Chemistry Form 6 Sem 1 03a
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PRE-UNIVERSITY SEMESTER 1 CHAPTER 3 CHEMICAL BONDING
Chemical Bonding can be generally divide to 5 main group
Electrovalent bonding (ionic) Covalent bonding Metallic bonding Hydrogen bonding Van der Waals bonding
To represent the types of bonding, a Lewis diagram (dot-andcross) is used. Each dot or cross represent one electron in valence shell and it’s a more convenient way in showing electrovalent. For both ionic & covalent bonding, octet rule must be fulfill where tendency of atoms to achieve noble gas configuration. Table 6.2 show some cation/anion with difference number of valence electron.
Electrovalent bond (ionic bond) Formed by transfering 1 or more e- from outer orbital to another. The atom ‘donate’ electron is name as cation and the atom who ‘receive’ electron is name as anion. The bond form when electrostatic attraction occur between 2 opposite charge ions. Formation of ionic compound involving a metal with low IE and a non-metal with high EA. Example for lithium fluoride (LiF). The electronic structure of the lithium and fluorine are : Lithium (Li) = 1s2 2s1 Fluorine (F) = 1s2 2s2 2p5
Practice : Draw the Lewis dot and cross diagram for these ionic compound Sodium chloride
Magnesium fluoride
2+
_
+ Na
Mg
Cl
_ F 2
Na
+
Cl
-
Mg
2+
F
2
Aluminium oxide
3+
2O
Al
Al
3+ 2
2
3
O
23
Covalent Bonding : Sharing of Electron Covalent bond is bond that formed in between atoms by sharing electron from its atoms in order to achieve a stable electronic configuration of ns2 np6 for atoms involve. (hydrogen achieve 1s2) Some non-metallic elements exist naturally as diatomic molecules like hydrogen, and halogens groups. Hydrogen molecule
Chlorine molecule
Oxygen molecule
Nitrogen molecule
From example above, we can see that in covalent bond, molecules may form single bond, double bond or triple bond in order to achieve stable valence electrons. Though, there are some molecules with the exceptions of achieving stable valence electrons.
Electron deficient compounds – compounds which the molecule (especially the center atom) does not achieve octet electron arrangement. Examples of these molecules are BeCl2 ; BF3 and AlCl3. Beryllium dichloride
Boron trifluoride
Aluminium trichloride
Electron rich compounds – compounds which have more than 8 electrons at center atom of molecules, such as PCl5, SF6 and ICl5.
Phosphorous pentachloride
Sulphur hexachloride
Iodine pentachloride
However, not all compounds can have more or less than 8 electrons in the center of the atom. There are certain limitation towards the application of the expansion of center atom
For example, nitrogen (N) and phosphorous (P) are both from Group
15phosphorous can exist as PCl3 and PCl5 while nitrogen can @@@ but only have NCl3 but not NCl5. This is because nitrogen which only have 2 shell, do not have empty d-orbital available, @@@@.....................@@@@@@@@@@@@@@@@@@@@@@@ but phosphorous contain d-orbital to fill in more electron @@@@@@@@@@@@@@@@@..................................
Same things occur when it come to hydrolysis of CCl4 and SiCl4. SiCl4 can undergoes hydrolysis with water according to the equation
SiCl
+ 2 H O SiO + 4 HCl
4 2 2 @@@@@@@@@@@@@@@@@@@@@@@@@@.
while CCl4 cannot. Despite the factors that they are from the same group
14 cannot undergoes hydrolysis as (Group @@@), CCl 4 carbon which only have 2 shell, do not have empty d-orbital available, so @@@@@@@@@@@@@@@@@@@@...@@@@@@@@@ water cannot form coordinative with carbon hence cannot undergoes @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ hydrolysis.
Examples : Draw the Lewis structure for the following molecules. CO2
HCN
CH3COOH
C2H2
NH3
CO32-
SO42-
C3H6
6.2.1 Dative bond Now, try drawing the Lewis structure for these molecules : SO2, SO3, NO3- or CO. SO2
SO3
NO3-
CO
Dative bond is formed when an atom that has lone pair electrons which can donate to molecule/ion that has empty unhybridise orbital. Following are a few applications of dative bond in covalent molecules
1. Dative bond in helping molecule to achieve octet.
NH4+
BF3.NH3
Dative bond in forming dimer ~ 2 monomer combine forming a dimer.
