H2 Chemistry Notes

Chemistry Notes General: -

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Room TP – 24000 cm3. (Standard conditions) Standard TP – 22400 cm3. If bond energy calculated (using data booklet) differs from actual answer, you can state ‘values in data booklet are average bond energy values./ or also because the compound is not in the gaseous state. Organic reactions are usually slow because need to break strong covalent bonds and heat is often required for reaction to proceed. NO in atmosphere causes photochemical smog, Ozone depletion. Hydrogen bonding is responsible for the open structure in ice, the reason why it has lower density than water. A radical contains odd number of electrons – because of the one unpaired single electron. Both SO2 and NO2 are reducing agents; they have donatable lone pairs/electrons. When drawing stereoisomers, when showing alkyl groups, write out the full structural formula, from left to right, DOES NOT NEED TO SUPERIMPOSE. Benzene DOES NOT CONDUCT ELECTRICITY because it exists as nonpolar covalent molecules and cant act as charge carriers. NH4+ is not an electrophile although it is positively charged because it cannot accept any more electrons in addition to its already existing 4 bonds ( cant expand its octet ). Before linking to ‘rate of reaction’, make explicit link to the activation energy required. Value of DeltaG does depend on the number of molar reactants, so multiply accordingly. All ammonium salts react with NaOH to give off ammonia. Dispersion forces = Van der waals forces. The basicity of compound is derived by how many hydrogen atoms are bonded to different electronegative atoms so that it can lose the H+. If you just see reagents like cold nitric acid/base, it means just acidification or basification; like NH2  NH3+. If a certain element coming from 2 different compounds reacts in different manners with the 3rd compound, look at the electronegativity of the element in the 2 compunds.

Gases – Ideality: -

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When there are more than 2 gases, and it’s a mixing scenario given, ALWAYS find individual partial pressures and then add up to find total pressure. Don’t just add or use proportion. At higher temperature, same gas and no.of mol  will deviate less from ideality because the molecules possess higher average KE  more able to overcome inter-molecular forces of attraction.

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Increase in Mr  larger electron cloud  strong IMFOA  higher boiling point  decrease in vapour pressure.

Equilibria: -

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pKa doesn’t always need to be smaller than pKw so calculate every time using the Ka equation. When the acidic and alkaline colours of an indicator is provided, know that the colour within the range itself will be the mixture of the two – only above or below the range would give the distinct colour provided. The equilibrium constant depends ONLY ON TEMPERATURE. The common ion effect causes equilibrium to shift left and thus, solubility to decrease compared to the solute in water. In some compounds, intramolecular hydrogen bonding may be able to stabilize the conjugate base, making it more basic. Pk2 is usually greater than Pk1 because it is energetically unfavourable to remove a positively charged H+ from an already negatively charged ion. Kw – basically, it is the ionic product of water. Its value is the product of [H+] and [OH-]. Unless otherwise stated, the concentrations of both these are equal and so, neutral solution. At a fixed temperature, Kw is fixed also. The percentage of gas (by volume) in the total volume determines the partial pressure of the gas itself. It can be then used to find Kp. An acidic buffer – need weak acid and its conjugate salt. It is possible to reduce/halogenate the 3 C=C bonds in benzene ring using H2 with Pt catalyst. To distinguish primary & secondary amines: Primary amines undergo hydrolysis to give off my NH3; Secondary amines do not. Kp will only be affected if you change the amount of gases in the mixture. Adding solids ( even if it is part of reaction ) will not affect Kp.

Chemical Energetics: -

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Entropy gives the measure of the degree of extent to which particles and energies are distributed within a system. Increase in Temperature  Rise in average KE of particles  more ways to distribute energy  increase in entropy. When a solution ( containing a solute ) becomes more diluted, entropy increases. When given a Temp-time graph, and asked to find max temp change of a reaction. Here are the steps: Measure the receding (cooling) gradient, extend backwards until the instant (time) when the mixture is added. A negative ‘deltaG’ value and large Kc value means the the reaction in the forward reaction is spontaneous will go to completion. Which compound releases most energy on combustion? The most oxidized compound will be least exothermic because it will contain more ‘O’ atoms -> because bonds involving ‘O’ are usually stronger, so RBBT is higher. So, less energy given out.

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The magnitude of Lattice Energy is proportional to the strength of electrostatic forces of attraction in the compound.

