8
8.1
8.2
8.3
8.4
1
Covalent Bonding
Formation of Covalent Bonds
Dative Covalent Bonds
Bond Enthalpies
Estimation of Average Bond Enthalpies using
Data from Energetics
8.5
8.6
8.7
8.8
8.9
2
Use of Average Bond Enthalpies to Estimate the
Enthalpy Changes of Reactions
Bond Enthalpies, Bond Lengths and Covalent
Radii
Shapes of Covalent Molecules and Polyatomic
Ions
Multiple Bonds
Covalent Crystals
8.1
3
Formation of
Covalent Bonds
Theory of covalent bond formation
1. Lewis Model (A classical treatment)
2. Valence Bond Theory (pp.31-35)
(by Linus Pauling)
3. Molecular Orbital Theory (pp.36-37)
(By Robert Mulliken)
4
1. Lewis Model
A covalent bond is formed by sharing of valence
electrons between atoms of non-metals.
Reasons : 1.
5
They have small/no difference in
E.N.
2.
They can achieve stable
electronic configurations.
(octet /duplet structures)
3.
The total energy of the system is
lowered when electrons are
shared between the two nuclei.
nucleus
electron
Force between the
electron and nucleus.
A
B
Force along the bond
axis.
If the electron is located between two
nuclei, there is an attractive force holding
the two nuclei together.
6
BB’ > AA’
A A’
B
B’
If the electron is not located between two
nuclei, there is a net force separating the
two nuclei.
7
1. Lewis Model
A Covalent bond is the result of shared
valence-electron pair(s) positioned between
two nuclei.
Sharing of electron pair(s) between
two nuclei.
8
Lewis(dot and cross) structure and Bond-line structure
• Electrons are represented by dots and/or crosses.
H
H


H CH
H
H

C H

H
H

H C H

H
• Lewis structures basically obey octet rule.
• Hydrogen adopts duplet structrue.
9
Lewis(dot and cross) structure and Bond-line structure
H
H
H
C H
H
Bond pairs
H O H
Lone pairs
10
H
C
H
H
H
O
H
Only lone pairs are shown as
dots/crosses in bond-line structure
O
O
N
O
O
Double
bond
N
N
Triple
bond
Multiple bonds
11
N
Rules for drawing Lewis structures
– A systematic way
Using CO2, CO32 and HClO4 as examples
1. Determine the arrangement of atoms
within a molecules/polyatomic ion
Central  least electronegative
Terminal  H & F (most electronegative)
12
CO2
O C O
CO32
O C O
O
HClO4
O
O
13
Cl
H
O
O
2. Determine the total no. of valence electrons(V)
in the species.
For neutral molecule,
V = sum of group numbers of atoms = m
For anion, An,
V=m+n
For cation, Cn+,
V=m–n
14
Species
CO2
V
4+26
=16
W
W–V
15
CO32
HClO4
4+36+2 1+7+46
=24
=32
3. Calculate the total no. of electrons that would
be needed if each atom obeys the octet rule(W).
2e for H
8e for other atoms
4. The difference (W – V) gives the no. of
electrons that have to be shared such that all
atoms in the species can obey octet rule.
16
17
Species
CO2
V
4+26
=16
CO32
HClO4
4+36+2 1+7+46
=24
=32
W
38=24
48=32
2+58
=42
W–V
24-16
=4 pairs
32-24
=4 pairs
42-32
=5 pairs
5. Assign a single bond to each terminal atom in the
species.
CO2
CO32
O C O
O C O
O
18
HClO4
O
O
Cl
H
O
O
5. Assign a single bond to each terminal atom in the
species.
If the resulting central atom has more than 4
single bonds, rearrange the terminal atoms such
that the central atom has 4 single bonds.
H
HClO4
O
O
19
Cl
H
O
O
O
O
Cl
O
O
5. Assign a single bond to each terminal atom in the
species.
If the resulting central atom has more than 4
single bonds, rearrange the terminal atoms such
that the central atom has 4 single bonds.
In SF6, W – V = 78 – (6+67) = 8
 4 single bonds are required
However, S has to form 6 single bonds.
Expansion of octet structure may happen for
elements in Period 3 and beyond.
20
6. If bonding electrons remain, assign them in pair
by making some of the bonds double or triple
bonds.
CO2
O
C
O
Two bond pairs left  3 ways
21
O
C
O
O
C
O
O
C
O
CO32
O
C
O
O
One bond pair left  3 ways
O
C
O
22
O
O
C
O
O
O
C
O
O
HClO4
H
O
O
Cl
O
23
O
No bond pair left
 Only one way
7. Assign lone pairs to terminal atoms to give them
octet.
If any electrons still remain, assign them to the
central atoms as lone pairs.
CO2
No. of lone pairs left = (16 – 8)/2 = 4
O
O
24
C
O
C
O
O
C
O
CO32 No. of lone pairs left = (24 – 8)/2 = 8
O
C
O
O
O
C
O
25
O
O
C
O
O
H
HClO4
O
O
Cl
O
O
No. of lone pairs left = (32 – 10)/2 = 11
26
8. Determine the formal charge on each atom.
Formal charge is the charge an atom in a species
would have if the bonding electrons are shared
equally between the atoms.
Formal charge of an atom in a species
1
 Gp no. - no. of lone pair e s - (no. of bond pair e-s)
2
-
Assume bond pairs are equally shared
27
CO2
O
C
O
1
C  4 - 0 - 8  0
2
1
O  64  4  0
2
28
CO2

O
C

O
1
C  4  0  8  0
2
1
Left O  6  6   2  -1
2
1
Right O  6  2   6  1
2
29
CO2

O

C
O
1
C  4  0  8  0
2
1
Left O  6  2   6  1
2
1
Right O  6  6   2  1
2
30

O
C


O
O
C

O
Separation of opposite formal charges
 Increase in potential energy of the species
 less stable Lewis structures
O
C
O
Most stable Lewis structure
31
CO3
1
C  4  0  8  0
2
2
O
C
O
O
O
C
O
32
O
O
C
O
O
1
O  64  4  0
2
CO32
O
C
O
O
O
C
O
33
O
O
C
O
O
1
O  6  6   2  1
2
CO32
O
C
O
O
O
C
O
34
O
O
C
O
O
There is no separation of opposite formal charges
 All three structures are stable
O
C

O
O


O

C
O

35
O
O

C
O
O
CO32
2
O
C
O
O
2
O
C
O
36
O
2
O
C
O
O
HClO4
O
H
1
H  1  0  2  0
2
O
1
O  64  4  0
2
Cl
O
O
1
Cl  7  0   8  3
2
1
O  6  6   2  -1
2
37
HClO4
H
O

O
3+
Cl

O
O

Substantial separation of opposite formal charges
 Very unstable
38
HClO4
O
H
H
O
O
Cl
O
O
O
Cl
O
O
Lone pair electrons must be rearranged to
minimize the separation of opposite formal charges.
39
HClO4
O
H
H
O
O
Cl
O
O
O
Cl
O
O
Cl can expand the octet structure to accommodate
14 valence electrons.
40
HClO4
O
H
H
O
O
Cl
O
O
O
Cl
O
1
Cl  7  0   7  0
2
41
O
HClO4
O
H
H
O
O
Cl
O
O
O
Cl
O
1
O  64  4  0
2
42
O
Breakdown of the Octet Rule (p.30)
1. Valence shell expansion (Q.21(g), (h) (k) to (n))
For elements from Period 3 and beyond can
expand the octet by utilizing
low-lying d-orbitals in bond formation.
Max. no. of bond pairs = Group no.
Gp 5
Gp 6
Gp 7
43
PF5
SF6
IF7
Breakdown of the Octet Rule (p.30)
2. Electron-deficient species (Q.21(o))
F
F
B
F
F
F
B
F
Dative bond (coordinate covalent bond) is a
covalent bond in which the bond pair electrons are
contributed solely by one of the bonding atoms.
44
Breakdown of the Octet Rule (p.30)
2. Electron-deficient species (Q.21(o))
F
F
B
F
F
F
B
F
The arrow(
)is pointing from the electron
donor from the electron acceptor.
45
F
F
B


F
unstable
Separation of opposite formal charges
Not favourable for the most electronegative F
to carry a positive formal charge
46
Q.21(d)
O




O
O
O
O
O
Separation of opposite formal charges is
unavoidable.