Forming Al2Cl6
Forming polymer of BeCl2
3.
Dative bond in formation of complex ion. Molecule / ion form dative bond (also known as coordinative bond) by ligand donating lone pair electron, which act as a @@@@.. in the formation of complex ions. For example hexaaquacopper (II) ion ; [Cu(H2O)6]2+
tetraamminenickel (II) ion ; [Ni(NH3)42+]
Hexacyanoferrate (III) ion ; [Fe(CN)6]3-
6.2.3 Resonance ~ a molecule/polyatomic ion in which two or more plausible Lewis structure can be written but the actual structure cannot be written at all Sulphur dioxide, SO2
Ethanoate ion, CH3COO–
Nitrogen dioxide, NO2
Sulphur trioxide, SO3
Carbonate ion, CO32-
Since the resonance structure cannot be determined as it does not have a permanent structure so it is expressed as a combined of resonance structure known as resonance hybrid
Resonance hybrid
Covalent Bonds : Overlapping of Orbitals 2 ways in explaining how covalent bond are attached :
Valence bond theory Valence-shell electron-pair repulsion theory (VSEPR)
Here we can explain and predict what type of molecular bond and shape will form through the bonding formation but it does not explain the stability of covalent bond. For valence bond theory, it used atomic orbital overlapping that result the formation of a new molecular orbital embracing both nuclei. The strength of covalent bond is proportional to the area where the atomic orbital overlap. Larger the area overlap, stronger the covalent bond.
Hybrid Atomic Orbitals 3 basic types of hybrid orbital
sp3 hybrid orbital (tetrahedral arrangement) sp2 hybrid orbital (trigonal planar arrangement) sp hybrid orbital (linear arrangement)
6.3.2 sp3 hybridisation The term sp3 gives an impression of the hybridisation involved _____ s 1 and _____ p orbitals 3 orbital Examples of molecules which give sp3 hybridisation are Methane sulphate ion
CH4 SO42-
silicon tetrachloride
SiCl4
Perchlorate ion
ClO4-
14 the For example, in methane, CH4, since carbon is in Group _____so 2s2 2p2 valance electron of C is _______
State of molecules
Orbital diagram
_____ _____ _____ 2p Ground state
____ 2s
Excited state ____ 2s
Hybridisatio n state
____ ____ ____ 2p
_____ _____ _____ _____ sp3
Illustration / Explanation
109.50 tetrahedral
6.3.3 sp2 hybridisation The term sp2 gives an impression of the hybridisation involved _____ s 1 and _____ p orbitals 2 orbital Examples of molecules which give sp2 hybridisation are
Sulphur trioxide
SO3
Boron trifluoride
BF3
Nitrate ion
NO3-
Carbonate ion
CO32-
13 element so the electron valance of B Since boron is Group ______ 2s2 2p1 is _________
State of molecules
Orbital diagram
_____ _____ _____ 2p ____ Ground state 2s
____ Excited state 2s
Hybridisatio n state
____ ____ ____ 2p
_____ _____ _____ sp2
____ pz
Illustration / Explanation
Formation of sp2 Hybrid Orbitals
Shape of molecule Trigonal planar Angle between bond pair 120o
6.3.4 sp hybridisation The term sp gives an impression of the hybridisation involved _____ s orbital 1 and _____ p orbitals 1 Examples of molecules which give sp hybridisation are Carbon dioxide
CO2
Beryllium chloride
Cyanic acid
HCN
Ethyne
BeCl2 C2H2
Let’s use beryllium chloride as example.
2 element so the electron valance of Since beryllium is Group ______ Be is ___________2s2
State of molecules
Orbital diagram
_____ _____ _____ 2p
Ground state ____ 2s
____ Excited state 2s
____ ____ ____ 2p
Hybridisatio _____ _____ sp n state
___ ___ py pz
Illustration / Explanation
Formation of sp Hybrid Orbitals
Shape of molecule Linear Angle between bond pair 180o
6.4 Hybridisation in organic molecules In this subtopic, we’re going to witness how is the formation of the bonding that exist in some organic molecules. The 3 organic molecules which will be discussed in this sub-topic are : methane, CH4 ethene, C2H4 ethyne, C2H2 All of the molecules above has carbon in it 14 element. It has the electronic configuration of Carbon is a group _____ 2s2 2p2 ______________ The orbital diagram Ground state of carbon : _____ _____ _____ _____ 2s 2p
Methane, CH4
Type of hybridisation :
Excited state of carbon
:
_____
_____ _____ _____
2s Hybridised state
:
2p
_____ _____ _____ _____ sp3
Molecular shape
:
tetrahedral @@@@@@@@@@@@@ Angle between the bonding pair : 0 109.5 @@@@@@@@@..