Reaction Kinetics: -

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Usually when they ask why the concentration of 1 or 2 compounds is kept relatively much higher than the variable reagent, it is so that the concentrations of these 2 compounds will be kept effectively constant so the rate of reaction will ONLY be affected by the variable reagents. When determining the half-life of a reagent, you must try to link it to the rate constant – which is determined by temperature, concentration of other reagents which influence rate, catalyst etc. In a first order reaction, half life is constant. So, rate of decomposition/formation would be independent of original concentration of reactants.

Organic Chemistry: General: -

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Dilute Sulphuric acid is not oxidizing agent; Only concentrated sulphuric acid is. Alcohols and carboxylic acids both dissolve in water as they are able to form hydrogen bonds. Traditionally, when a compound reacts with sodium metal to give off hydrogen gas. Only 0.5 mol of H2 is given off for EACH functional group that reacts with Na. I.E: If 1 mol of H2 is formed, that means 2 functional groups (same/different) has reacted with Na. Similarly, when a compound reacts with a carbonate, only 0.5 moles of CO2 is given off for each functional group. In a compound with many chiral carbons, the number of optical isomers = 2^(no. of chiral carbons + no. of cis-trans isomers) When you notice that the product has lost H2O – one possibility is intra molecular condensation that can lead to formation of amide bond. In the production of compounds, AlCl3 can be used – it acts as a Lewis Acid catalyst because it only has 6 electrons around it and is electron – deficient; it can accept 2 more electrons from lone pair of Cl- to achieve an octet configuration. If a given complex compound contains a lone pair, it will probable be able to react with dilute acids even at room temperature as it can accept protons. [AlH4]- acts as a source of H- rather than H+ ions because the hydrogen atoms is more electronegative then the Al atom, so it is partially negative.

Alkanes:

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Iodoalkanes cannot be made from FRS as the first propagation step is endothermic i.e less energy evolved from the forming of weak H-I bond. CFCs are non-toxic and inert. How they deplete ozone layer? UV radiation from the sun breaks the C-Cl bond in CFCs to form chlorine radicals. These radicals can initiate a chain reaction of destruction of ozone layer. For FRS, excess alkane need to be used to prevent multiple substitution. In the termination step of FRS with the reaction of two organic radicals, there is still no sharing of carbon atoms – just a bond between the two only. The termination step is the least likely to occur steps in FRS.

Alkenes: -

If an alkene is fixed into a ring ( cyclo compound ), the C=C double bond does not cause geometric isomerism.

Alcohols & Phenols: -

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Why milder conditions required for phenol than methylbenzene? Lone pair of electrons in phenol s delocalized into benzene ring  benzene ring more electron rich  more susceptible to electrophilic attack. The electron-donating effect of the alkyl group not that strong. Phenols are weaker acids than Carboxylic acids  conjugate base of phenol is a stronger base than conjugate base of carboxylic acid. Alcohols are not acidic enough to react with NaOH

Arenes: -

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In the reaction between conc. HNO3 and benzene, the intermediate formed will have a ‘H atom’ and ‘NO2’ attached to carbon making it SP3. In the production of compounds, AlCl3 can be used – it acts as a Lewis Acid catalyst because it only has 6 electrons around it and is electron – deficient; it can accept 2 more electrons from lone pair of Cl- to achieve an octet configuration. When there is a substituent added to a benzene ring of a compound which is symmetrical, the 3- and 5- are the same compound so don’t count them as different. When question ask about isomers of benzene ring compounds – if CH3 is only 2,4 directing. CH3 with the substituent on 3 is still an isomer but it is just the minor product. If both NH2 and –Br are present as substituents in a aromatic ring, the directing of NH2 will take precedence.

Carbonyl compounds: -

NaBH4 only reduces aldehydes and ketones.

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Also, in an organic compound contains a C atom with two –OH bonds attached to it, it is unstable, so it automatically becomes C=O instead.

Carboxylic Acids: -

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Acid chlorides are very acidic – they dissociate to give HCL. They are even more acidic than the corresponding carboxylic acid. A cyclic compound with C-C single bond can exhibit cis-trans isomerism because if the bond is pointing into the ring/outside the ring is different; cuz you cannot rotate about the ring so it is not the same. Amides are neutal compounds. If PCl5 is there in the reactant but Cl is not in the product, it must have formed an acid chloride which then formed an amide, thus, losing the Cl.