O
S
O
O
S
O
Separation of opposite formal charges can be
avoided by expansion of octet
47
Q.21(a)
H
H
H
+
H
N
H

H
H
N
H
48
H
H

N
H
H
Once formed, dative covalent
bond cannot be distinguished
from normal covalent bond.
The four N – H bonds are
identical.
Q.21(a)
H
H
B
H
-
H
H
H

H
H
B
H
49
H

B
H
H
Breakdown of the Octet Rule (p.30)
3. Odd-electron species
Q.22
N
O


N
O
Less stable
Separation of opposite formal charges
Positive formal charge on the more
electronegative atom
50
Q.21(j)
O


N
O
2NO2(g)  N2O4(l)
O
O
N
O
51
O
O
N
N
O
O
N
O
Q.21(p)
O
Bond length
O
0.121 nm
Bond strength
strongest

O
O


O
O
52
0.133 nm
0.149 nm
weakest
Q.21(p)

Bond length
Bond strength
O
O
0.133 nm
stronger


O
O
0.149 nm
weaker
Oxygen atoms have to be separated further apart
to minimize repulsion between negative formal
charges(lone pairs).
 Weaker bond
53
8.2 Dative Covalent Bonds (SB p.218)
Dative Covalent Bonds
A dative covalent bond is formed by the
overlapping of an empty orbital of an atom
with an orbital occupied by a lone pair of
electrons of another atom.
Remarks
(1) The atom that supplies the shared pair of electrons is
known as the donor while the other atom involved in the
dative covalent bond is known as the acceptor.
(2) Once formed, a dative covalent bond cannot be
distinguished from a ‘normal’ covalent bond.
54
8.2 Dative Covalent Bonds (SB p.218 – 219)
A. NH3BF3 molecule
55
8.2 Dative Covalent Bonds (SB p.219)
B. Ammonium Ion (NH4+)
56
8.2 Dative Covalent Bonds (SB p.219 – 220)
D. Aluminium Chloride Dimer (Al2Cl6)
Al: relative small
atomic size; high
I.E.’s required to
become a cation
of +3 charge.
AlCl3
57
8.2 Dative Covalent Bonds (SB p.219 – 220)
D. Aluminium Chloride Dimer (Al2Cl6)
Why doesn’t Al form ionic compounds with Cl?
(a dimer of
AlCl3)
Check Point 8-2
58
8.1 Formation of Covalent Bonds (SB p.213)
A. Electron Sharing in Covalent Bonds
Attraction between
oppositely charged nuclei
and shared electrons
electrostatic
( _____________
in nature)
H H
Shared
electrons
ee-
The shared electron
pair spends most of the
time between the two
nuclei.
Overlapping of atomic orbitals  covalent bond formation
59
8.1 Formation of Covalent Bonds (SB p.213)
A hydrogen molecule is achieved by
partial overlapping of 1s orbitals
60
8.1 Formation of Covalent Bonds (SB p.214)
Electron density map for covalent compounds
There is
substantial
electron density
at all points along
the internuclear
axis.
Thus electrons
are shared
between the
two atoms.
61
Compare electrondensity-map for
ionic compounds:
8.1 Formation of Covalent Bonds (SB p.214)
B. Covalent Bonds in Elements
•
Hydrogen molecule
Dot and cross diagram
62
8.1 Formation of Covalent Bonds (SB p.215)
•
Chlorine molecule
•
Oxygen molecule
63
8.1 Formation of Covalent Bonds (SB p.215)
•
64
Nitrogen molecule
8.1 Formation of Covalent Bonds (SB p.216)
C. Covalent Bonds in Compounds
65
8.1 Formation of Covalent Bonds (SB p.216)
66
VB theory vs MO theory
VB : 1. Electrons distribute in localized bonds
2. Only valence electrons are involved in bonding
MO :1. Electrons distribute in delocalized bonds
2. All electrons are involved in “bonding”
bonding and anti-bonding
67
2
O
C
O
O

2
2
O
C
O
O

O
C
O
O
VB : the real distribution of electrons in CO32 is described
by a combination of three resonance structures in
which all bonds are localized.
68
2
MO : The six electrons are delocalized over the entire
structure as described by molecular orbitals.
2
O
C
O
69
O

2
2
O
C
O
O

O
C
O
O
Four principles of molecular orbital theory
1. The total number of molecular orbitals produced
is equal to the total number of atomic orbitals
contributed by the atoms that have combined.
For H2, there are two modes of overlap between
the two 1s orbitals
1. Addition
ψσ1s  ψA,1s  ψB,1s
Waves combine constructively (in phase)
2. Subtraction ψσ1s*  ψA,1s  ψB,1s
Waves combine destructively (out of phase)
70
Antibonding orbital : the probability of finding the
electrons between the two nuclei 
 nuclei repel each other
 A,1s
 B,1s
*
1s
Head-on overlap
Bonding orbital :  probability of finding electrons
between the two nuclei.
 Nuclei are pulled together
71

1s
Electron density map of 1s molecular orbital of H2
    A,1s  B,1s
1s
72
Head-on overlap
73
Side-way overlap
Build-up of electron cloud above and below
the plane on which the bonding atoms lie.
74
2. Energy : - Bonding MO < Parent AO < Antibonding MO
75
76
3. Electrons of molecules fill up the MOs in
increasing order of energy according to the
Pauli exclusion principle and Hund’s rule
1s*
1s
77
Q.26
There is no gain of stability when the AOs of two
helium atoms overlap.
2 e involved in bonding
He(A)
He2
1s*
He(B)
2 e involved in antibonding
Overall : No e involved in bonding
Bond order = 0
1s
78
8 e involved in bonding
4 e involved in antibonding
Overall : 4 e involved in bonding
O=O
Bond order = 2
79
10 e involved in bonding
4 e involved in antibonding
Overall : 6 e involved in bonding
NN
Bond order = 3
80
4. AOs combine to give MOs most effectively
when AOs are of similar energy.
Effectiveness of overlap : 1s-1s > 1s-2s > 1s-2p > 1s-3s > 1s-3p > 1s-3d > …
The energy level diagram can be substantially
simplified by considering only the effective
overlaps. E.g. 1s-1s, 2s-2s, 2p-2p
81
Only 1s-2px overlap is considered.
82
Energetics of
formation of covalent
species
83
8.3 Bond Enthalpies (SB p.221)
Bond Dissoication Enthalpy
Bond dissociation enthalpy is the enthalpy
change for breaking one mole of a bond in a
particular environment with the reactants
and products in the gaseous state under
standard conditions.
B.D.E. / kJ mol1
Cl – Cl(g)  Cl(g) + Cl(g)
84
242
B.D.E. / kJ mol1
Cl – Cl(g)
Cl(g) + Cl(g)
242
Br – Br(g) 
Br(g) + Br(g)
193
I – I(g)

I(g) + I(g)
151
O = O(g)

O(g) + O(g)
498
N  N(g)

N(g) + N(g)
945
85

Q.27
o
ΔHatm
/ kJ mol1
1
2
Cl2(g)  Cl(g)
121
B.D.E. / kJ mol1
Cl – Cl(g)  Cl(g) + Cl(g)
ΔH
o
atm
86
=
1
B.D.E.
2
242
1/2Br2(l)
Hatm
Br(g)
1/2B.D.E.
Hv
1/2Br2(g)
Hv = enthalpy change of vapourization
By Hess’s law,
Hatm = Hv + 1/2B.D.E. > 1/2B.D.E.
87
1/2I2(s)
Hatm
I(g)
1/2B.D.E.
Hs
1/2I2(g)
Hs = enthalpy change of sublimation
By Hess’s law,
Hatm = Hs + 1/2B.D.E. > 1/2B.D.E.
88
Notes : 1. B.D.E. values are always > 0
Bond breaking processes are endothermic
2. B.D.E. values depend on the environment of the bond.
CH3H(g)
CH2H(g)
CHH(g)
CH(g)
89




CH3(g)
CH2(g)
CH(g)
C(g)
+
+
+
+
H(g)
H(g)
H(g)
H(g)
B.D.E.
B.D.E.
B.D.E.
B.D.E.
= 423 kJ mol1
= 480 kJ mol1
= 425 kJ mol1
= 334 kJ mol1
8.3 Bond Enthalpies (SB p.222)
Bond Enthalpy
Bond enthalpy is the average of the bond
dissociation enthalpies taken from a large
number of species containing a particular
chemical bond.
90
CH3H(g)
CH2H(g)
CHH(g)
CH(g)




CH3(g)
CH2(g)
CH(g)
C(g)
+
+
+
+
H(g)
H(g)
H(g)
H(g)
B.D.E.
B.D.E.
B.D.E.
B.D.E.
= 423 kJ mol1
= 480 kJ mol1
= 425 kJ mol1
= 334 kJ mol1
The average B.D.E. values of C – H bond in CH4

1
(423  480  425  334) kJ mol 1  415.5 kJ mol 1
4
Bond enthalpy of C – H bond = E(C – H) = 413 kJ mol1
An average B.D.E. values from CH4, C2H6, C3H8…etc.
91
o
Estimation of B.E. from Hfo and Hatm
4E(C – H)
CH4(g)
C(g) + 4H(g)
o

H
 atm
ΔHfo[CH4 (g)]
C(graphite) + 2H2(g)
o
ΔHfo[CH4 (g)] + 4E(C – H)  Hatm
NaCl(s)
 ΔHLo[NaCl(s)]
o
ΔH
 atm  1st IE  1st EA
H [NaCl(s)]
o
f
Na(s) +
92
Na+(g) + Cl(g)
1
2
Cl2(g)
Example 1(a)
4E(C – H)
CH4(g)
C(g) + 4H(g)
ΔHfo[CH4 (g)]
o