Ethene, C2H4
Type of hybridisation :
Excited state of C : _____ 2s
Hybridised state
_____ _____ _____ 2p
: _____ _____ _____
sp2 Molecular shape Trigonal planar
Angle between bond pair – bond pair 120o
sp2
_____
pz
Ethyne, C2H2
sp
Type of hybridisation :
Excited state of C : _____ _____ _____ _____ 2s 2p Hybridised state : _____ _____ _____ _____
sp Molecular shape Linear
Angle between bond pair – bond pair 180o
py pz
As a conclusion, the formation of double bond one bond (σ) and (C=C) is due to ______sigma one (π) _____pi bond While the formation of triple bond (C≡C) is due one two to ______sigma bond (σ) and _____pi bond (π)
3.5 Hybridisation in water, H2O and ammonia, NH3 The hybridisation of ammonia is similar to that in methane (sp3 hybridisation). Nitrogen, N which has the electron valence as @@@@@@@@. where the ground state can be stated in the orbital diagram below
Ground state : ____ 2s
____ ____ ____ 2p
Excited state : ____ 2s
____ ____ ____ 2p
Hybridised state : ____ ____ ____ ____ sp3 *Compare the angle between the bonding pair of N–H to N–H in ammonia and C–H to C–H in methane Angle between H–N–H < Angle between H–C–H Shape :
Same goes to the hybridisation of water (sp3 hybridisation). Oxygen, O, which has the electron valance as @@@@@@@.., where the ground state can be stated in the orbital diagram below Ground state : ____ ____ ____ ____
2s Excited state : ____ 2s
2p ____ ____ ____ 2p
Hybridised state : ____ ____ ____ ____ sp3 *Compare the angle between the bonding pair of H–O–H in water and H–C–H in methane Angle between H–O–H Angle between H–C–H
From the 2 examples above, we can tell how the lone pair electrons affecting the angle between the bonding pair and bonding pair. In ammonia, not only that there is the repulsion between bonding pair and bonding pair but there’s also the repulsion between bonding pair and lone pair. Since the angle between the bonding pair and bonding pair decrease, there’s a probability that its due to the effect of stronger repulsion between the bonding pair and lone pair electron. This statement is supported as in the repulsion between the H–O–H in water is smaller than in ammonia, NH3. as a conclusion, we can conclude that
lone-pair vs. lone pair repulsion
>
lone-pair vs. bonding pair repulsion
>
bonding-pair vs. bonding pair repulsion
Valence Shell Electron Pair Repulsion (VSEPR) Theory ~ state that the electron-pair repulsion stated that electron pairs around central atom repel each other 3 main rules
Bonding pairs and lone pairs of electrons arrange themselves to be as far apart as possible. The order of repulsion strength of lone pair and bond pair are lone-pair & lone-pair > lone-pair & bond-pair > bond-pair & bond-pair Double / triple bond are considered as 1 bonding pair when predicting the shape of molecules or ions
Diagram below shows the type of bonding and the molecular shape predicted.
