SK015

 Elements in the same group have the Lewis symbol of elements in period 2
same valence electronic configurations
thus, similar Lewis symbols. Group Number of valence Element Lewis
electrons symbol
7 1 1 Li
2 2 Be 8
13 B
3

14 4 C

15 5 N

16 6 O

17 7 F

18 8 Ne

4.1.2 Octet Rule  Types of electronic configurations stabilities
of cations and anions are:
 Octet rule states that atoms tend to form
bonds to obtain 8 electrons in the valence  a) noble gas configuration
shell (as in atom of a noble gas)  b) pseudo-noble gas configuration
 c) half-filled orbital configuration
 Atoms combine by gaining, losing or
sharing electrons to achieve stability as the
noble gas configuration.

9 10

Electronic Configuration of Cations and Anions  Group 15, 16 and 17 elements accept electrons to
form anions with noble gas configurations
a) Noble gas configuration Example:
Group 1, 2 and 13 elements donate valence O : 1s22s22p4
electrons to form cations with noble gas O2─: 1s22s22p6
configurations (isoelectronic with neon)

Example: Cl : 1s22s22p63s23p5
Na : 1s22s22p63s1 Cl─ : 1s22s22p63s23p6
Na+ : 1s22s22p6 (isoelectronic with Ne)
(isoelectronic with Ar)
Ca : 1s22s22p63s23p64s2
Ca2+ : 1s22s22p63s23p6 (isoelectronic with Ar1)1 12

98

b) Pseudo-noble gas configuration c) Half-filled orbital configuration

 d block elements donate electrons from 4s  d block elements donate electrons to achieve
orbitals to form cations with pseudo-noble gas the stability of half-filled orbitals
configuration. Example:
Mn : 1s22s22p63s23p63d54s2
Example: Mn2+ : 1s22s22p63s23p63d5
Zn : 1s22s22p63s23p63d104s2 (stability of half-filled 3d orbital )
Zn2+ : 1s22s22p63s23p63d10
(pseudo-noble gas configuration ) Fe : 1s22s22p63s23p63d64s2
Fe3+ : 1s22s22p63s23p63d5
13
(stability of half-filled 3d orbital)

14

4.1.3 Bonds formation using Lewis a) Ionic bond (Electrovalent bond)
Symbols
 Ionic bond (or electrovalent bond) is the
a) Ionic bond / Electrovalent bond electrostatic attraction between
b) Covalent bond positively and negatively charged ions
c) Dative bond / Coordinate covalent bond in an ionic compound.

15  Ionic compound formed when valence
electrons are transferred between atoms
(metal to nonmetal) to give two oppositely
charged ions that attract each other.

16

Example 1: NaCl  The electrostatic forces between Na+ and
Cl- produce ionic bond
Sodium, an electropositive metal, tends to remove
its valence electron to form Na+ ion with the stable  These two processes occur simultaneously
noble gas configuration.
 The formation of ionic bond can be shown
Chlorine, an electronegative element (non-metal), using Lewis symbol:
tend to accept electron from Na to obtain stable
noble gas configuration. +

17 18

99

Example 2: CaCl2 Exercise:

++ 2 By using Lewis symbol, show how the ionic
bond is formed in the compounds below.
 Ca atom (metal) transfer 2 electrons, one to each
chlorine atom, it become a Ca2+ ion with the stable (a) MgO
electronic configuration of noble gas. (b) KF
(c) Na2O
 At the same time each chlorine atom (non-metal) gained
one electron becomes a Cl- ion to achieve noble gas 20
configuration

• The electrostatic attraction between Ca2+ and Cl─ ions
formed ionic bond.

19

Answer: Answer :
(a) MgO

(b) KF + -

Mg + O Mg 2+ O 2- K +F KF

 Mg atom transfer 2 electrons to achieve octet  K atom remove 1 electron to achieve octet
configuration and become a Mg2+ ion. configuration and become a K+ ion.