H
 atm
C(graphite) + 2H2(g)
o
ΔHfo[CH4 (g)] + 4E(C – H)  Hatm
4E(C – H) = [715 + 4  218  (75)] kJ mol1
= 1662 kJ mol1
E(C – H) = 415.5 kJ mol1
93
Enthalpy / kJ mol1
C(g) + 4H(g)
4218 kJ mol1
C(g) + 2H2(g)
715 kJ
C(graphite) + 2H2(g)
75 kJ mol1
94
mol1
CH4(g)
4E(C – H)
Example 1(b)
C2H6(g)
o
ΔHatm
[C2H6 (g)] = 6E(C – H) + E(C – C)
2C(g) + 6H(g)
o
ΔHatm
[compound]
= the enthalpy change when one mole of the compound in
gaseous state is broken down into its constituent atoms
in gaseous state under standard conditions.
o
ΔHatm
[element]
= the enthalpy change when one mole of atoms of the
element in gaseous state are formed from the
element in its normal and most stable state under
standard conditions.
C(graphite)  C(g)
95
Example 1(b)
C2H6(g)
o
ΔHatm
[C2H6 (g)] = 6E(C – H) + E(C – C)
2C(g) + 6H(g)
H
ΔHfo[C2H6 (g)]
o
atm
2C(graphite) + 3H2(g)
o
ΔHfo[C2H6 (g)] + 6E(C – H) + E(C – C)  Hatm
E(C – C) = [2715 + 6218  (85) 6415.5] kJ mol1
= 330 kJ mol1
96
E(C – C) in C2H6
Estimated value
E(C – C) in C2H6
Experimental value
330 kJ mol1
386 kJ mol1
The discrepancy is due to the fact that
the average B.D.E. values of C – H bonds in
CH4 and C2H6 are not the same.
97
Q.28
H
H
H
H
H
C
C
C
C
H
H
H
H
4C(g) + 10H(g)
H
o
ΔHatm
[C4H10 (g)]  5165kJ mol-1 = 3E(C – C) +10E(C – H)
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
5C(g) + 12H(g)
o
ΔHatm
[C5H12(g)]  6337kJ mol-1 = 4E(C – C) +12E(C – H)
Solving (1) and (2)
E(C – C) = 348 kJ mol1
98
(1)
E(C – H) = 412 kJ mol1
(2)
Use of bond enthalpies to estimate enthalpy changes of
reactions
Assumption : The bond enthalpies are transferable from
one molecule to another.
99
In each series,
adjacent members
differ by 1C & 2H
Linear plot shows that
bond enthalpies are
transferable.
100
Use of bond enthalpies to estimate enthalpy changes of
reactions
Assumption : The bond enthalpies are transferable from
one molecule to another.
o
ΔHreaction
 E(bonds broken) - E(bonds formed)
Bond forming is always exothermic
101
Example 1
Given : E(ClCl) = 242 kJ mol1 E(CH) = 413 kJ mol1
E(CCl) = 339 kJ mol1 E(HCl) = 431 kJ mol1
Estimate the H of the following reaction.
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
Bond broken / B.E.(kJ mol1) Bond formed / B.E.(kJ mol1)
C–H
Cl – Cl
413
242
C – Cl
H – Cl
339
431
o
ΔHreaction
 (242 + 413 – 339 – 431) kJ
= 115 kJ
102
Q.29(a)
N2H4(g) + 2F2(g)  N2(g) + 4HF(g)
Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
N–N
ΔH
163
4N – H
4390
2F – F
2158
o
reaction
NN
4H – F
945
4565
 (163+ 4390+2158– 945–4565) kJ
= 1166 kJ
Highly exothermic, used as rocket fuel
103
Q.29(b)
CH4(g) + H2O(g)  CO(g) + 3H2(g)
Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
ΔH
4C – H
4413
CO
2O – H
2463
3H – H
o
reaction

3436
(4413 + 2463 – 1072 – 3436) kJ
= 198 kJ
104
1072
Q.29(c)
2O3(g)  3O2(g)
Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
ΔH
2O – O
2146
2O = O
2498
o
reaction

(2146 + 2498 – 3498) kJ
= 206 kJ
105
3O = O
3498
What assumptions have been made ?
1. Bond enthalpies are transferable.
2. O3 has one O – O single bond and one O = O double
bond
O
O
O

O
O
O
In fact, there is delocalization of electrons making the
two O – O bonds identical with a bond order of 1.5
106
Q.29(d)
4CH3NHNH2(g) + 5N2O4(g)  4CO2(g) + 12H2O(g) + 9N2(g)
Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
12C – H
107
12413
8C = O
4C – N
4305
24O – H
9N – N
9163
9N  N
12N – H
12390
10N – O
10163
10N = O
10594
8740
24463
9945
Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
12C – H
12413
8C = O
4C – N
4305
24O – H
9N – N
9163
9N  N
12N – H
12390
10N – O
10163
10N = O
10594
o
ΔHreaction

8740
24463
9945
(19893 – 25537) kJ
= 5644 kJ Experimental value = -5116 kJ
Energetics, Q.10
108
Q.10
4CH3NHNH2(l) + 5N2O4(l)
H
5Hf[N2O4(l)]
4Hf[CH3NHNH2(l)]
4CO2(g) + 9N2(g) + 12H2O(l)
4Hf[CO2(g)]
12Hf[H2O(l)]
4C(graphite) + 12H2(g) + 4N2(g) + 5N2(g) + 10O2(g)
By Hess’s law,
H = [4(393) + 12(286) - 4(+53) – 5(20)] kJ
= 5116 kJ
109
What assumptions have been made ?
1. Bond enthalpies are transferable.
2. N2O4 has two N – O single bonds and two N = O
double bonds
N
O
O
O