No of Class surround atoms
AB2
AB3
AB4
2
3
4
No of lone pair electron
0
0
0
Molecular shape
Diagram of the molecular shape
Example of molecules
Linear
BeCl2 CO2 HCN
Trigonal Planar
CO32AlCl3 BF3
Tetrahedral
CH4 SiCl4 SO42-
No of Class surround atoms
No of lone pair electron
Molecular shape
Diagram of the molecular shape
Example of molecules
AB5
5
0
Trigonal bipyramidal
PCl5 BiCl5
AB6
6
0
Octahedral
SF6 TeCl6
VSEPR
Class
# of atoms bonded to central atom
# lone pairs on central atom
AB3
3
0
AB2E
2
1
Arrangement of electron pairs
Molecular Geometry
trigonal planar trigonal planar
trigonal planar bent
10.1
VSEPR
Class
# of atoms bonded to central atom
# lone pairs on central atom
AB4
4
0
AB3E
3
1
Arrangement of electron pairs
Molecular Geometry
tetrahedral
tetrahedral
tetrahedral
trigonal pyramidal
10.1
VSEPR
Class
# of atoms bonded to central atom
# lone pairs on central atom
AB4
4
0
Arrangement of electron pairs
Molecular Geometry
tetrahedral
tetrahedral
AB3E
3
1
tetrahedral
trigonal pyramidal
AB2E2
2
2
tetrahedral
bent O H
H
10.1
VSEPR
Class
AB5 AB4E
# of atoms bonded to central atom
5 4
# lone pairs on central atom
Arrangement of electron pairs
Molecular Geometry
0
trigonal bipyramidal
trigonal bipyramidal
1
trigonal bipyramidal
see - saw
10.1
VSEPR
Class
AB5
# of atoms bonded to central atom
5
# lone pairs on central atom
0
AB4E
4
1
AB3E2
3
2
Arrangement of electron pairs
Molecular Geometry
trigonal bipyramidal
trigonal bipyramidal
trigonal bipyramidal trigonal bipyramidal
distorted tetrahedron T-shaped F F
Cl F 10.1
VSEPR
Class
AB5
# of atoms bonded to central atom
5
# lone pairs on central atom
0
AB4E
4
1
AB3E2
3
2
AB2E3
2
3
Arrangement of electron pairs
Molecular Geometry
trigonal bipyramidal
trigonal bipyramidal
trigonal bipyramidal trigonal bipyramidal
distorted tetrahedron
trigonal bipyramidal
T-shaped linear I I I
10.1
VSEPR
Class
# of atoms bonded to central atom
# lone pairs on central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
square pyramidal F F F
Arrangement of electron pairs
Molecular Geometry
Br F
F
10.1
VSEPR
Class
# of atoms bonded to central atom
# lone pairs on central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
AB4E2
4
2
octahedral
square pyramidal square planar
Arrangement of electron pairs
Molecular Geometry
F
F Xe F
F
10.1
5 GENERAL STEPS TAKEN WHEN WRITING LEWIS STRUCTURE FOR MOLECULES AND IONS Calculate the total number of valence electrons from all atoms Arrange all the atoms surrounding the central atom by using a pair of electron per bond Assign the remaining electrons to the terminal atoms so that each terminal atom has 8 electrons (H = 2 e-) Place any left-over electron on the central atom. @ Form multiple bonds if there are not enough electrons to give the central atom an octet of electrons.
i)
PCl3
1 P => 1 (5) 3 Cl => 3 (7)
= 5 electrons = 21 electrons Total = 26 electrons Step 2 : place 1 bond from surround to center atom e- used = 3 (2) = 6 e- remained = 26
Step 3 : placed each surround atom with 6 ee- used = 3 (6) = 18 e- remained = 2 Step 4 : place remained e- at center of atom
ii) SF6 1 S => 1 (6) 6 F => 6 (7)
= 6 electrons = 42 electrons Total = 48 electrons Step 2 : place 1 bond from surround to center atom e- used = 6(2) = 12 e- remained = 36
Step 3 : placed each surround atom with 6 ee- used = 6 (6) = 36 e- remained = 0
SO42-
b)
4 Surround Atom + 0 Lone pair eArrangement : tetrahedral Shape : tetrahedral
d)
SF6
6 Surround Atom + 0 Lone pair eArrangement : octahedral Shape : octahedral
c)
POCl3
4 Surround Atom + 0 Lone pair eArrangement : tetrahedral Shape : tetrahedral
e)
I3 -
2 Surround Atom + 3 Lone pair eArrangement : trigonal bipyramidal Shape : linear
f)
ICl3
g)
3 Surround Atom + 2 Lone pair eArrangement : trigonal bipyramidal Shape : T shape
i)
PCl5
5 Surround Atom + 0 Lone pair eArrangement : trigonal bipyramidal Shape : trigonal bipyramidal
SbCl52-
5 Surround Atom + 1 Lone pair eArrangement : octahedral Shape : square pyramidal
j)
CO32-
3 Surround Atom + 0 Lone pair eArrangement : trigonal planar Shape : trigonal planar
6.6 Electronegativity and Polar Molecules Electronegativity are measurement of ability of an atom in molecules to attract a pair of electron For 2 identical atoms, since they have same electronegativity so they have no difference in electronegativity. These molecules are called polar molecules While if 2 not identical form a covalent bond, the bonding electrons will attracted more strongly by more electronegative element. We can indicate the polarity of hydrogen chloride molecules in 2 ways.
