter: (1b) Cu2S(s) + O2(g) ==> 2Cu(l) + SO2(g) Note: Sulphur dioxide itself is a useful chemical in its own right: o It is used as a bleach in the manufacture of wood pulp for paper manufacture o and its toxic nature makes it useful as a food preservative by killing bacteria. (2) The Contact Process of sulphur trioxide production must be economically efficient for the manufacture of the important industrial chemical sulphuric acid. o In the Contact Process reactor the sulphur dioxide is mixed with air (the required stoichiometric volume/mole SO2:O2 ratio is 2:1, in practice 1-2:1 is used) and the mixture passed over a catalyst of vanadium(V) oxide V205 at a relatively high temperature of about 450°C and at a pressure of between 1-2 atm. In the reactor the sulphur dioxide is oxidised in the reversible exothermic reaction ... o o

(2) 2SO2(g) + O2(g)

2SO3(g)

2

pSO3 Kp =

------------------2

pSO2 pO2



 



The reaction forms sulphur trioxide and the equilibrium is very much to the right hand side ... o The reaction forms sulphur trioxide and the equilibrium is very much to the right hand side because despite the reaction being exothermic a relatively high temperature is used which favours the reverse reaction R to L, from the energy change equilibrium rule, i.e. increasing temperature shifts the equilibrium in the endothermic direction. However the value of Kp is high enough to give a 99% yield. o The reaction is favoured by high pressure (pressure equilibrium rule, 3 => 2 gas molecules, LHS ==> RHS), but only a small increase in pressure is used to give high yields of sulphur trioxide, because the formation of SO3 on the right hand side is so energetically favourable (approx. 99% yield, i.e. only about 1% SO2 unreacted). o The use of the V2O5 catalyst ensures a fast reaction without having to use too a higher temperature which would begin to favour the left hand side too much (energy change equilibrium rule), but remember a catalyst does not affect the % yield or equilibrium concentration of SO 3, you just get there more economically faster. o Multiple reactor beds are used to ensure the maximum % conversion and heat exchange systems are used to control the temperature, and pre-heat incoming reactant gases. (3) The sulphur trioxide is dissolved in concentrated sulphuric acid to form fuming sulphuric acid (oleum). o SO3(g) + H2SO4(l) ==> H2S2O7(l) (4) Water is then carefully added to the oleum to produce concentrated sulphuric acid (98%). o H2S2O7(l) + H2O(l) ==> 2H2SO4(l) o If the sulphur trioxide is added directly to water an acid mist forms which is difficult to contain because the reaction to form sulphuric acid solution is very exothermic! o If you 'add' equations (3) + (4) you get  (5) SO3(g) + H2O(l) ==> H2SO4(l) Good anti-pollution measures need to be in place since the sulphur oxides are harmful and would cause local acid rain! To help this situation AND help the economics of the process, any unreacted sulphur dioxide is recycled through the reactor.



Concentrated sulphuric acid can be used in the laboratory as a dehydrating agent. o Dehydration is the removal of water or the elements of water from a compound and can be described as an elimination reaction. Usually and adjacent H and OH in a molecule are removed to form the water. o When added to some organic compounds containing hydrogen and oxygen, e.g. sugar, concentrated sulphuric acid removes the elements of water from the compound leaving a 'spongy' black carbon residue. o If alcohols are heated with conc. sulphuric acid, they are dehydrated to alkenes.  e.g. ethanol ==> ethene + water  CH3CH2OH ==> CH2=CH2 + H2O o When added to blue copper sulphate crystals concentrated sulphuric acid removes the water of crystallisation leaving white anhydrous copper sulphate. In this case the water already exists BUT not in a mixture and so the following reaction is classified as a chemical change.  CuSO4.5H2O(s) CuSO4(s) + 5H2O(absorbed into the H2SO4 which it reacts with) o Conc. H2SO4 catalyses the reaction between an alcohol and carboxylic acid to form an pleasant smelling ester liquid but it isn't considered a dehydration reaction (H comes from one molecule and OH from the other).  e.g. the esterification ethanoic acid + ethanol ==> ethyl ethanoate + water  CH3COOH + CH3CH2OH ==> CH3COOCH2CH3 + H2O o Concentrated sulphuric acid can be used as a drying agent e.g. in the preparation of gases.  The prepared gas is bubbled through a dreschel/dreschler bottle (illustrated on the right), containing the concentrated sulphuric acid. In this case the water vapour is just a component in a gaseous mixture. Most gases can be dried in this way except the alkaline gas ammonia which will exothermically react to form a solid salt. In this case the water vapour is just a component in a gaseous mixture.