O

N
N
N
O
O
O
O
O
O
N
O
N
O
In fact, there is delocalization of electrons making all
four N – O bonds identical with a bond order of 1.5
110
Estimate the enthalpy change of the following reaction.
Given : ΔHfo[NH4ClO4 (s)] = 295 kJ mol1
2NH4ClO4(s)
o
ΔHreaction
2(295)
N2(g) + Cl2(g) + 2O2(g) + 4H2O(g)
2E(O=O) + 4E(H-H) 8E(O-H)
N2(g) + Cl2(g) + 2O2(g) + 2O2(g) + 4H2(g)
Hreaction = 2E(O=O) + 4E(H-H) 8E(O-H) 2(295)
= 374 kJ
111
Estimate the lattice enthalpy of NH4ClO4.
NH4+(g) + ClO4(g)
ΔHLo[NH4ClO4 (s)]
Given : ΔHfo[NH4ClO4 (s)] = 295 kJ mol1
First IE of H = 1312 kJ mol1
First EA of O = 141 kJ mol1
E(Cl=O) = 533 kJ mol1
E(Cl-O) = 209 kJ mol1
E(Cl-Cl) = 242 kJ mol1
112
NH4ClO4(s)
Estimate the lattice enthalpy of NH4ClO4.
NH4+(g) + ClO4(g)
ΔHLo[NH4ClO4 (s)]
NH4ClO4(s)
½E(Cl-Cl)+2E(O=O)
-E(Cl-O)-3E(Cl=O) + 1st EA[O]
½E(NN)+2E(H-H)
ΔHfo
= 295
-4E(N-H) +1st IE[H]
½N2(g) + 2H2(g) + ½Cl2(g) + 2O2(g)
ΔHLo = -295-½E(NN)-2E(H-H)+4E(N-H)-1st IE[H]
- ½E(Cl-Cl)-2E(O=O)+ E(Cl-O)+3E(Cl=O)- 1st EA[O]
= 556 kJ
113
Relationship between bond strength and bond length
Bond strength :
- determined by B.D.E. or bond enthalpy
- depends on the balance between
attractive forces and repulsive forces within the
molecule
Attractive forces between bond pairs and bonding
nuclei strengthen a covalent bond.
Repulsive forces between lone pairs on adjacent
bonding atoms weaken a covalent bond.
114
Relationship between bond strength and bond length
Bond length :
- inter-nuclear distance between bonding atoms
- determined by electron diffraction or
X-ray diffraction
115
Relationship between bond strength and bond length
Bond length depends on
1. the number of electron pairs involved in the
covalent bond (bond order)
2. the sizes (covalent radii) of the bonding atoms
116
Bond lengths listed in data books are average
values
 Bond lengths depend on environment.
Deviation of actual bond lengths from the average
ones are always small
 Bond lengths are transferable
117
Molecule
Bond length / nm
HO – H
0.0958
H–O–O–H
0.0960
O
H
0.0950
C
O
CH3O – H
118
H
0.0956
Molecule
Bond length / nm
diamond
0.15445
CH3 – CH3
0.1536
CH3 – CHF2
0.1540
O
H3C
0.1500
C
H
119
Molecule
Bond length / nm
CH3 – H
0.1091
CH3CH2 – H
0.1107
CH2=CH – H
0.1087
C
120
H
0.1084
Relationship between bond strength and bond length
1. Single Bond vs Multiple Bond (between atoms
of the same kinds)
Bond
CC
CC
CC
Bond enthalpy
(kJ mol-1)
348
612
837
Bond length/nm
0.154
0.134
0.120
Bond length : - Since the bonding atoms are of the same kind,
bond length only depends on bond order.
121
Bond
CC
CC
CC
Bond enthalpy
(kJ mol-1)
348
612
837
Bond length/nm
0.154
0.134
0.120
Bond length : - Since the bonding atoms are of the same kind,
bond length only depends on bond order.
- As the bond order increases, the
electrostatic attraction between bond pairs
and nuclei increases
 bonding atoms are drawn closer together
 shorter bond length
122
Bond
CC
CC
CC
Bond enthalpy
(kJ mol-1)
348
612
837
Bond length/nm
0.154
0.134
0.120
Bond strength : Since there is no lone pair on C,
 bond strength mainly depends on the
attraction between bond pairs and nuclei
which in turn depends on bond order.
123
Bond
CC
CC
CC
Bond enthalpy
(kJ mol-1)
348
612
837
Bond length/nm
0.154
0.134
0.120
 bond
 bond
E(C=C) < 2E(C – C)
 bond
C – C  bond is stronger than C – C  bond
124
Bond
CC
CC
CC
Bond enthalpy
(kJ mol-1)
348
612
837
Bond length/nm
0.154
0.134
0.120
 bond
 bond
E(CC) < 3E(C – C)
Two  bonds
C – C  bond is stronger than C – C  bond
125
Bond
NN
OO
FF
Bond enthalpy
(kJ mol-1)
944
496
158
Bond length/nm
0.110
0.121
0.142
Bond length : NN < O=O < F-F (atomic size : N > O > F)
- Bond order is the dominant factor
126
Bond
NN
OO
FF
Bond enthalpy
(kJ mol-1)
944
496
158
Bond length/nm
0.110
0.121
0.142
Bond strength :  greatly from N2 to F2 because
1. bond order  from N2 to F2
2. repulsive forces between lone pairs on
adjacent bonding atoms  from N2 to F2
127
N
N
O
O
F
F
Increasing repulsive forces between lone
pairs on adjacent bonding atoms
128
2. Halogens
Bond
FF
ClCl
BrBr
II
Bond enthalpy
(kJ mol-1)
158
242
193
151
Bond length/nm
0.142
0.199
0.228
0.267
Bond length : Depends mainly on size of bonding atoms
Shorter bond length doesn’t mean stronger bond.
129
2. Halogens
Bond
FF
ClCl
BrBr
II
Bond enthalpy
(kJ mol-1)
158
242
193
151
Bond length/nm
0.142
0.199
0.228
0.267
Bond strength : ClCl > BrBr > II
It is because the attractive force between bond
pair and nuclei increases as the size of the
bonding atoms decreases.
130
2. Halogens
Bond
FF
ClCl
BrBr
II
Bond enthalpy
(kJ mol-1)
158
242
193
151
Bond length/nm
0.142
0.199
0.228
0.267
Bond strength : FF < ClCl > BrBr > II
It is because F atoms are so small that the
repulsion between lone pairs on adjacent F atoms
becomes dominant
131
3. Hydrogen halides
Bond
HF
HCl
HBr
HI
Bond enthalpy
(kJ mol-1)
565
431
366
299
Bond length/nm
0.109
0.135
0.151
0.171
Bond length : -
Depends mainly on size of bonding atoms
132
3. Hydrogen halides
Bond
HF
HCl
HBr
HI
Bond enthalpy
(kJ mol-1)
565
431
366
299
Bond length/nm
0.109
0.135
0.151
0.171
Bond strength : HF > HCl > HBr > HI It is because
1. the attraction between nuclei and bond pair 
as the size of bonding atoms 
2. there is no lone pair on H atom,
thus no repulsion between lone pairs.
133
Q.30
Bond
CH
CC
OO
Bond enthalpy
(kJ mol-1)
415
612
496
Bond length/nm
0.109
0.154
0.121
(a) Bond length : C–H < C=C
because H atom is smaller than C atom
Bond strength : C=C > C–H
because more bond pairs are involved in C=C.
134
Q.30
Bond
CH
CC
OO
Bond enthalpy
(kJ mol-1)
415
612
496
Bond length/nm
0.109
0.154
0.121
(b) Bond length : O=O < C=C
because O atom is smaller than C atom
Bond strength : O=O < C=C
because there are repulsive forces between
lone pairs on adjacent oxygen atoms.
135
Q.31
NN < CC
N is smaller than C
136
Q.31
NN < CC < O=O
CC has a greater
bond order
137
Q.31
NN < CC < O=O < N=N < C=C
Atomic size : O < N < C
138
Q.31
NN < CC < O=O < N=N < C=C < O–O
C=C has a greater
bond order
139
Q.31
NN < CC < O=O < N=N < C=C < O–O < N–N < C–C
Atomic size : O < N < C
140
Covalent Radius
•
141
Half the internuclear distance between two
singly and covalently bonded atoms of the
same kind in a molecule
The covalent radii (in nm) of some elements
142
Van der Waals’ Radius
•
Half the internuclear distance between two
atoms in adjacent molecules which are not
chemically bonded.
I2
143
Atomic radius  from left to right across a period because
ENC experienced by bonding electrons  from left to right.
144
Atomic radii of noble gases were obtained by
extrapolation (smaller values)
145
Calculated
value
0.031
0.038
0.071
0.088
0.108
The covalent radii (in nm) of some elements
146
Alternatively, van der Waals’ radii of noble gases
are taken as the atomic radii (larger values).
0.140
0.154
0.188
0.202
0.216
The covalent radii (in nm) of some elements
van der
Waals’ radius
147
Additivity rule of covalent radii
Assumption : Electrons are equally shared between A and B
Pure covalent bond
148
Bond
CBr in
CBr4
CF in
CF4
CO in
CH3OH
CO in
CO2
Experimental
value/nm
0.1940
0.1320
0.1430
0.1160
Estimated bond
length/nm
0.1910
0.1480
0.1510
0.1275
% deviation
-1.54%
12.12%
5.59%
9.91%
Failure of additivity rule indicates formation of
covalent bond with ionic character due to polarization of
shared electron cloud to the more electronegative atom.
149
Bond
CBr in
CBr4
CF in
CF4
CO in
CH3OH
CO in
CO2
Experimental
value/nm
0.1940
0.1320
0.1430
0.1160
Estimated bond
length/nm
0.1910
0.1480
0.1510
0.1275
% deviation
-1.54%
12.12%
5.59%
9.91%
Polarization of a covalent bond always results in the
formation a stronger bond with shorter bond length.
+

C
150
F
Benzene : Kekule structure
H
H
H
0.134 nm
C
H
C
C
C
C
C
0.154 nm
H
H
In fact, all six carbon – carbon bonds are identical with
a bond length equal to 0.139 nm which is intermediate
between C – C bond length and C=C bond length
151
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)
(a) Predict the approximate bond lengths of
Si – H, P – H, S – H and H – Cl from the following data:
Bond
H–H
Si – Si
P – P (P4)
S – S (S4)
Cl – Cl
Bond length (nm)
0.074
0.235
0.221
0.207
0.199
(Hint: Assume that covalent radii are additive.)
152
Answer
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)
0.235
0.074
nm

nm
(a) Bond length of Si – H =
2
2
= 0.154 5 nm
0.221
0.074
nm 
nm
2
2
= 0.147 5 nm
Bond length of P – H =
0.207
0.074
nm

nm
Bond length of S – H =
2
2
= 0.140 5 nm
0.074
0.199
nm

nm
Bond length of H – Cl =
2
2
= 0.136 5 nm
153
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)
(b) The bond enthalpies of Si – H, P – H, S – H and H – Cl are
given in the following table:
Bond
Si – H
P–H
S–H
Cl – H
Bond enthalpies (kJ mol–1)
+318
+322
+338
+431
Assume the actual bond lengths are very close to that
calculated in (a), describe the relationship between bond
length and bond enthalpy.
Answer
154
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)
(b) The bond enthalpy of a covalent bond is related to the length. The
larger the bond length, the weaker the attractive force between the
two bonded atoms and the smaller is the bond enthalpy.
Back
155
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)
Why does the covalent radius of a given element
change from one compound to another compound?
The covalent radius of an atom is determined by the
size of its bonding electron cloud which may vary due to
the presence of different electron clouds and bonding
atoms in different compounds.
Back
156
8.7
Shapes of Covalent
Molecules and
Polyatomic Ions
157
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)
Valence Shell Electron Pair Repulsion
theory (VSEPR theory)
Developed by Gillespie and Nyholm
5 rules to predict the shapes of molecules
and polyatomic ions
158
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)
Rule 1
The bond pairs and lone pairs in the valence
shell arrange around the central atom in
such a way as to minimize the electrostatic
repulsion .
 Maximum angular separation between
valence electron pairs around the
central atom
159
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)
A. Molecules and Polyatomic Ions without
Lone Pair Electrons on the Central Atom
•
Examples:
1. Beryllium Chloride (BeCl2) Molecule
2. Boron Trifluoride (BF3) Molecule
3. Methane (CH4) Molecule
4. Ammonium Ion (NH4+)
5. Phosphorus Pentachloride (PCl5) Molecule
6. Sulphur Hexafluoride (SF6) Molecule
160
Step 1
Draw the bond-line / Lewis structure
Step 2
Predict the shape using VSEPR theory
161
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)
1. Beryllium Chloride Molecule (BeCl2)
Electronic Diagram
Cl
Be
Shape in word
162
Shape in Diagram
Cl
Bond angle
= angle between
2 bonds
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)
1. Beryllium Chloride Molecule (BeCl2)
Electronic Diagram
Cl
Be
Shape in word
163
Shape in Diagram
Cl
Linear
180 is the maximum
possible angular
separation
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
2. Boron Trifluoride Molecule (BF3)
Electronic Diagram
Shape in Diagram
F
B
F
164
F
120 is the maximum
possible angular
separation
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
2. Boron Trifluoride Molecule (BF3)
Electronic Diagram
Shape in Diagram
F
B
F
F
Shape in word
Trigonal planar
165
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
3. Methane (CH4) Molecule
Electronic Diagram
Bond-line structure
H
H
90
H
H
C
C
H
H
H
H
166
Not the maximum
angular separation
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
3. Methane (CH4) Molecule
Electronic Diagram
Shape in Diagram
H
H
C
H
H
Shape in word
Tetrahedral
167
H
C
H
H
H
bond in plane of the paper
bond coming out of the paper
bond going behind the paper
168
The three-dimensional shape of a
methane molecule can be represented as:
169
mirror images of each other
identical
170
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
4. Ammonium Molecule (NH4+)
Electronic Diagram
Shape in Diagram
H
H
N
H
H
Shape in word
Tetrahedral
171
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
5. Phosphorus Pentachloride (PCl5) Molecule
Electronic Diagram
Bond-line structure
Cl
Cl
Cl
Cl
Cl
P
Cl
Cl
P
Cl
Cl
72
Cl
Not the maximum
angular separation
172
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
5. Phosphorus Pentachloride (PCl5) Molecule
Electronic Diagram
Shape in Diagram
Cl
Cl
Cl
P
Cl
Cl
Shape in word
Trigonal bipyramidal
173
stronger bond
weaker bond
Bond pairs occupying the axial positions have less angular
separation.
They experience stronger repulsion
P – Cl bonds at axial position are weaker and have a
greater bond length (214 pm).
174
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)
6. Sulphur Hexafluoride (SF6) Molecule
Electronic Diagram
Shape in Diagram
F
F
F
S
F
F
F
Shape in word
Octahedral
175
Making balloon models
176
Rule 2
Double bond and triple bond are
considered as one electron pair or one
negative centre.
177
Q.33
O
Linear
C
O