Halogen derivatives: -

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Compounds with C-F bond usually chemically inert cuz C-F bond energy very large. Excess NH3 needs to be added to halogenoalkane to increase the chance of the alkyl halide reacting with the NH3 rather than the amines already formed. This prevents forming of secondary amines and so on. Why is chlorobenzene inert ? The p orbital on the Cl atom overlaps with the pi orbitals of the benzene. Hence, lone pair is delocalized into the benzene ring by resonance. This results in C-Cl bond having partial double bond character and hence, harder to cleave. Thus, chlorobenzene inert towards nucleophilic sub. The by-product of the base hydrolysis of nitriles is ammonia. For Organic compounds containing halogens attached to C atoms. If it undergoes alkaline hydrolysis ( NaOH ), all the Cl atoms are converted to OH by Nucleophilic substitution. In a halogenoalkene, in a C-X bond , if the same C is involved in a C=C bond, then the electrons from the X atom is delocalised into the pie C=C bond, strengthening the C-X bond, so making it unreactive. Pie orbital overlap.

Reduction & Oxidation: -

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LiAlH4 is very reactive so it must be kept dry ( dry ether ) and there is no need for heating or reaction may become too vigorous. Moreover, H2 with Ni/Pt should not be used to reduce COOH as it does not happen easily. LiAlH4 in dry ether is the best option for COOH. It can also reduce esters (hydrolysis then –COOH  -OH) and amides ( similar way ) but cannot reduce C=C. If you want to oxidize an aldehyde to ketone which is on a sidechain of a benzene ring, do not use KMnO4 as that will lead to complete oxidation and give benzoic acid instead.

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Whether complete side chain oxidation occurs or not, depends on the first carbon atom attached to the benzene ring. If the first carbon, is a tertiary carbon ( attached to 4 c atoms ) then complete side chain oxidation does not occur. The no. of mol of CO2 given out from the complete oxidation of side chain in benzene is calculated by totally how many mole of C atoms is lost when each side chain is oxidised to –COOH. Reduction has NO effect on ester linkages When KMnO4 is used as an oxidising agent in alkaline medium, the solution first changes to dark green ( MnO4 2- ions) and then brown ppt is seen (MnO2). In acidic medium, changes from purple to colourless (Mn2+) Once C6H5NO2 is reduced to C6H5NH2, it cannot be oxidized back.

Distinguishing Tests: -

Alkaline aqueous Iodine can also be used to distinguish 1-bromo and 2-bromo butane. Because, the NaOH produces 1- and 2- hydroxyl butanes – which will then differ in the iodoform test.

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Nitrogen Compounds: -

Normal haemoglobin molecules exist in anionic form and repulsion will occur between the negatively charged glutamate. Amides are less basic than amines because amides have a resonance structure involving the movement of a pair of electrons from the nitrogen atom to the oxygen atom.

Chemical Bonding: -

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When Fluorine is the central atom, F cant form dative bonds because it is very electronegative. When question says why a compound can exist with Cl but not with F, a possible reason could be that Cl is a period 3 element and can expand its octet to accommodate more electrons. In an ionic lattice structure, each cation is surrounded by all anions. Each anion is surrounded by all cations. When considering bond length, (especially between carbon atoms), look at the SP hybridisation. If one of the C atoms is SP1 hybridised, then it has more significant S character and will make the bond stronger  bond length smaller. A co-ordinate ( dative ) bond is a covalent bond and if it’s the first pair of bonding electron between two atoms, it is also a sigma bond. If a compound, got functional groups containing ‘O’ and ‘H’ closeby each other, then it is likely to have INTRAmolecular hydrogen bonds. More intramolecular H bonding  less INTERmolecular bonding  lower Melting point.

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The number of hydrogen bonds depends on the number of lone pairs on the electronegative atom also. In sp-sp bonding, the sp is responsible for the first bond ( sigma ). The other 2 bonds in the triple bond ( pie bond ) is p-p orbital overlap not sp-sp. Benzene ring has delocalised pie bond because each pair of electrons is shared between more than 2 adjacent carbon atoms; it exits in two resonant structures. All the bond lengths between all the C atoms is equal so, supports the benzene ring. P4 is highly reactive due to strained bond angles caused due to interelectronic repulsion. As long as there is a lone electron in the central atom which has two other bonds, it is bent ( NOT LINEAR ANYMORE ) The theory of – difference in electronegativity affecting bond angles only apply when lone pairs are present around the central atom. For enantiomers of certain compound with two chiral carbons, if the 2 chiral carbons reverse plane-polarized light in opposite directions, so it cancels out so only 1 isomer.

Atomic structure: -

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When question asks about electron arrangements of atoms and ions – write is this way: 1s2, 2s2 2p6 ….. CCl4 is inert to water – because Carbon is from period 2. It does not have energetically accessible d orbitals to accept lone pair of electrons from water molecules to form dative bonds. Tetrahedral can never display cis-trans.

