CHAPTER 4 - CHEMICAL BONDING

Solution 5) VSEPR theory :
repulsion of lone pair- lone
5. H2O pair electrons > lone pair –
bonding pair electrons
1) Lewis structure : >bonding pair-bonding pair
electrons.
Valence e- O = 6
6) The shape :
2H=2 H O H
104.5o
8
O
bonding -4 H

4 H

2) No of electron pair: V-shaped
4 electrons pair and 2 bonding pair Bond angle H-O-H: 104.5°

electrons and 2 lone pair electron 101

3) Class of molecule : AB2E2

4) Electron-pair arrangement :
tetrahedral

Solution

6. SO2 5) VSEPR theory :
repulsion of lone pair –bonding

1) Lewis structure : pair electrons > bonding pair-
Valence e- S = 6 bonding pair electrons

2O = 12 O- S+
18 O

bonding - 4

14 6) The shape :
2) No of electron pair:
O- S+ O
2 bonding pair electrons and 1 <120o
lone pair

3) Class of molecule : AB2E

V-shaped

4) Electron-pair arrangement : Bond angle O-S-O: <120 °

Trigonal Planar

102

Solution 5) VSEPR theory :
repulsion of bonding pair-
7. PCl5 bonding pair electrons
are equal.
1) Lewis structure :
6) The shape :
Valence e- P = 5 Cl
Cl
5 Cl = 35 Cl P Cl Cl P Cl
40
Cl120o Cl
bonding - 10 Cl Cl
Trigonal bipyramidal
30 Bond angle Cl-P-Cl: 120°, 90o

2) No of electron pair: 103

5 electrons pair and all are bonding

pair electrons

3) Class of molecule : AB5

4) Electron-pair arrangement :
Trigonal bipyramidal

Solution 5) VSEPR theory :
repulsion of bonding pair-
8. SF6 bonding pair electrons
are equal.
1) Lewis structure :
6) The shape :
Valence e- S = 6 F
FF
6 F = 42 FS F F SF
48
FF
bonding - 12 F FF
Octahedral
36 Bond angle F-S-F: 90o

2) No of electron pair: 104

6 electrons pair and all are bonding

pair electrons

3) Class of molecule : AB5

4) Electron-pair arrangement :
octahedral

Solution 5) VSEPR theory :
repulsion of lone pair –bonding
9. SF4 pair electrons > bonding pair-
bonding pair electrons
1) Lewis structure :
6) The shape :
Valence e- S = 6 F
F SF
4 F = 28 S F FF
34 F
See-saw
bonding - 8 F Bond angle F-S-F: <120°, <90°
26

2) No of electron pair:

5 electrons pair there are 4 bonding

pair electrons and 1 lone pair

electrons

3) Class of molecule : AB4E

4) Electron-pair arrangement :
trigonal bipyramidal

105

Solution F 5) VSEPR theory :
repulsion of lone pair- lone
10. ClF3 Cl F pair electrons > lone pair –
F bonding pair electrons
1) Lewis structure : >bonding pair-bonding pair
Valence e- Cl = 7 electrons.
3 F = 21
28 6) The shape :
bonding - 6

octet terminal 22 F
4 - 18
90o

F Cl F

2) No of electron pair: T-shaped
5 electrons pair , there are 3 bonding pair

electrons and 2 lone pair electrons Bond angle Cl-P-Cl: ~90o

3) Class of molecule : AB3E2

4) Electron-pair arrangement : 106
Trigonal bipyramidal

Solution F 5) VSEPR theory :
F Br F repulsion of lone pair –bonding
11. BrF5 pair electrons > bonding pair-
bonding pair electrons
1) Lewis structure : F F
Valence e- Br = 7 6) The shape :

5 F = 35 F
FF
42
Br <90o
bonding - 10
FF
32
Square pyramidal
octet terminal - 30 Bond angle F-Br-F: <90o

2
2) No of electron pair:

6 electrons pair, there are 5 bonding

pair electrons and 1 lone pair

electrons

3) Class of molecule : AB5E

4) Electron-pair arrangement : 107
octahedral

Solution 5) VSEPR theory :
repulsion of lone pair- lone
12. I3- - pair electrons > lone pair –
bonding pair electrons
I I+ I >bonding pair-bonding pair
electrons.
1) Lewis structure :
Valence e- 3I = 21 6) The shape :
charge = 1
22 180O
bonding - 4
18 I I+ I
linear
octet terminal - 12
6 Bond angle I-I-I: 180o