 The 2 electrons are accepted by an oxygen atom  The electron is accepted by a fluorine atom as
to achieve the octet configuration and formed the to achieve the octet configuration and formed
oxide ,O2- ions. chloride,F- ion.

• The electrostatic attraction between Mg2+ and O2- • The electrostatic attraction between K+ and F-
ions created ionic bond.
ions created ionic bond.
21 22

Answer : Properties of Ionic compounds

(c) Na2O + 2- Ionic bond is very strong, therefore ionic
Na + O + Na compounds:
2 Na O

 Each Na atom remove 1 electron to achieve octet a) Have very high melting and boiling points
configuration and become a Na+ ion.
b) Solid at room temperature
 The two electrons are accepted by a fluorine atom
as to achieve the octet configuration and formed c) Hard and brittle
oxide,O2- ion.
d) Soluble in water
• The electrostatic attraction between Na+ and O2-
ions created ionic bond. e) Can conduct electricity when they are in
molten form or aqueous solution because of
23
the moving ions 24

100

b) Covalent Bond The formation of covalent bonds can be represented
using Lewis symbol.
• A chemical bond formed by the sharing of one or
more electron pairs between two atoms. Example 1: H H or H H

• Involve non-metal atoms (atoms of the same or H +H
similar electronegativity)
By sharing the electron, each hydrogen atom achieve
Why should two atoms share electrons? the stable noble gas electronic configuration of
To gain stability by forming noble gas helium (eight electrons in its valence shell)

configuration 26

25

Example 2: F F or F F c) Coordinate Covalent Bond (Dative Bond)

F +F  Dative bond is a bond in which the pair of
shared electrons is supplied by one of
• By sharing the electrons, each chlorine atom has the two bonded atoms
eight electrons in its valence shell.
 Involve overlapping of a full orbital and an
• A bonding pair electrons can be represented by empty orbital
two dots or a single line.
28
• lone pairs electrons are electron pairs that are not
involved in bonding

27

Requirement to form a dative bond: Example 1: H3O+
i. Donor atoms should have at least one
H+ + OH H +
lone pair electrons H
ii. The acceptor atoms should have empty dative bond OH
H
orbitals in the valence shell.
• H+ is the acceptor atom that has empty orbital in
29 its valence shell.

• O is the donor atom that has lone pair.

• A dative bond is shown by an arrow() pointing
from the donor atom to the acceptor atom.
30

101

Example 2: Exercise :
Use the Lewis symbol to show the formation of NH4+ Use the Lewis symbol to show the formation of the
Solution following compound:
i) F3BNH3
31
answer :

ii) Al2Cl6

answer :

32

4.1.4 Lewis Structure of Covalent Species Example

Steps in Writing Lewis Structures Draw the Lewis structure for each of the

1. Count total number of valence electron of atoms involved. following compounds:
- Add one electron for each negative charge.
- Subtract one electron for each positive charge. i. HF ii. CH4

2. Draw skeletal structure of the compound. iii. CHCl3 iv. NH3
- the least electronegative atom occupies the central atom.
- H atom always occupies terminal atom v. H2O vi. HCN
- draw a single covalent bond between the central atom and each of the vii. PO43 viii. NH4+
surrounding atoms. ix. C2H4

3. Complete the octet (or duplet of hydrogen atom) for all terminal atoms.

4. Place any remaining electrons on the central atom.If the central

atom is not octet, form multiple bonds by converting lone pairs from the

surrounding atoms into bonding pair with the central atom. 33 34

Solution Solution

Draw the Lewis structure for each of the following vii. PO43 viii. NH4+
compounds: H+
O-
i. HF ii. CH4 iii. CHCl3 O- P O 3- H N+ H
H
HF H Cl O-
36
HCH H CH

HH

iv. NH3 v. H2O vi. HCN ix. C2H4
HNH
OH HC N HH
H H
35 HC CH

102

 Compounds may have these kind covalent 4.1.5 Bond Length
bonds:
The distance between the nuclei of two bonded
i. Single bond. Example : ClCl atoms is called bond length.

ii. Double bond (two atoms share two C C C C CC
pairs of electrons). Example : OO 1.54 Å 1.34 Å 1.20 Å

iii. Triple bond (two atoms share three pairs • As the number of bonds between the carbon atoms
of electrons). Example : NN increase, they are held more closely and tightly
together and this causes the bond length
37 decreases.