Periodicity: -

Be forms an amphoteric oxide ( like Al ) due to its high charge density – high degree of covalency. When seeing trends down group, melting point is different from thermal stability. Melting point of group II compounds decrease down the group because of ‘lattice energy’.

Transition elements: -

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When copper ions and iodide ions are involved, and a brown ppt is formed. It is a white ppt of CuI and is brown because it is in brown Iodine. SiCl4 cant be used to produce ‘high charge density cations’ in water as it will hydrolyse to give SiO2 and HCl. High charge density ions have high polarising power. They weaken the O-H bonds in water molecules to release H+ ions. Acid base reaction then can occur with other compounds. The main definition of T.E: They have partially filled D-orbitals. When water is added to a solution containing a transition element complex ion, it does now always need to be Ligand Exhange; can just cause a shift in equilibrium if – water is part of a reaction in the previous step. Sometimes, you might need to investigate the entropy of different complex ions. In this case, you write out the equation of formation of the

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ions. i.e; from the central ion + 6 water ligands + the reacting ligand  to form the new product. See the total number of molecules before and after and observe entropy change. For example: Let en represent H2NCH2CH2NH2 [Ni(H2O)6]2+(aq) + 6NH3(aq) [Ni(NH3)6]2+(aq) + 6H2O(l) 7 particles 7 particles [Ni(H2O)6]2+(aq) + 3en(aq) [Ni(en)3]2+(aq) + 6H2O(l) 4 particles 7 particles To form [Ni(en)3]2+(aq), three en molecules replace 6 water molecules, hence ΔS is more positive. If the question gives you the colour of the complex, and asks you about the energy gap in the complex, to compare the wavelengths, you should compare the wavelength of the complement of its colour ( because you want the color THAT IS ABSORBED ) When trying to find the co-ordination number of an ion, you should know the dentatity ( type of ligand ) first. i.e: so if bidentate ligand, if ratio of ligand to central ion is 2:1 – then co-ordination number is 4. When asks about presence of dipoles, you must answer based on the structure and shape of the complex ion to see if the dipoles cancel each other or not. Transition metals are the only metals which give coloured ions in aqueous solutions. Phenols and all its derivatives are acidic. When colour change occurs in the complex ion of transition element, its because: in the presence of ligands, the energy gap between the d-orbitals change. So, different ligands lead to different energy gap so different region of visible light spectrum absorbed so different colour. Usually, if need to remove excess OH- ions in solution with transition metals, Conc. HCl is used because addition of Dil. HCl adds H2O which can cause ligand exchange. - Cr3+ is a green solution; Cr(OH)3 is a grey-green ppt. CrO4 2- is a yellow solution. MnO4 2- is a green solution. MnO2 is a brown solid. MnO4 2- is a purple solution.

Na2CO3 in (aq) acts as a base and can provide OH- ions for acid-base reaction.

Reaction Kinetics: -

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The units of ‘Rate’ in the rate equation is always – ‘mol dm^-3 s^-1 When using multiple-step reaction to determine the rate equation, things to note of: 1) The equation is proportional to the reactants in the slow step. 2) If any of the reactants is present as a product in the previous ‘fast’ step (reversible), then the reaction is proportional to the reactants of the ‘fast’ step and inversely proportional to the products of the ‘fast reversible’ step. Catalysts increase the value of rate constant by decreasing activation energy and providing an alternative reaction pathway.

Electrolysis: -

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When both the half cells of an electrochemical cell are given ( x/y ) and (p/q), use the data booklet to determine which is the anode and cathode first by looking at E values. Once determine already, then you can see the observations caused by making changes to the setup. Temperature does affect Ecell – because we do not know enthalpy change of half-cell so can be endo/exo; and POE may shift accordingly. When they give you the electrolysis cell setup and give u the battery terminals, remember that electron flow is opposite to conventional current flow. That is how you will determine the anode and cathode. If the question asks – why is the electrode ( NOT MEMBRANE ) porous ? It is to increase total surface area of electrodes to allow reactions at electrode to occur more rapidly. The membrane is porous to allow selective ions to pass through to maintain electrical neutrality. When question asks why a final product is produced at an electrode, you CANNOT use the electrolyte as a reactant. It is usually the PRODUCTS of the electrolysis that react together to give the final product. Usually, the concentration of H+ is kept constant, because in many cases, it is part of the half equations of redox reaction in the electrolysis. When dilute NaOH is added to an half cell, it can act as a source of OHions and react with cations to form hydroxides. In electrolysis of sodium halide, a higher concentration of halide ion will lower its reduction potential and in most cases, make it more likely to be oxidized, unless its fluorine.