2) No of electron pair: 108
5 electrons pair, there are 2 bonding
pair electrons and 3 lone pair
electrons

3) Class of molecule : AB2E3

4) Electron-pair arrangement :
trigonal bipyramidal

Solution

13. NO2- 5) VSEPR theory :
repulsion of lone pair –bonding
1) Lewis structure :
pair electrons > bonding pair-
Valence e- N = 5
- bonding pair electrons

2O = 12 O- N O
charge = 1

18

bonding - 4 6) The shape :

14

octet terminal - 12 O- N
<120o
2 O
2) No of electron pair:

2 bonding pair electrons and 1

lone pair V-shaped
Bond angle O-N-O: <120 °
3) Class of molecule : AB2E

4) Electron-pair arrangement : 109
Trigonal Planar

Solution 5) VSEPR theory :
repulsion of lone pair- lone
14. XeF2 pair electrons > lone pair –
bonding pair electrons
1) Lewis structure : >bonding pair-bonding pair
electrons.
Valence e- Xe = 8
6) The shape :
2 F = 14 F Xe F
22 180O

bonding - 4 F Xe F
linear
18
Bond angle F-Xe-F: 180o
octet terminal - 12
110
6
2) No of electron pair:

5 electrons pair, there are 2 bonding

pair electrons and 3 lone pair

electrons

3) Class of molecule : AB2E3

4) Electron-pair arrangement :
trigonal bipyramidal

Solution F F 5) VSEPR theory :
FI repulsion of lone pair –bonding
15. IF5 pair electrons > bonding pair-
bonding pair electrons
1) Lewis structure : F F
Valence e- I = 7 6) The shape :

5 F = 35 F
FF
42
I <90o
bonding - 10
FF
32
Square pyramidal
octet terminal - 30 Bond angle F-I-F: <90o

2
2) No of electron pair:

6 electrons pair, there are 5 bonding

pair electrons and 1 lone pair

electrons

3) Class of molecule : AB5E

4) Electron-pair arrangement : 111
octahedral

Solution 5) VSEPR theory :
repulsion of bonding pair
16. CH2Cl2 electrons are equal

1) Lewis structure : Cl 6) The shape :
Valence e- C = 4
2Cl = 14 Cl C H Cl 109.5o
2H= 2 H
20
bonding - 8 H C
12 Cl

2) No of electron pair: H
4 electron pairs and all are bonding
pair electrons tetrahedral
Bond angle Cl-C-Cl: 109.5 °

3) Class of molecule : AB4

4) Electron-pair arrangement :
Tetrahedral

112

MOLECULAR SHAPE

Central atom WITHOUT lone Central atom WITH lone pair
pair (BASIC SHAPE)
• BENT / V-SHAPED
• LINEAR • TRIGONAL PYRAMIDAL
• TRIGONAL PLANAR • DISTORTED TETRAHEDRAL
• TETRAHEDRAL • T-SHAPED
• TRIGONAL BIPYRAMIDAL • LINEAR
• OCTAHEDRAL • SQUARE PYRAMIDAL
• SQUARE PLANAR

113

Bond angles in
i. methane, CH4
ii. ammonia, NH3
iii. water , H2O

114

H

H CH
H

Methane, CH4
 Central atom C has 4 bonding pair electrons

 Repulsions between bonding pairs are equal

 Every H-C-H bond angle is 109.5

115

H NH
H

Ammonia, NH3
 Central atom N has 3 bonding pairs and 1 lone pair

electrons.
 According to VSPER theory, repulsion between lone pair-

bonding pair > bonding pair-bonding pair.
 Lone-pair repels the bonding-pair more strongly, the

three bonds are pushed closer together.
 Every H-N-H bond angle decreases to 107.3

116

OH
H

Water, H2O
 Central atom O has 2 bonding pairs and 2 lone pairs

 Repulsion between lone pair-lone pair > lone pair-
bonding pair > bonding pair-bonding pair electrons

 The lone pairs tend to be as far from each other as
possible. The two O-H bonds are pushed toward each
other.