178
O
N
O
H
C
N
Q.33
O
Trigonal planar
S
O
O
Expansion of octet to form more bonds
179
2
Q.33
O
C
O
O
All bonds are identical due to delocalization
of electrons
2
O
C
O
180
O

2
2
O
C
O
O

O
C
O
O
Q.33
2
Trigonal planar
181
Q.33
All bonds are identical due to delocalization
of electrons
Trigonal planar
182
Q.33
O
S
O
183
All bonds are identical due to
 delocalization of electrons
O

Tetrahedral
O
Q.33
O
> 109.5
Cl
P
< 109.5
Tetrahedral (distorted)
Cl
Cl
Repulsion between double bond and single bond
> Repulsion between single bond and single bond
184
Q.33
+
H
109.5
N
H
H
H
Tetrahedral
185
Rule 3
Repulsion between electron pairs
follows the order : lone pair–lone pair > lone pair–bond pair
> bond pair–bond
pair
186
Lone pairs are attracted by one nucleus
only,
 they occupy larger space or
they are more diffused,
 they produce more repulsion.
Lone pairs are not considered in the shapes
of molecules and polyatomic ions.
Lone pairs are represented by
187
Q.34(1)
Cl
Sn
Sn
Cl
Cl
Cl
Trigonal planar
with respect to
electron pairs
188
Q.34(1)
Sn
Cl
<120
>120
Cl
Bent or V-shaped
with respect to
atoms
189
AX2E
Q.34(2)
H
N
H
N
H
H
190
H
H
Tetrahedral
with respect to
electron pairs
Q.34(2)
>109.5
N
H
H
107 H
Trigonal pyramidal
with respect to
atoms
191
AX3E
192
Q.34(3)
H
O
H
H
O
H
Tetrahedral
with respect to
electron pairs
193
Q.34(3)
AX2E2
H
104.5
O
H
b
a
a > b > 104.5
b
Bent or V-shaped
with respect to
atoms
194
195
NH2
H

H
N
O
H

N
O
H
AX2E2
Tetrahedral
w.r.t. electron pairs
Bent or V-shaped
w.r.t. atoms
196
a
N
H
b
P
H
107 H
H
90
H
H
Larger lone pair  stronger repulsion
b > a > 109.5
197
180 > a > b
Q.35
O
O
N
N
O
O
y>x a>b
O
N
180
O
N
x
O
O
198
N
a
Y
O
O
O
N
b
O
Rule 4
Single electron exerts less repulsive
forces on other electrons than
electron pair does.
Repelling power : lone pair > triple bond > double bond
> single bond > single electron
199
Q.36(1)
Cl

Cl
I
Cl
-
I
Cl
Rule 5
To minimize the great repulsion between
lone pairs,
(1) lone pairs should occupy positions with
the largest angular separation.
i.e. the equatorial position
200
Q.36(1)
Cl
I
Cl
AX2E3
201
-
Trigonal bipyramidal
w.r.t. electron pairs
Linear
w.r.t. atoms
Q.36(2)
-
Cl
Cl
I
Cl
Cl
Cl
I
Cl
Cl
Cl
Rule 5
To minimize the great repulsion between
lone pairs,
(2) lone pairs are located as far apart from
one another as possible.
202
Q.36(2)
Cl
I
Cl
Cl
Cl
AX4E2
203
Octahedral
w.r.t. electron pairs
Square planar
w.r.t. atoms
Q.37
F
F
S
F
90
F
S
F
<120
F
F
F
S
F
F
204
F
F
Q.37
S
F
F
F
F
AX4E
205
Trigonal bipyramidal
w.r.t. electron pairs
Seesaw
w.r.t. atoms
Q.37
Cl
90
Cl
I
Cl
206
Cl
Cl
I
Cl
Q.37
Cl
Cl
I
Cl
AX3E2
207
Trigonal bipyramidal
w.r.t. electron pairs
T-shaped
w.r.t. atoms
Q.37
I

I
I
180
I
I
AX2E3
I
Trigonal bipyramidal
w.r.t. electron pairs
Linear
w.r.t. atoms
208
Q.37
F
F
< 90
I
F
F
F
F
F
F
I
F
F
AX5E
Octahedral
w.r.t. electron pairs
Square pyramidal
w.r.t. atoms
209
Q.37
O
O
Xe
F
F
F
F
F
F
Xe
F
F
Lone pair and double bond must be located
as far apart from each other as possible
210
Q.37
O
< 90
Xe
F
F
F
F
F
F
O
Xe
F
F
AX5E
Octahedral
w.r.t. electron pairs
Square pyramidal
w.r.t. atoms
211
Q.37
F
F
Xe
F
F
F
Xe
F
90
F
AX4E2
F
Octahedral
w.r.t. electron pairs
Square planar
w.r.t. atoms
212
Summary : -
213
No. of negative
centre
Pattern
Shape
2
AX2
Linear
3
AX3
Trigonal planar
3
AX2E
Bent / V-shaped
4
AX4
Tetrahedral
4
AX3E
Trigonal pyramidal
4
AX2E2
Bent / V-shaped
214
No. of negative
centre
Pattern
Shape
5
AX5
Trigonal bipyramidal
5
AX4E
Seesaw
5
AX3E2
T-shaped
5
AX2E3
Linear
6
AX6
Octahedral
6
AX5E
Square pyramidal
6
AX4E2
Square planar
Hybridization and hydrid orbitals(by Linus Pauling)
Problems identified : For CH4
Experimental findings : 1. Four identical C – H bonds
2. Bond angle = 109.5
215
By VB Theory,
2s
C
2p




Only 2 single bonds can be formed.

Promotion of a 2s electron to a 2p
orbital.
2s
C*
216

2p



2s
C*

2p


1s

4H

Two types of orbital overlaps : 2s–1s head-on overlap  stronger C – H bond
2p–1s head-on overlap  weaker C – H bond
217
2s
C*

2p


1s

4H

The three 2p orbitals of carbon are at right
angles, 90, and fail to match the tetrahedral
angle of 109.5.
The spherical 2s and 1s orbitals could overlap in
any direction.
218
Solutions to problems : Atomic orbitals (2s, 2px, 2py, 2pz ) first mix to
give four identical hybrid orbitals (sp3) arranged
tetrahedrally.
2s
2p
C*




sp3 hybridization



sp3
219

Solutions to problems : sp3




1s
4H

sp3 – 1s
head-on overlaps
four identical C – H single bonds arranged
tetrahedrally.
220
sp3
hybridization
Electron density of hybrid orbital is more
concentrated in the bonding direction than
s and p orbitals
 better overlap of orbitals and stronger bond.
221
sp3
hybridization
Each sp3 hybrid orbital has
25% s character and
75% p character
222
223
Basic Concepts of the Theory of Hybridization
1.
Hybridization is a mathematical treatment of
the wavefunctions of atomic orbitals to give
new wavefunctions representing hybrid
orbitals.
2. Only orbitals of similar energies can be mixed
together to form stable hybrid orbitals.
224
3. The number of hybrid orbitals produced equals
the number of atomic orbitals mixed together.
4. Hybridization is assigned only after the
structure of a species is known.
In other words, the theory of hybridization
has no predicting power,
it is just a backward statement to
rationalize the experimental findings.
225
Interpretation of shapes of molecules and
polyatomic ions using the theory of hybridization
1. BeCl2
Experimental findings : A linear molecule in gas state (over 520oC) with
two identical Be – Cl bonds
226
2s
Be
Be*
2px

2s
2px


excitation
sp hybridization

227

Two identical sp hybrid
orbitals lying along the x-axis
sp
hybridization
Each sp hybrid orbital has
50% s character and
50% p character
228
sp
Be*

3px

2Cl

sp-3px
head-on overlaps
Two identical Be – F  bonds along the
x-axis with angular separation of 180
229
Interpretation of shapes of molecules and
polyatomic ions using the theory of hybridization
2. BF3
Experimental findings : A trigonal planar molecule with three identical
B – F bonds
230
2s
B
2px 2py 2pz


excitation
2s
B*
2px 2py 2pz



sp2 hybridization

231


Three identical sp2 hybrid
orbitals lying on the x-y plane
sp2
hybridization
Each sp2 hybrid orbital has
1/3 s character and
2/3 p character
232
2px
sp2
B*



3F

sp2-2px
head-on overlaps
Three identical B – F  bonds on the x-y
plane with angular separation of 120
233
2s
B*

2px 2py 2pz


The empty unhybridized 2pz
orbital of B is perpendicular
to the x-y plane
234
Interpretation of shapes of molecules and
polyatomic ions using the theory of hybridization
4. NH3
Experimental findings : A trigonal pyramidal molecule with three
identical N – H bonds
235
2s
N
2px 2py 2pz




sp3 hybridization
N
 


Four identical sp3 hybrid
orbitals arranged tetrahedrally
236
sp3
hybridization
237
1s
sp3
N
 