δ+
H
δ–
Cl
The separation of charge (between δ+ and δ– ) in a poplar molecule is called dipole When 2 electrical charges of opposite sign are separated by small distance, dipole moment is established
Molecules that are polar have large dipole moments. Molecules that are non polar have zero dipole moment. Still, for some molecules, even there are different in electronegativity but it doesn’t mean that these molecules there are polar molecules. When the surrounding atom are symmetrically surrounded by identical (same) atom, they are non-polar Example of molecules which are non polar
Dipole Moments and Polar Molecules
electron poor region
electron rich region
H
F
δ+
δ−
µ=Qxr Q is the charge r is the distance between charges 1 D = 3.36 x 10-30 C m
Which of the following molecules have a dipole moment? H2O, CO2, SO2, and CH4 O
S
dipole moment polar molecule
dipole moment polar molecule H
O
C
O
no dipole moment nonpolar molecule
H
C
H
H no dipole moment nonpolar molecule
Nitrogen dioxide, NO2
Methane, CH4
Ethene, C2H4
Benzene, C6H6
Boron trifluoride, BF3
Cyanide acid, HCN
Sulphur dioxide, SO2
Sulphur trioxide, SO3
Ammonia, NH3
Ammonium ion, NH4+
Ethane, C2H6
Chloroethane, C2H5Cl
Cyclohexane, C6H12
Chlorocyclohexane, C6H11Cl
Carbon dioxide, CO2
Carbonate ion, CO32-
Phosphorous trichloride, PCl3
Phosphorous pentachloride, PCl5
cis–but-2-ene
trans–but-2-ene
A simple experiment which can be used to determine either a molecule is polar or non polar is illustrated below By using the liquid form of the compound, it is flow out slowly from burette while a negative charged rod is bring close to the flow of the liquid. If the liquid is deflected to the direction of negative charged, this polar liquid is @@@@ If it remain undeflected, this liquid non-polar is @@@@@.
From the example above, classified which compounds can be deflected and which cannot
Compound which can be deflected by charged rod
Compound which cannot be deflected by charged rod
Nitrogen dioxide, Cyanide acid, Sulphur dioxide, Ammonia, Chloroethane, Chlorocyclohexane, Phosphorous trichloride Cis-but-2-ene
Methane, ethene, benzene, Sulphur trioxide, Ammonium ion, Ethane, cyclohexane, Carbon dioxide, Carbonate ion, Phosphorous pentachloride, Trans-but-2-ene
Electronegativity and Type of Chemical Bond. Actually, the type of bond that would form can be tell by using the difference of electronegativity (∆EN). More larger the difference, the more tendency of electron form low EN move an electron to higher EN atom and ionic compound is formed. The relationship between the ionic character and the difference in the electronegativity of the bonded atom is shown on next slide (or page 220). The presence of dipoles gives ionic character to polar covalent molecules. When the polarity of the covalent molecule increases, the ionic character also increase. An ionic bond is formed if – the cation has a small ionic radius – anion has a large ionic radius – both cation & anion carries a low electrical charge. Polarisation ~ the distortion of the charge cloud of the negative ion by a neighbouring positive ion.
Fig. 9.18
3.6.1
Covalency Properties in Ionic Molecules
From the graph above, the dotted line represent the arbitrary line between ionic and covalent characteristic of a molecule. To be more specific, there more likely an ionic compound may have high covalent characteristic (exemplified by LiI), or conversely covalent compound having high ionic characteristic (exemplified by HF). The covalent characteristic of a molecule is dependent on the ability of a cation to polarise an anion. Polarisation indicates the ability of a cation to attract the electron density of an anion when put next to the cation involved. When a cation is able to pull the electron density of the anion closer to it, as if the anion wanted to share electron with cation, hence increase the covalency of the molecule
A+
X–
Highly ionic compound Large cationic size Small anionic size
B+
Y–
Highly covalent compound small cationic size large anionic size
• The covalency properties of a molecule is dependent on the cation and anion where they can be explained qualitatively via •
Polarisation power of cation
•
Polarisability of anion
3.6.1.1 Polarisation Power of Cation Polarisation Power of Cation – measure the ability of a cation to polarise the electron cloud of the anion. 2 factors determining the polarisation power of cation Charge of cation ⇒ Greater the charge of ion, higher the effective nuclear charge of cation, hence it will be able to attract the neighboring electron density of anion. This will caused the polarization power of cation increase, hence increase the covalent characteristic of cation.