• Thus, multiple bonds become shorter and stronger.

38

Example: Bond length, Å Formal Charge and the most plausible Lewis
1.43 structure
C─N 1.38
C=N 1.16 • Formal charge is the charge on a certain
CN 1.47 atom in a Lewis Structure.
N─N 1.24
N=N 1.10 • It can be counted as :
NN 1.43
C─O 1.23 Formal = Number of ― Total number of ― Number of
C=O 1.13 Charge valence non-bonding bonds
CO
electrons electrons involved

39 40

• Formal charge is used to find the most EXAMPLE 1:
stable/most plausible Lewis structure a) Draw all the possible Lewis structure of COCl2.
b) Predict the most plausible structure.
• The most stable/most plausible Lewis Structure
may has: 42

i. Formal charge on each atom equal/closest
to zero

ii. Formal charge as small as possible
iii. Negative formal charge on a more

electronegative atom and positive formal
charge on a more electropositive atom.

41

103

Answer: Structure 2  The sum of the formal charge in a Lewis Structure
00 0 must :
Structure 1
i. Equal zero for a neutral molecule
-1 0 +1
Example : PCl3 (formal charge = 0)
00
The most plausible structure is structure 2 because ii. Equal the magnitude of the charge for a
it shows zero formal charge for all atoms. polyatomic ion.

Example: NH4+ (formal charge = +1)

43 44

EXAMPLE 2: ANSWER: - Structure 3 -

a) Draw all the possible Lewis structure of SCN. a) Structure 1 Structure 2 S C N(0) (-1)
b) Predict the most plausible structure. -
(-1) (0) (+1) (-2) (0)
45
S- C N SCN -
(0) + 2-
(0)

b) The most plausible structure is structure 3 46
because more electronegative atom, N has
negative formal charge and formal charge of
each atoms closest to zero.

Exception to the Octet Rule a) Incomplete octet
Three conditions:
a) Incomplete octet  Occurs when central atom has less than 8
b) Expanded octet electrons.
c) Odd number electrons
 Elements that can form incomplete octet are:
Boron(B) , Beryllium(Be) & Aluminium(Al)

 This atoms form covalent compounds with non-
metal, although they are metals.

 This is because their ions have smaller size and
higher charge, which enable them attracting the

electrons of the anions strongly towards
themselves, resulting in sharing of bonding

47 electrons. 48

104

Example : Solution
i. Draw the Lewis structures for these compounds:
a) BeCl2 (+1) (+ 1)
b) BF3
c) AlBr3 a. (-2) Even though Be is octet, formal charge of
ii. Choose the most plausible Lewis structures Cl Be is -2 , Be is an electropositive element.
for the respective compounds. Explain Cl Be

49 (0) (0) Be is not octet, however formal
(0) charge of all atoms are zero, it is the
Cl most plausible Lewis structure.
Cl Be

• A plausible Lewis structure need to fulfilled
the appropriate formal charge of every atom.

• Thus, Be form incomplete octet

50

Solution Solution

(0) (-1) (+1) (0) (-1) (+1) Even though Al is octet, formal charge of Al
is -1, Al is an electropositive element.
b. F B- F+ Even though B is octet, formal charge of B is c. Br Al- Br+
Al is not octet, however formal charge
-1, B is an electropositive element. Br of all atoms are zero, it is the most
F plausible Lewis structure.
(0)
(0)
(0) (0) (0)
(0) (0) (0) B is not octet, however formal charge of
F B F all atoms are zero, , it is the most F Al F

F plausible Lewis structure. F

(0) (0)

• A plausible Lewis structure need to fulfilled the • A plausible Lewis structure need to fulfilled the
appropriate formal charge of every atom. appropriate formal charge of every atom.