3H

sp3-1s
head-on overlap
Three identical N – H  bonds with
angular separation of 107
238
sp3
N
 


Lone pair in sp3 hybrid
orbital
239
Q.38
H2O
H
H
O
H
VSEPR
O
H
Tetrahedral
 sp3 hybridization
Backward statement !!!
240
O
2s
2px 2py 2pz

 

sp3 hybridization
1s
O
  
Lone pairs

2H

sp3-1s
head-on overlaps
Two identical O – H  bonds with
angular separation of 104.5
241
CH3NH2
H
H
C
H
H
N
N
VSEPR
C
H
H
H
H
H
H
Tetrahedral
 sp3 hybridization
242
sp3 – 1s
head-on overlap
H
N
Lone pair in
sp3 orbital
H
sp3 – sp3 head-on overlap
C
H
H
H
sp3 – 1s
head-on overlap
243
Hybridization Involving s, p, and d Atomic Orbitals
PF5
Experimental findings : A trigonal bipyramidal molecule with 3 stronger
and identical P – F bonds at equatorial positions
and 2 weaker and identical P – F bonds at axial
positions
244
3s
P
3px 3py 3pz




excitation
3s
P*
3px 3py 3pz




3d

sp3d hybridization

245




Five identical sp3d hybrid
orbitals in trigonal
pyramidal arrangement
F
F
P
F
F
F
Five identical P – F sigma bonds formed by
sp3d – 2p head-on overlaps
246
Q.39
SF6
Experimental findings : An octahedral molecule with 6 identical S – F bonds
247
S
3s
3px 3py 3pz

 

excitation
3s
S*
3px 3py 3pz



3d



sp3d2 hybridization

248





Six identical sp3d2
hybrid orbitals in
octahedral arrangement
F
F
F
S
F
F
F
Six identical S – F sigma bonds formed by
sp3d2 – 2p head-on overlaps
249
Rules for Identifying the
Type of Hybridization of
a Central Atom in a Given
Species
1. Work out the most
stable Lewis structure
of the species.
2. Predict the molecular
shape of the species
with respect to
electron pairs around
the central atom using
VSEPR theory.
250
Rules for Identifying the
Type of Hybridization of
a Central Atom in a Given
Species
3. Match the predicted
molecular shape with the
type of hybridization
according to the
following table.
251
Q.40
BH4

H
B
H
252
H
H
sp3
Q.40
SF5
F
F
F
253

S
F
F
sp3d2
Q.40
Cl
BCl3
B
Cl
Cl
sp2
F
ClF3
F
Cl
F
254
sp3d
Q.40
OSF4
F
O
S
F
F
F
XeO6
4-
O
4
O
Xe
O
O
O
O
255
sp3d
sp3d2
Multiple Bonds
256
Ethene (CH2 = CH2) Molecule
Experimental findings :
A planar molecule with 4 identical C – H bonds and
one C  C bond. All bond angles are approximately
equal to 120o
H
257
H
C
C
H
H
VSEPR
H
>120
C
H
C
H
< 120
H
trigonal planar
sp2 hybridized
2s
C
2px 2py 2pz



excitation
2s
C*
2px 2py 2pz




sp2 hybridization

258


Three identical sp2 hybrid
orbitals lying on the x-y plane
sp2


2pz

x- y plane
259

2pz orbital
sp2 – 1s head-on overlaps
 4 identical C – H  bonds
260
sp2 – sp2 head-on overlap
 C – C  bond
261
Side-way overlap
262
E(C=C) < 2E(C–C)
kJ mol1
612
2(348)
Head-on overlap is better than side-way overlap
 sigma bond is stronger than pi bond
263
The 2pz orbitals cannot come closer to each
other to have a better overlap since it will
weaken the sigma bond.
264
The electron cloud in a pi bond concentrates in regions
above and below the nodal plane (x – y plane) where there
is zero probability of finding the electrons.
265
Orbital overlaps for C2H4
266
Ethyne (CH  CH) Molecule
Experimental results :
A linear molecule with two identical C – H bonds
and one CC bond
H
C
C
H
VSEPR
H
C
C
H
Linear
sp hybridized
267
2px 2py 2pz
2s
C



excitation
2s
C*
2px 2py 2pz




sp hybridization

268

Two identical sp hybrid
orbitals lying along the x-axis
sp

269

2py
2pz


sp – sp head-on overlap
 C – C  bond
sp

270

2py
2pz


sp – 1s head-on overlaps
 two identical C – H  bonds
sp

271

2py
2pz


2py – 2py side-way overlap
 C – C  bond
sp

272

2py
2pz


2pz – 2pz side-way overlap
 C – C  bond
273
Bond
B.E.(kJ mol1)
C–C
348
C=C
612
CC
837
Less energy is given out
when the second pi bond
is formed due to overcrowded electrons
264 first pi bond
225 second pi bond
Q.40
O
C
O
VSEPR
O
C
O
Linear
C : sp hybridized
274
Q.40
O
O
C
O
VSEPR
2s
2px 2py 2pz




sp2 hybridization
sp2
  
275
2pz

O
C
O
Linear
C : sp hybridized
O : sp2 hybridized
sp
C


2py
2pz


sp2
2O   
2p

sp – sp2 head-on overlaps  Two identical C – O  bonds
276
sp
C


2py
2pz


sp2
O   
2py – 2py side-way overlap  C – O  bond
277
2py

sp
C


2py
2pz


sp2
O   
2pz – 2pz side-way overlap  C – O  bond
278
2pz

N
N
Each N is surrounded by two negative centers
 N is sp hybridized
2s
N

2px 2py 2pz



sp hybridization
 
279
N
sp
2py
2pz
 


N
sp
2py
2pz
 


sp – sp head-on overlap  N – N  bond
280
N
sp
2py
2pz
 


N
sp
2py
2pz
 


Lone pairs are in sp hybrid orbitals
281
N
sp
2py
2pz
 


N
sp
2py
2pz
 


2py – 2py side-way overlap  N – N  bond
282
N
sp
2py
2pz
 


N
sp
2py
2pz
 


2pz – 2pz side-way overlap  N – N  bond
283
Benzene (C6H6)
The simplest aromatic hydrocarbon
Experimental Findings of Benzene : A cyclic and planar structure with six identical
C – H bonds and six identical C – C bonds.
Bond
C–C
Benzene
C=C
Bond length
(nm)
1.54
1.39
1.34
Bond order
1
1.5
2
284
Benzene (C6H6)
Electron density map
Symmetrical distribution of electrons
285
Interpretation in Terms of Valence Bond Theory
Step 1 : Most stable Lewis structure
H
H
H
H
H
H
H
H
H
H
H
286
H
Q.42
H
H
H
H
H
H
H
H
H
H
H
H
The real structure is the resonance hybrid of the
two resonance structures.
 Each C – C bond has 50% single bond character
and 50% double bond character.
287
Kekulé structure(1865)
Thiele
structure
(1899)
288
Step 2 : Type of hybridization of C
H
H
H
H
H
H
H
H
H
H
H
H
Three negative centers around each C atom
 Trigonal planar
 sp2 hybridization
289
Q.43 Sigma framework
sp2 – 1s head-on overlaps
 six identical C – H  bonds
290
Q.43 Sigma framework
sp2 – sp2 head-on overlaps
 six identical C – C  bonds
291
Q.43 Pi framework
The six unhybridized 2pz orbitals are ⊥ to the
molecular x – y plane
292
Q.43 Pi framework
2pz – 2pz side-way overlaps
 Three identical C – C pi bonds
293
294
Problems Leading to a Modification of Valence
Bond Theory
1. More resonance structures can be drawn
for benzene although they make little
contribution to the real structure of benzene.
Kekulé structure(1865)
More stable
More contribution to
the real structure
295
Dewar structure(1867)
Less stable
Less contribution to
the real structure
Q.44
Less stable
Least
stable
2pz orbitals are far apart from each other
 poor overlaps
296
Problems Leading to a Modification of Valence
Bond Theory
2. All the resonance structures drawn are based
on the assumption that bonding electrons are
localized between two atoms.
Better approach (MO theory) : Delocalization of  electrons
The six  electrons are shared by all six C atoms
297
Delocalization of Pi Electrons – A simplified
MO Theory
The six 2pz orbitals overlap with one
another to give a continuous Pi electron
cloud above and below the molecular plane.
298
In fact, the combination of six 2pz orbitals
gives rise to six molecular orbitals.
299
Anti-bonding
orbitals
     