Size of cation ⇒ Smaller the size of cation, closer the neighboring anion to the nucleus of cation, hence easier for the cation to polarise the anion and result an increment in the polarization power of cation, and increase the covalent characteristic of cation.
♦ Both factors can be explained in another term called as charge density where Charge Density = Charge / Ionic Radius ♦ From the equation above, Charge Density will have a greater value, provided that cation has a high charge and small cationic radius. ♦ Greater the charge density, higher the polarization power, greater the covalent characteristic of the cation.
3.6.1.2 Polarisability of Anion Polarisability of an anion ~ ability of the anion to allow the electron density to be polarised by cation. 2 factors determining the polarisability of an anion Charge of anion
Size of anion
⇒ Greater the charge of anion, lower the ⇒ Larger the size of anion, further the effective nuclear charge of anion. This will outermost electron from the nucleus of weakened the electrostatic attraction forces the anion, easier for the cation to between nucleus and the outermost electron in polarise the anion, and cause the anion, and increase the polarisability of the polarisability to increase, hence increase anion, hence increase the covalent the covalent characteristic of anion. characteristic of anion
Unlike cation, anion does not have a term that combined both factors of charge and ionic radius. However, information of polarisability of anion enable the prediction of the covalent characteristic of a molecule, since in order to form a covalent bond, it depend on both polarisation power of cation and polarisability of the anion
3.6.2 Prediction of Chemical Bond :Fajans’ Rule In 1923, Kazimierz Fajans formulated an easy guidance to predict whether a chemical bond will be covalent or ionic, and depend on the charge on the cation and the relative sizes of the cation and anion. They can be summarized in the following table Ionic compound
Low positive charge
Large cation
Small anion
Covalent compound
High positive charge
Small cation
Large anion
Based on these guidance, the bonding of a few compounds shall be discussed to understand the application of Fajans’ Rule in the chemical bonding
Lithium halide (LiX) Lithium ion, Li+ (1s2) has a small size due to only 1 shell present in its ion. But since it has a low charge, so its charge density is not too high. That is why, all lithium halide are ionic compound. The covalency of lithium halide varies from a highly ioniccharacteristic to highly covalency, depending on the polarisability of the anion next to Li+ When a group of halide, F– ; Cl–; Br–; I– is put close to Li+, the covalency of lithium halide increase when going down to Group 17 halide. LiF is highly ionic, since the fluoride ion has small ionic size and low charge, hence has low polarisability. Ionic size increase with the increasing shell when going down to Group 17 halide, hence increase the polarisability, which allowed lithium ion to polarise the anion’s electron density, hence increase the covalency
Cl– Br–
F–
Li+
Aluminium halide (AlX3) and aluminium oxide (Al2O3) Aluminium ion (Al3+) has high charge density, due to its high charge unit and its small ionic radius. So, depending on the anion, aluminium has a high tendency to form covalent compound. For example, when going down to Group 17 halide, aluminium fluoride (AlF3) forms ionic compound (since F- has a low polarisability), while aluminium trichloride (AlCl3), aluminium tribromide (AlBr3) and aluminium iodide (AlI3) form covalent compound (since chloride, bromide and iodide have high polarisability). This explained why aluminium fluoride has a high melting point (10400C), while aluminium trichloride and tribromide are 1920C and 780C respectively. As for aluminium oxide (Al2O3), it is an ionic compound with high covalent characteristic, as aluminium ion has high covalent characteristic due to its high charge density. This explained the high melting point of Al2O3 (20500C) yet it is insoluble in water. It also explained the amphoteric properties of aluminium oxide where aluminium oxide can act as an acid (covalent characteristic), as well as a base (ionic characteristic).