 H-O-H bond angle decreases to 104.5

117

Bond and Molecules polarity

• A covalent bond can be polar or non-polar.
• A non-polar bond will form a non-polar

molecule however, molecules with polar
bond may form polar or non-polar
molecule.

118

Cl Cl

Non-polar covalent bond
• Involves two atoms of same electronegativity

(Cl-Cl).
• It is formed when bonding electrons are

shared equally between the two atoms.

119

Polar Covalent Bond
• Polar covalent bond forms when the bonding pair

electrons are not equally shared between two
atoms of different electronegativity .

• The atom with greater electronegativity draws the
electron pair closer to it, thus forms a polar
covalent bond.

120

Example
The polar bond of HCl

(δ+) H Cl (δ-)

• Bonding electrons are shared unequally between
two atoms of different electronegativity (H-Cl).

• Cl is more electronegative than H, thus attracts the

bonding electrons to it and bears a partial negative

charge (δ-).

• The hydrogen atom experiences electron deficient

and has a partial positive charge (δ+). 121

(δ+) H Cl (δ-)

electron density

• The shift of the electrons density is shown by a
cross arrow, ( ) from the δ+ to the δ- end,
which indicate the bond dipole.

(δ+) (δ-)

H Cl

The bond dipole is a vector quantity which has
both magnitude and direction

122

Polar Molecules and Dipole Moments
• A quantitative measures of the polarity of a bond

is a dipole moment (μ)
μ = Qr

where ,
Q = the magnitude of the charges
r = the distance that separates the
charges.

• A dipole moment, μ is measured in debye (D)
units.

123

• A polar molecule has a dipole moment which
is (μ0).

• A non-polar molecule has no dipole moment
which is (μ=0). It occurs when the bond dipole or
the resultant dipole counterbalance each other.

124

• The dipole moment can be determine
qualitatively from the nett or resultant of bond
dipole.

Example :
 2 bond dipoles, a and b in the water molecule
produces a net or resultant bond dipole upwards.

 therefore, μ0 so, water molecule is polar.

a a
b resultant bond dipole

b

125

Polarity of molecules
• Molecules can exist as :

a) diatomic ( only two atoms involved) or
b) polyatomic (more than two atoms) molecules.

• For a diatomic molecule, the atoms concerned
could be of same or different electronegativity.

126

• Diatomic molecules of the same element :
 do not have bond dipole
 form non-polar molecules

e.g : H2, N2 and Cl2

Cl2 : Cl  Cl
 = 0,  non-polar bond
 non-polar molecule

127

• Diatomic molecules of different electronegativity

(different elements ) :

 have bond dipoles (polar bonds)
 have resultant bond dipole
 thus, dipole moments (μ0)
 so, the molecules are polar .

• e.g: HF, HCl, NO and CO

μ0

128

• For a polyatomic molecule,

the dipole moment is determine by the polarity
of the bonds and the molecular geometry.

even if the molecule has polar bonds, the
molecules will not necessarily have a dipole
moment.

129

Example

Determine the polarity of the following molecules
i. carbon dioxide, CO2
ii. carbon tetrachloride, CCl4
iii. chloromethane, CH3Cl
iv. ammonia, NH3

130

Solution

i. CO2

• The CO2 molecular geometry is linear.

• Oxygen is more electronegative than carbon so
there is a bond dipole in each carbon-oxygen
bond.

• The bond dipoles are equal in magnitude and
they pointed at the opposite direction, leading to
a cancellation of their effects.

• The dipole moment is zero (μ=0), CO2 is 131
nonpolar.

ii) CCl4

Cl resultant dipole bond

bond dipole C

Cl Cl
Cl

• Carbon tetrachloride, CCl4 is a symmetrical
molecule with tetrahedral shaped.

• Cl is more electronegative than C, therefore each of
C-Cl bond is polar.

• Due to its symmetrical structure the resultant bond
dipoles cancelled each other.