Analysis of some REDOX reactions 





The oxidation of sulphur dioxide with bromine 2+ o SO2(aq) + Br2(aq) + 2H2O(l) ==> SO4 (aq) + 2Br (aq) + 4H (aq) 2+ o (i) the oxidation half reaction is: SO2(aq) + 2H2O(l) ==> SO4 (aq) + 4H (aq) + 2e 2 The sulphur changes from ox. state +4 to +6 (SO2 to SO4 ). o (ii) the reduction half-reaction is: Br2(aq) + 2e ==> 2Br (aq)  Two bromine atoms (as molecule) change from ox. state 0 to -1. o The hydrogen (+1) and oxygen (-2) do not change oxidation state.  (i) + (ii) equals the balanced equation, 2 electrons gained and lost or an ox. state rise and fall of 2 units.  Bromine is the oxidising agent (gain/accept e s, lowered ox. state),  and sulphur dioxide is the reducing agent (loses e s, inc. ox. state of S). o Sulphur dioxide does ionise to a small extent in water to give the sulphite ion, and adding a strong nonoxidising acid like dilute hydrochloric acid to sodium metabisulphite produces the ion, which means another equation can also adequately describe the redox change in terms of sulphur and bromine.  e.g. if the sulphite ion acts as the reducing agent the reaction with chlorine would be written as: 22+  SO3 (aq) + Cl2(aq) + H2O(l) ==> SO4 (aq) + 2Cl (aq) + 2H (aq) The oxidation of hydrogen sulfide by iron(III) ions o If an iron(III) salt (old name, ferric salt) is added to hydrogen sulphide solution a precipitate of sulphur forms and the orange-brown solution turns pale green. 3+ 2+ + o H2S(aq) + 2Fe (aq) ==> 2Fe (aq) + 2H (aq) + S(s) o Oxidation: 1 S at (-2) change to 1 S at (0), H2S ==> S, a loss of 2 electrons, inc. 2 ox. state units. o Reduction: 2 Fe at (+3) change to 2 Fe at (+2), gain in total of 2 electrons, decrease in 2 ox. state units. o No change in the oxidation state of the 2H's (+1) involved. o The iron(III) ion acts as the oxidising agent (gains/accepts e s, lowered ox. state of Fe) and the hydrogen sulphide is the reducing agent (loses/donates e s, inc. ox. state of S). The decomposition of hydrogen peroxide o Hydrogen peroxide decomposition, catalysed by the black solid manganese(IV) oxide, MnO 2.  2H2O2(aq) ==> O2(g) + 2H2O(l)  Ox. state changes: 4O at (-1) change to 2O at (0) in O2 and 2O at (-2) in H2O  and H is unchanged at (+1).

o

A case of disproportionation where an element in a species simultaneously changes into a higher and lower oxidation state i.e. here two oxygen atoms increase their oxidation state and two oxygen atoms decrease their oxidation state. It also means that hydrogen peroxide simultaneously acts as a reducing agent and oxidising agent.

o

Part 8. The p-block elements: 8.5 Group 0/18 The Noble Gases

Group 0/18 The Noble Gases down group 0/18 ===> property\Z symbol, name Period

2He

helium

10Ne

neon

18Ar

argon

36Kr

krypton

54Xe

86Rn radon (radioactive)

xenon

1

2

3

4

5

6

o

-270

-249

-189

-157

-112

-71

o

-269

-246

-186

-152

-108

-62

0.12

1.21

1.40

2.16

3.50

na

2370

2080

1520

1350

1170

1040

melting point/ C boiling point/ C -3

density/gcm (liquid) 1st IE/kJmol

-1

electron configuration electron configuration

2

2.8 2

1s

2

2

1s 2s 2p

2.8.8 6

2

[Ne]3s 3p

2.8.18.8 6

10

2

[Ar]3d 4s 4p

2.8.18.18.8 6

known oxidation states non stable non stable non stable an unstable +2

10

2

[Kr]4d 5s 5p

2.8.18.32.18.8 6

14

10

+2,4,6,8

na

electronegativity

5.50

4.84

3.20

2.94

2.40

na

atomic covalent radius/pm

He 49

Ne 51

Ar 94

Kr 109

Xe 130

Rn 136

         

   