Metallic Bonding The properties of metals cannot be explained in terms of the ionic / covalent bond. In ionic / covalent compound, electron are not free to move under the influence of applied potential (charge) difference. Therefore, ionic solid and covalent compound are insulator. In metal, electron are delocalised and metal atoms are effectively ionised. Metallic bond ~ electrostatic attraction between the positively charged metal ion and the electron delocalised. Because of this, electron now can freely move from cathode to anode when a metal is subjected to an electrical potential. The mobile electron can also conduct heat by carrying the kinetic energy from a hot part of the metal to a cold part. This electron delocalised can also use to explain the electrical and thermal conductivities of metal
The Band Theory : Overlapping of Orbital The number of molecular orbitals produced is equal to the number of atomic orbitals that overlap. In a metal, the number of atomic orbitals that overlap is very large. Thus the number of molecular orbital produced is also very large. The energy separations between these metal orbitals are extremely small. So, we may regard the orbital as merging together to form a continuous band of allowed energy state. This collection of very closed molecular orbital energy levels is called an energy band. This theory for metal is called band theory
Electrical Conductors Molecular orbital model == 2 group of energy level.
Lower energy level – valence band → form from overlap of outer most orbital containing valence electron of each atom. Higher energy level – conduction band → energy level filled with mobile electron
But there are some case where valence band can also serve as conduction band (caused by the movement of delocalised molecular orbital) Electrical conductivities decrease when temperature increase – vibration of the lattice of ion impedes the free movement of electron in conduction band. conduction band valence band
Insulator Difference between conductors, semi-conductors, and insulator depend on the energy gap between the 2 bands. Conductor – 2 bands overlaps so conduction band always partly filled. Insulator – gap between the band is large and no electron exist in the conduction band. E.g. insulator – diamond When 2s and 2p orbital of C is combine to form 2 energy bands, valence band is filled with electron. In insulator, the energy gap between the band is large. Under normal condition, few electrons in valence band can jump across to conduction band. If electron cannot reach conduction band across the gaps, the electrical conduction cannot take place.
Semiconductor There’s still energy gaps between 2 bands in semiconductor, but it is smaller than insulator. In semiconductor, some electrons have sufficient energy to jump across the energy gaps and electron can move freely in conduction band thus enable electrical conduction. Still, the electrical activity is not as good as metal (conductor) Increasing temperature can help to improve the conductivity because electron gain thermal energy and are able to reach conduction band. It can also improve its effectiveness by adding small amount of substance. This adding is what we called doping. It can help to increase electrons to fill in valence band. Example of doping is Si dope P (n-type). Si dope Ge (p-type) Depend on the needs, this process can help to create the various type of semiconductor in electronic characteristic.
7.1 Van der Waals forces Van Der Waals forces are the intermolecular forces formed between covalently bond molecules which exist as simple molecules. There are 2 types of Van Der Waals forces namely ♥ Permanent Dipole – Permanent dipole forces ♥ Temporary dipole – induced dipole forces
7.1.1 Dipole-dipole attraction forces 1. Polar molecule possessed dipole moment. Each of the polar molecules have an overall magnitude. For example in hydrogen chloride H –––– Cl δ+ δ– 2. The dipole inside polar molecules is permanent and the forces between the molecule form as the positive end of dipole will attract to the negative end of another molecule’s dipole.
3. This kind of forced are called permanent dipole-dipole forces. 4. The strength of the attraction depends on two factors : dipole moment and relative molecular mass
5. Higher the dipole moment – the more polar the molecule – stronger the Van Der Waals forces 6. Comparisons were made between 4 molecules that have nearly equaled of molecular mass, but with different dipole moment Compounds Propane , CH3CH2CH3 Methyl methoxide, CH3–O–CH3 Chloromethane Methyl cyanide, CH3CN
RMM 44 44 50.5 41
DM 0.1 1.3 1.9 3.9
Boiling point (°C) - 18.0 4.0 6.0 56.0
7. Methyl cyanide exhibit the highest boiling point among the 3 molecules as it has the highest dipole moment among these molecules, which makes the attraction between the dipole-dipole attraction become stronger, and required a higher temperature to break the attraction forces among CH3CN-----CH3CN.