• Therefore, the molecule has no dipole moment
(μ=0); CCl4 is a nonpolar molecule.
132

iii) CH3Cl Cl

C H

H
H

• Molecular geometry: tetrahedral

• Cl is more electronegative than C, C is more

electronegative than H 133
• Has resultant bond dipole.
• Dipole moment (μ0)
• Therefore, CH3Cl is a polar molecule.

iv) NH3

• Molecular geometry : trigonal pyramidal
• N is more electronegative than H
• Form resultant bond dipole
• Dipole moment μ0
• NH3 is a polar molecule.

134

Steps to determine the polarity of a molecule:

Draw the correct Lewis structure find out
the class of molecule draw the correct
molecular shape draw the respective
bond dipoles find out the resultant bond
dipole state the dipole moment, 
state the polarity

135

Example
Determine the polarity of the following molecules :

HBr ; SO3 ; ClF3 ; CF4 ; H2O ; SO;2
XeF4 ; NF3 ; Cis-C2H2Cl2 ; trans-C2H2Cl2

136

ANSWER
i) HBr

H Br

• Molecular shape : linear
• Br is more electronegative than H
• Form resultant bond dipole
• Dipole moment μ0
• HBr is a polar molecule.

137

ANSWER O-
ii) SO3
S
O- O

• Molecular shape : trigonal planar

• O is more electronegative than S

• Due to its symmetrical structure the resultant
bond dipoles cancelled each other.

• Dipole moment μ=0

• SO3 is a nonpolar molecule. 138

iii) ClF3

F
F Cl F

• Molecular shape: T-shaped
• F is more electronegative than Cl
• Has resultant bond dipole.
• Dipole moment (μ0)
• Therefore, ClF3 is a polar molecule.

139

iv) CF4

F resultant dipole bond

bond dipole C

FF
F

• Carbon tetrafluoride, CF4 is a symmetrical
molecule with tetrahedral shaped.

• F is more electronegative than C, therefore each of
C-F bond is polar.

• Due to its symmetrical structure the resultant bond
dipoles cancelled each other.

• Therefore, the molecule has no dipole moment
(μ=0); CF4 is a nonpolar molecule.
140

v) H2O

O
H

H

• Molecular shape: V-shaped
• O is more electronegative than H
• Has resultant bond dipole.
• Dipole moment (μ0)
• Therefore, H2O is a polar molecule.

141

vi) SO2

S+
O- O

• Molecular shape: V-shaped
• O is more electronegative than S
• Has resultant bond dipole.
• Dipole moment (μ0)
• Therefore, SO2 is a polar molecule.

142

vii) XeF4

XeF F2-

FF

• Xenon tetrafluoride, XeF4 is a unsymmetrical
molecule with see-saw shaped.

• F is more electronegative than Xe, therefore each
of Xe-F bond is polar.

• Due to its unsymmetrical structure the resultant
bond dipoles cannot cancelled each other.

• Therefore, the molecule has dipole moment (μ≠0);
XeF4 is a polar molecule.

143

viii) NF3

N F
F

F

• Molecular shape : trigonal pyramidal
• F is more electronegative than N
• Form resultant bond dipole
• Dipole moment μ0
• NF3 is a polar molecule.

144

ix) Cis-C2H2Cl2 Cl

C Cl
H C

H

• Molecular shape : trigonal planar
• Cl is more electronegative than C and C is

more electronegative than H
• Form resultant bond dipole
• Dipole moment μ0
• Cis-C2H2Cl2 is a polar molecule.

145

x) trans-C2H2Cl2

Cl H
C
C Cl
H

• Molecular shape : trigonal planar

• Cl is more electronegative than C and C is more
electronegative than H

• The resultant bond dipoles cancelled each other.

• Dipole moment μ=0

• Trans-C2H2Cl2 is a polar molecule.

146

Conclusion

Factors that affect the polarity of molecules
are:

• molecular shape.
• electronegativity of the bonded atoms.

147

BOND

POLAR BOND 2 type of bonds NON-POLAR BOND

may formed NON-POLAR MOLECULES

NON-POLAR POLAR
MOLECULES MOLECULES

Symetrical molecules (has the Non-symetrical molecules
same terminal atom) i. basic shape with different

i. basic shape terminal atom
ii. molecules with lone pairs : ii. molecules with lone pairs
* linear (from trigonal

bipyramidal)
* square planar

148

149

150