Bonding
orbitals
300
2pz
The most stable bonding molecular orbital
which can accommodate only two electrons
301
Simple Rules for Recognizing Delocalization
of  Electrons
1. Two or more resonance structures (with the
same/similar stability) can be drawn to
represent  bond arrangement of a species
302
Simple Rules for Recognizing Delocalization
of  Electrons
2. There are more than two adjacent atoms
having p orbitals that are parallel to one
another.
p orbitals parallel to one another
 Maximum side-way overlap
p orbitals ⊥ to one another
 Minimum side-way overlap
303
Q.45
O
C
stable
O
O
C
O
O
C
O
unstable
Only ONE stable structure can be drawn
 No significant delocalization of pi electrons
304
Q.45
There are only TWO adjacent p orbitals
parallel to each other : 2py // 2py ; 2pz // 2pz
 No significant delocalization of pi electrons
305
Using Curly Arrows to illustrate the
Delocalization of  Electrons (p.60)
Rule 1 : A curly arrow represents the movement
of an electron pair (pi electron pair or
lone pair)
306
Using Curly Arrows to illustrate the
Delocalization of  Electrons (p.60)
Rule 2 : If an electron pair moves in on a new
atom, another pair must leave that atom
so that the octet rule is observed.
307
Driving Force for Delocalization of Pi Electrons
1. There is more space for electrons to distribute within
the molecule.
Repulsion between electrons is minimized.
2. Electrons can be attracted by more than two nuclei at
the same time.
308
Evidence for the Stabilization of Benzene by
Delocalization of Pi Electrons
Q.46
H
+ H–H
H
ΔHo = 120 kJ mol1
 E(C=C) + E(H–H) – E(C–C) – 2E(C–H)
309
H
+ H–H
H
ΔHo = 120 kJ mol1
 E(C=C) + E(H–H) – E(C–C) – 2E(C–H)
H
H
H
H
H
+ 3H – H
H
ΔHo  3E(C=C) + 3E(H–H) – 3E(C–C) – 6E(C–H)
= 3(120) kJ mol1 = 360 kJ mol1
310
+ 3H2(g)
Delocalization enthalpy
Enthalpy
H = 152 kJ mol1
+ 3H2(g)
H = 360 kJ mol1
H = 208 kJ mol1
Reaction coordinate
311
More examples of delocalization of pi electrons
NO3
Rule 1 : -
N
O
312
O
O
O
N
O
O
N
O
O
O
More examples of delocalization of pi electrons
N
O
O
O
O
N
O
N
O
O
O
O
N and O are sp2 hybridized such that
the 2pz orbital of N and the 2pz orbitals of adjacent O
atoms are parallel to one another
 maximum side-way overlap
313
More examples of delocalization of pi electrons
Rule 2 : -
N
N
O
O