8. Another factor which influence the strength of permanent dipoledipole forces, are the factor of relative molecular mass. 9. Higher the mass, stronger the forces of attraction ( Van Der Waals forces ), higher the boiling point or melting point of the substance
Hydrogen chloride, H – Cl
36.5
Melting point (°C) - 114
Hydrogen bromide, H – Br
81.0
- 87
- 66
Hydrogen iodide, H – I
128
- 51
- 35
RMM
Boiling point (°C) - 85
7.1.2 Temporary dipole – induce dipole forces Non-polar molecules have a dipole moment = 0. Basically, they won’t have any attraction between the molecules as there are no significant poles with charge in the molecule, so how they interact ??!!! For non-polar molecules, they may have a chance to form asymmetrical structure, as the distribution of electron within the molecule are not even, giving the atom a temporary dipole moment. During the formation of temporary dipole moment, induction process takes place where the distribution of electron are uneven and give the atom which are temporary rich of electron to form dipole. These dipoles also known as induce dipole. When induced dipole is formed , a temporary interaction between the molecules formed and produces weak forces among them.
This theory is introduced by Frite London in 1930. It is known as London dispersion forces.
In (a) the non-polar molecule which does not have a dipole within the molecule begin to fluctuate and thus forming a “temporary” dipole as in (b). Thus the forces of attraction will formed between the temporary dipole and this forces is named as London Forces
7.2 Effect of the intermolecular forces ( Van der waals ) on the physical properties of the molecules H vapourisation give a quantitative measurement of strength of attractive forces present in liquid. So, H vapourisation , the boiling point , the intermolecular forces among its molecules. When a molecule increase in size, the number of electron also increase, so the attraction between the electron valence and nucleus become less. This distortion of electron cloud can easily occur and increase the polarisability of the negative ion. This can be relating with the dispersion forces among molecules therefore H vapourisation , e.g. : Value of boiling point of halogen gas increase. ( from F2 I2 ) In hydrocarbon, boiling point increase with relative molecular mass (RMM). Molecule with higher RMM will have a higher boiling point. The effect of branched chain in hydrocarbon will also affect the boiling point of hydrocarbon involved
Structure
RMM
Boiling point (°C)
2,2–dimethyl propane
72
4
2-methylbutane
72
18
72
36
n–pentane
CH3 – CH2 – CH2 – CH2 – CH3
This is due to a larger surface area in a straight chain of hydrocarbon, and allows greater forces between the molecules – giving larger Van der Waals forces – compare to branch chain hydrocarbon
7.3 Hydrogen Bonding Hydrogen bond is a special dipole–dipole interaction between H atom with other atom with high electronegativity. ( N, O, F )
It is extra stable than normal Van der waals forces and required a high energy to break the bond. This explained why the boiling point of NH3, H2O and HF are higher than other hydrogen compound from each of their particular group.
Decreasing
molar mass Decreasing boiling point
Hydrogen bond can also be used to explain the different of boiling point of some organic compound. In the diagram above, the trend of the compound in the same group deviates for N, O and F, as it form hydrogen bond among themselves. Hydrogen bond can be compared among NH3 , H2O and HF. HF has a higher boiling point than NH3 due to higher electronegativity of fluorine compare to nitrogen. So the dipole moment of H–F is greater than N–H, which results greater hydrogen bond. Though, O has a lower electronegativity than F, but H2O has a greater boiling point compare to HF because in between H2O ---- H2O molecules, they can form 2 hydrogen bond between the molecule but between HF --- HF can only form one hydrogen bond. So, the more the hydrogen formed, greater the forces, higher the boiling point. The factors of hydrogen bonding can also use to explain the solubility of some organic compound in water, like example, ethane cannot dissolve in water but ethanol can dissolve in water, due to the hydrogen bonding.
Some of the molecules gain more stability by forming dimer with its molecules. E.g. : When ethanoic acid is brought to mass spectrometer for detection and it gives a peak at m/e at 120. This indicates the shows that ethanoic acid (CH3COOH) has a RMM of 120, as CH3COOH , RMM = 60.
This indicate ethanoic acid exist as dimer where interaction of hydrogen bonding between end of each functioning group – COOH occur.
There is another application of hydrogen bond, which is the intermolecular forces and intramolecular forces. In 2-nitrophenol and 4-nitrophenol, the boiling point of the 2 compounds can be explain below :
Since 2-nitrophenol form strong hydrogen bond as intramolecular forces, the interaction between 2-nitrophenol molecules are weaker among each other, compare to 4-nitrophenol, which used hydrogen bond as their intermolecular forces. With stronger hydrogen bond which act as the intermolecular forces, the boiling point of 4-nitrophenol is expected to be higher than 2-nitrophenol
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