2

[Xe]4f 5d 6s 6p

6

GENERAL COMMENTS and TRENDS The p-block Group of Noble Gases are the last group in the Periodic Table i.e. they form the last elements at the end of a period and are all non-metals. They are all non-metallic elements and all are colourless gases at room temperature and pressure with very low melting points and boiling points. They form 1% of air, and most of this is argon. All the noble gases, except radon, are separated by the fractional distillation of liquified air. % in air by volume: 0.0005% He, 0.0018% Ne, 0.93% Ar, 0.0001% Kr, 0.00001% Xe, ?% Rn - impossible to be zero, but an extremely minute trace hopefully! (varies with local geology) Helium can also be obtained from natural gas wells where it has accumulated from radioactive decay (alpha particles become atoms of helium gas when they gain two electrons). They are very unreactive elements because the highest occupied electron level shell is completely full, meaning they have a full shell of outer electrons! They have no 'wish' electronically to share electrons to form a covalent bond or to lose or gain electrons to form an ionic bond. In other words, they are electronically very stable. They exist as single atoms, that is they are monatomic He Ne Ar etc. (NOT diatomic molecules as with many other gases - reasons given above). This is because of their electronic stability. Their very inertness is an important feature of their practical uses. Down the Group with increasing atomic number ... o The melting point and boiling point steadily increase as the number of electrons in the atoms increases so does the 'intermolecular forces' - increase in instantaneous dipole - induced dipole forces still exist, even between individual atoms. o The density steadily increases. o They are more likely to react and form a compound with very reactive elements like fluorine. Stable compounds of xenon are now known and synthesised BUT not before 1961! The first 3 Noble Gases, showing their electron arrangements (in various styles) with full very stable outer shells. Helium, with one full shell only (outer = inner !) has the highest ionisation energy of any element and is chemically the most stable and least reactive of any element in the periodic table and has no meaningful chemistry.

2

1s

2

2

1s 2s 2p

6 2

[Ne]3s 3p 

6

USES of Noble Gases

o

HELIUM The gas is much less dense than air (lighter) and is used in balloons and 'airships'. Because of its inertness it doesn't burn in air UNLIKE hydrogen which used to be used in large balloons with 'flammable' consequences e.g. like the R101 airship disaster! Helium is also used in gas mixtures for deep-sea divers.

o

NEON Neon gives out light when high voltage electricity is passed through it, so its used in glowing 'neon' advertising signs and fluorescent lights.

o

ARGON Argon, like all the Noble Gases, is chemically inert. It used in filament bulbs because the metal filament will not burn in Argon and it reduces evaporation of the metal filament. It is also used to produce an inert atmosphere in high temperature metallurgical processes, eg in welding where it reduces brittle oxide formation reducing the weld quality. Its bubbles are used to stir mixtures in steel production. Argon is the cheapest to produce. KRYPTON Not used by superman! BUT is used in fluorescent bulbs, flash bulbs and laser beams. XENON used in fluorescent bulbs, flash bulbs and lasers.

o o

o

RADON Rocks, e.g. granite, can contain uranium metal compounds which are radioactive. When they 'decay' radioactively, radioactive and harmful radon gas can be formed. Radon has almost no uses, but does have dangers! Radio-isotopes of radon are produced by radioactive decay of heavy metals (e.g. uranium) in the ground. Can build up in cellars. Like all radio-isotopes it can



cause cell damage (DNA) and ultimately cancer (see link below). However it is used in some forms of cancer treatment. NOBLE GAS COMPOUNDS - yes they do exist! o From the early 1960's compounds have been made, but only xenon compounds are stable and usually combined with oxygen and fluorine, which, not surprisingly, are the more reactive non-metals e.g. o o Xe(g) + 2F2(g) => XeF4(g) (using Ni catalyst 60 C)  The molecule has a square planar shape. o There is now quite an extensive chemistry of xenon e.g.  xenon(II) fluoride XeF2 (linear), xenon(VI) fluoride XeF6  xenon(VI) oxide (xenon trioxide) XeO3 which has a trigonal pyramid shape  xenon oxytetrafluoride XeOF4 (Xe ox. st. +6) 4 the xenonate(VIII) ion XeO6 ion exists in salts such as Na4XeO6.8H2O which is stable and can be crystallised as a hydrated salt from aqueous solution. o I don't know of any stable compound of helium and argon, but argon(II) fluoride ArF 2 has been prepared at o low temperatures (
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