O
314
N
O
O
O

O
O
O


N
O


O

O

O
O


N
O


O


O



N
O
Q.47
HNO3
O
O
N
O
H
O
N
O
H
O
Structures with equal stability
315
The ionizable hydrogen atom must bond to the
more electronegative oxygen
O
H
O
S
O
H
O
O
Hydroxyl group
H
O
H
316
O
S
O
H
O
N
O
H
O
O
H
O
Cl
O
C
O
H
O
O
O
H3C
C
O
317
H
O
O
N
O
H
O
N
O
H
O
O
Unstable due to greater
separation of opposite
O
formal charges
318
N
H
O
Bond order = 1.5
130
0.122 nm
0.122 nm
Bond order = 1.5
0.141 nm
Bond order = 1
319
NO3 Structures with equal stability
N
O
320
O
O
O
N
O
O
N
O
O
O
NO3
All ∠ONO = 120
All N – O bonds are identical with
a bond length of 0.121 nm and
1
a bond order of 1
3
321
CH3COO Structures with equal stability
O
H3C
O
H3C
C
O
C
O
Two identical C – O bonds
with a bond order of 1.5
322
CO32
O
O
O
O
C
C
C
O
O
O
O
Structures with equal stability
All C – O bonds are identical with a bond
1
order of 1
3
323
O
O3
Structures with equal stability
½
½
All O – O bonds are identical with a bond
order of 1.5
324
NO2
Structures with equal stability
½
½
All N – O bonds are identical with a bond
order of 1.5
325
Covalent Crystals
326
8.9 Covalent Crystals (SB p.240)
Covalent Crystals
1. Simple molecular structures
2. Molecular structures (e.g. C60)
3. Giant covalent structures
327
8.9 Covalent Crystals (SB p.240)
Substances with Simple Molecular
Structures
•
Consist of discrete molecules held
together by weak intermolecular forces
•
Atoms in a molecule are held together by
strong covalent bonds
•
Examples: H2 , O2 , H2O, CO2, I2
p.82
328
8.9 Covalent Crystals (SB p.240)
Substances with Giant Covalent
Structures
329
•
Consist of millions of atoms bonded
covalently together to give a
3-dimensional network
•
No simple molecules present
•
Examples: diamond, graphite and quartz
(silicon(IV) oxide)
8.9 Covalent Crystals (SB p.240)
1. Diamond
A diamond crystal
330
8.9 Covalent Crystals (SB p.241)
The structure of diamond
331
Similar to the structure of ZnS, p.18)
Unit cell
Each C atom is tetrahedrally
bonded to four other C
atoms
All the C – C bonds are
strong and identical
Bond angle = 109.5
Bond length = 0.154 nm
Bond order = 1
332
8.9 Covalent Crystals (SB p.241)
2. Graphite
Graphite
333
Each C atom is covalently bonded to 3 other C
atoms in the same layer.
A network of coplanar hexagons is formed
334
Weak van der Waals’ forces hold the layers
together
335
A
B
A
3D structure
336
Q.48
Why is the C – C bond length of graphite(0.142 nm)
shorter than that of diamond (0.154 nm) ?
337
C : sp2 hybridized
Firstly,
s character : sp2 > sp3
sp2-sp2 overlap is better than sp3-sp3 overlap
Strength of  bond : graphite > diamond
338
Secondly,
Each C atom has a half filled 2pz orbital ⊥ to the plane.
Side-way overlaps of 2pz orbitals give extensively
delocalized  bonds.
339
C=C > C–C > C-C
1.34 nm
340
1.42 nm
1.54 nm
8.9 Covalent Crystals (SB p.242)
Comparison of the properties of
diamond and graphite
Property
Density (g cm-3)
Hardness
Melting point (C)
Colour
Electrical conductivity
Diamond
3.51
10 (hardest)
3 827
Colourless
None
Graphite
2.27
< 1 (very soft)
3 652 (sublime)
Shiny black
High
Diamond : extremely hard
 used in drills and cutting tools
341
8.9 Covalent Crystals (SB p.242)
Comparison of the properties of
diamond and graphite
Property
Density (g cm-3)
Hardness
Melting point (C)
Colour
Electrical conductivity
Diamond
3.51
10 (hardest)
3 827
Colourless
None
Graphite
2.27
< 1 (very soft)
3 652 (sublime)
Shiny black
High
Graphite : soft and slippery
 used as pencil ‘lead’ and lubricant
342
8.9 Covalent Crystals (SB p.242)
Comparison of the properties of
diamond and graphite
Property
Density (g cm-3)
Hardness
Melting point (C)
Colour
Electrical conductivity
Diamond
3.51
10 (hardest)
3 827
Colourless
None
Graphite
2.27
< 1 (very soft)
3 652 (sublime)
Shiny black
High
Graphite : high electrical conductivity
 Used as electrode
343
Q.49
The layers in graphite are held by weak van der
Waals’ forces.
The layers can slip over each other when a force is
applied to them.
Graphite is slippery and can be used to make
lubricant
344
Q.49
The high electrical conductivity of graphite is due
to the 2pz electrons which delocalize extensively
throughout the whole structure.
Graphite is used to make electrodes.
345
8.9 Covalent Crystals (SB p.242)
Comparison of the properties of
diamond and graphite
Property
Density (g cm-3)
Hardness
Melting point (C)
Colour
Electrical conductivity
Diamond
3.51
10 (hardest)
3 827
Colourless
None
Density : diamond > graphite
C.N. :
346
4
3
Graphite
2.27
< 1 (very soft)
3 652 (sublime)
Shiny black
High
8.9 Covalent Crystals (SB p.242)
3. Quartz
347
6B
8.9 Covalent Crystals (SB p.242)
Quartz
348
No. of Si atoms = 8(⅛) + 6(½) + 4 = 8
No. of O atoms = 16
Si : O = 1 : 2
349
The END
350
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)
Example 8-7A
Example 8-7B
Check Point 8-7
351
8.1 Formation of Covalent Bonds (SB p.218)
(a) How many lone pair and bond pair electrons are present
in NH3 and H2O molecules respectively?
Answer
(a) Ammonia has one lone pair and three bond pairs of electrons.
Water has two lone pairs and two bond pairs of electrons.
352
8.1 Formation of Covalent Bonds (SB p.218)
(b) Nitrogen can only form one chloride, NCl3, while
phosphorus can form two chlorides, PCl3 and PCl5.
Explain briefly.
Answer
(b) The electronic configuration of nitrogen is 1s22s22px12py12pz1. Its
outermost shell electrons are filled in the second quantum shell.
There are no lowlying d orbitals available for nitrogen to expand octet.
It has a maximum of three half-filled p orbitals to form three bonds,
i.e. NCl3.
353
8.1 Formation of Covalent Bonds (SB p.218)
(b) The ground state electronic configuration of phosphorus is
1s22s22p63s23px13py13pz1. It has three half-filled p obitals for bond
formation. Thus, three chlorine atoms can form bonds with it to give
PCl3.
After promoting one 3s electron to the low-lying d orbitals, the excited
state electronic configuration of phosphorus becomes
1s22s22p63s13px13py13pz13d1. It now has five half-filled orbitals
available for bond formation. Therefore, five chloride atoms can form
bonds with it to give PCl5.
354
8.1 Formation of Covalent Bonds (SB p.218)
(b) Phosphorus has low-lying d orbitals which allow it to expand octet
(contain more than eight outermost shell electrons) whereas nitrogen
has not.
Back
355
8.2 Dative Covalent Bonds (SB p.220)
(a) Draw a “dot and cross” diagram for the product formed in
the reaction between an ammonia molecule and a
hydrogen chloride molecule.
Answer
(a)
356
8.2 Dative Covalent Bonds (SB p.220)
(b) There is a dative covalent bond present in a HNO3
molecule. Draw a “dot and cross” diagram of the
molecule.
Answer
(b)
357
8.2 Dative Covalent Bonds (SB p.220)
(c) State the major difference between an ordinary and a
dative covalent bond.
Answer
(c) A dative covalent bond is covalent bond in which the shared pair of
electrons is supplied by only one of the bonded atoms, whereas
electrons in an ordinary covalent bond come from both bonded atoms.
Back
358
8.3 Bond Enthalpies (SB p.222)
Why do two atoms bond together? How does covalent
bond strength compare with ionic bond strength?
Answer
They are of similar strength. For example, the lattice enthalpy of NaCl
is 771 kJ mol–1 while the H–H bond enthaly is 436 kJ mol–1. It is a
misconception that ionic bond must be stronger (or weaker) than
covalent bond.
Back
359
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) Referring to Table 8-2 on page 222, calculate the enthalpy
change for the following reactions and state whether the
reactions are endothermic or exothermic.
(i) Reaction between nitrogen and hydrogen.
N2(g) + 3H2(g) → 2NH3( g)
Answer
(a) (i)
Sum of average bond enthalpies of reactants
= E(N  N) + 3 E(H – H)
= [+944 + 3  (+436)] kJ mol–1
= +2 252 kJ mol–1
360
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (i) Sum of average bond enthalpies of products
= 6 E(N – H)
= 6  (+388) kJ mol–1
= +2 238 kJ mol–1
H = [+2 252 – (+2 328)] kJ mol–1
= –76 kJ mol–1
 The reaction is exothermic.
361
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (ii) Reaction between hydrogen and chlorine.
H2(g) + Cl2(g) → 2HCl(g)
Answer
(a) (ii) H – H + Cl – Cl → 2H – Cl
Sum of average bond enthalpies of reactants
= E(H – H) + E(Cl – Cl)
= (+436 + 242) kJ mol–1
= +678 kJ mol–1
Back
362
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (ii) Sum of average bond enthalpies of products
= 2 E(N – Cl)
= 2  (+431) kJ mol–1
= +862 kJ mol–1
H = [+678 – (+862)] kJ mol–1
= –184 kJ mol–1
 The reaction is exothermic.
363
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (iii) Complete combustion of hydrogen.
(a) (iii)
Sum of average bond enthalpies of reactants
= 2 E(H – H) + E(O = O)
= [2  (+436) + 496] kJ mol–1
= +1 368 kJ mol–1
364
Answer
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (iii) Sum of average bond enthalpies of products
= 4 E(O – H)
= 4  (+463) kJ mol–1
= +1 852 kJ mol–1
H = [+1 368 – (+1 852)] kJ mol–1
= –484 kJ mol–1
 The reaction is exothermic.
365
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (iv) Complete combustion of ethanol.
(a) (iv)
Answer
Sum of average bond enthalpies of reactants
= E(C – C) + E(C – O) + E(O – H) + 5 E(C – H) + 3 E(O = O)
= [+348 + 360 + 463 + 5  (+412) + 3  (+496)] kJ mol–1
= +4 719 kJ mol–1
366
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (iv) Sum of average bond enthalpies of products
= 4 E(C = O) + 6 E(O – H)
= [4  (+743) + 6  (+463)] kJ mol–1
= +5 750 kJ mol–1
H = [+4 719 – (+5 750)] kJ mol–1
= –1031 kJ mol–1
 The reaction is exothermic.
367
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (v) Complete combustion of octane.
(a) (v)
Answer
Sum of average bond enthalpies of reactants
= 14 E(C – C) + 36 E(C – H) + 25 E(O = O)
= [14  (+348) + 36  (+412) + 25  (+496)] kJ mol–1
= +32 104 kJ mol–1
368
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(a) (iv) Sum of average bond enthalpies of products
= 32 E(C = O) + 36 E(O – H)
= [32  (+743) + 36  (+463)] kJ mol–1
= +40 444 kJ mol–1
H = [+32 104 – (+40 444)] kJ mol–1
= –8 340 kJ mol–1
 The reaction is exothermic.
369
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
(b) Calculate the enthalpy change for the reaction
CH4(g) + H2O(g) → CO(g) + 3H2(g)
using the following bond enthalpies.
E(C – H in CH4) = +435 kJ mol–1
E(C  O in CO) = +1 078 kJ mol–1
E(H – H in H2) = +436 kJ mol–1
E(H – O in H2O) = +464 kJ mol–1
370
Answer
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)
Back
(b) CH4(g) + H2O → CO(g) + 3H2(g)
Sum of average bond enthalpies of reactants
= 4 E(C – H) + 2 E(O – H)
= [4  (+435) + 2  (+464)] kJ mol–1
= +2 668 kJ mol–1
Sum of average bond enthalpies of products
= E(C  O) + 3 E(H – H)
= [+1 078 + 3  (+436)] kJ mol–1
= +2 386 kJ mol–1
H = [+2 668 – (+2 386)] kJ mol–1 = +282 kJ mol–1
371
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237)
(a) Explain why a molecule of CCl4 is tetrahedral, but a
molecule of NCl3 is trigonal pyramidal in shape.
(a) In a CCl4 molecule, there are four bond pairs of electrons Answer
on the central
carbon atom. The bond pairs have to stay as far away as possible. They
take up the shape of a tetrahedron and thus the molecule is tetrahedral
in shape. The four electron pairs in a NCl3 molecule take up the shape
of a tetrahedron as well. However, one of the electron pairs is a lone
pair and the other three are bond pairs. The shape of a NCl3 molecule is
thus trigonal pyramidal.
372
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237)
(b) Deduce the shape of a molecule of BCl3.
Answer
(b) A BCl3 molecule has six outermost shell electrons around the central
boron atom, forming three bond pairs. The shape of the BCl3 molecule
is thus trigonal planar.
373
7.5 Ionic Radii (SB p.208)
(c) Draw the structures of molecules of XeF2, XeF4 and XeF6
where Xe is a noble gas element with eight electrons in
its outermost shell.
Answer
(c)
Back
374
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)
The following data refer to the molecules NH3, H2O and HF.
375
Molecule
Bond length (nm)
Bond angle
NH3
0.101
107 
H2O
0.096
104.5 
HF
0.092
–
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)
(a) Briefly explain the variation in bond length.
Answer
(a) The bond lengths of the three molecules decrease as follows:
H–N
H–O
H–F
0.101 nm
0.096 nm
0.092 nm
The atomic radius of H is the same in the three molecules, so the bond
lengths of the molecules depend on the size of the N, O and F atoms.
N, O and F are in the same period in the Periodic Table. Since atomic
sizes decrease across a period owing to the increase in effective
nuclear charge, the bond lengths of the three molecules decrease
accordingly.
376
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)
(b) Explain why the bond angle of H2O is less than that of
NH3.
Answer
(b)This can be explained by the valence shell electron pair repulsion theory.
The central oxygen atom in H2O has two lone pairs and two bond pairs
of electrons while the central nitrogen atom in NH3 has one lone pair
and three bond pairs of electrons.
The electrostatic repulsion between electron pairs decreases in this
order:
lone pair and lone pair > lone pair and bond pair > bond pair and bond
pair
Thus, the bond angle of H2O is less than that of NH3.
377
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)
(c) Match the following bond enthalpies to the bonds in the
above three molecules:
+562 kJ mol–l, +388 kJ mol–l, +463 kJ mol–l
Answer
(c) The bond enthalpies are:
H–N
H–O
H–F
+388 kJ mol–l
+463 kJ mol–l
+562 kJ mol–l
The bond enthalpies increase as shown owing to the decrease in bond
length and increase in polarity of bonds.
Back
378
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.238)
What are the shapes of a H2S molecule and a H3O+ ion?
Explain their shapes in terms of the valence shell electron
pair repulsion theory.
Answer
H2S molecule is V-shaped. In H2S molecule, there are two bond pairs and
two lone pairs of electrons in the outermost shell of the central sulphur atom.
All three types of electrostatic repulsion (lone pair – lone pair, lone pair –
bond pair, bond pair – bond pair) are present. The two lone pairs will stay
the furthest apart and the separation between the lone pair and a bond will
be greater that that between the two bond pairs. Therefore, the H – S – H
bond angle in the H2S molecule is about 104.5 instead of 109.5 in
tetrahedron.
379
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.238)
H3O+ ion has a trigonal pyramidal shape. In H3O+ ion, the central oxygen
atom forms two covalent bonds with two hydrogen atoms respectively. Also,
one dative covalent bond is formed between the oxygen atom and the
remaining hydrogen ion. We can regard the central oxygen atom has three
bond pairs and one lone pair of electrons. According to the valence shell
electron pair repulsion theory, the lone pair will stay further away from the
three bond pairs. The three bond pairs are in turn compressed closer
together. Thus, the H – O – H bond angles in the H3O+ ion are about 107
instead of 109.5 in tetrahedron.
Back
380
8.8 Multiple Bonds (SB p.240)
(a) Does sulphur obey the octet rule in forming a SO2
molecule? Explain your answer.
Answer
(a) In the formation of SO2 molecule, sulphur does not obey the octet rule
because sulphur has 10 electrons in its outermost shell.
381
8.8 Multiple Bonds (SB p.240)
Back
(b) Draw a “dot and cross” diagram of the hydrogen cyanide
molecule (HCN). Describe and explain the shape of the
molecule.
Answer
(b)
HCN molecules has a linear shape as the central carbon atom does not
have any lone pair electrons. In order to minimize electrostatic
repulsion, the two electron clouds of the central carbon atom are
separated at a maximum with bond angles of 180.
382