• Thus, B form incomplete octet • Thus, Al form incomplete octet

51 52

b) Expanded Octet Example:
Draw the Lewis structure for:
 Occurs when central atom has more than 8
electrons. i) PCl5 Cl Cl
Cl P Cl
 Atoms of the elements from period 3 or higher
can expand their valence shell due to the Cl
existence of empty d orbitals.
ii) SF6 FF
 Examples of element that can form expanded
octet are phosphorous (P), sulphur (S), iodine FSF
(I), selenium (Se) FF

53 54

105

c) Odd number electrons Resonance
• The central atom of some molecule may contain
• Resonance structures are two or more plausible
an odd number of valence electrons. Lewis structures for a molecule/ion whereby cannot
• Nitrogen may form compounds that contain odd be represented accurately by only one Lewis
structure.
number electrons.
Example: Example: ozone (O3)

Nitric oxide, NO O OO OO O
Nitrogen dioxide, NO2
 These structures have the same relative placement of
55 atoms but differ in the distribution of valence electrons.

• Each structure is called a resonance structure

• Double headed arrow (↔) indicates that the structures
are resonance to each other.
56

Example: Solution
Draw the resonance structures of nitrate ion,
NO3

57 58

106

4.2 Molecular Shape Learning Outcomes
and Polarity
At the end of this lesson, students should be able to
i. VSEPR theory
ii. Electron pair arrangements a) explain Valence Shell Electron Pair Repulsion
(VSEPR) theory.
and Molecular shape
iii. Bond and molecule’s polarity b) draw the basic molecular shapes : linear, trigonal
planar, tetrahedral, trigonal bipyramidal and
59 octahedral.

c) predict the shapes of molecule and bond angles in
a given species.

d) explain bond polarity and dipole moment.
e) deduce the polarity of molecules based on the

shapes and the resultant dipole moment.

60

Valence-Shell Electron-Pair Repulsion • The electron repulsions at the central atom
(VSEPR) Theory are between lone pair – lone pair , lone pair-
bonding pair and bonding pair-bonding pair.
• Valence-shell electron-pair repulsion
(VSEPR) theory emphasizes that each group bonding pair – A lone pair
of valence electrons around the central bonding pair X
atoms is located as far away as possible from lone pair – lone pair
the others in order to minimize repulsions. repulsionA repulsion

• Central atom is surrounded by bonding pair bonding pair lone pair – bonding pair
electrons or/and lone pair electrons, thus create repulsion
repulsions between them. A

61 62

• Lone-pair electrons spread out more than • Due to these electron pair - electron pair
bonding-pair electrons, so the repulsion of lone- repulsion, molecules have different shapes
pair electrons is greater than other electron- and bond angles.
pairs.
• Thus, the VSEPR model is used to predict
• Therefore, the order of the repulsive forces the shapes of covalent molecules and
strength is polyatomic ions.

Lone pair-lone pair > lone pair- bonding pair > 64
bonding pair-bonding pair

63

107

Electron-pair arrangement and molecular • The distribution of electron-pairs around the
shape central atom are two, three, four, five and six
electron pairs with certain bond angles.
• Electron-pair arrangement is the distribution
of electron-pair about the central atom. • Thus, result five geometric shapes : linear,
trigonal planar, tetrahedral, trigonal
• It is used to predict the molecular shape of a bipyramidal and octahedral respectively.
molecule or ion.
• The electron-pairs arrangements and the bond
• Molecular shape is the geometric arrangement angles are shown in the table below :
of the atomic nuclei (the actual molecular
shape) 66

65

Number of Electron-pair Arrangement Number of Electron-pair Arrangement
electron-pairs electron-pairs

2

4

electron
pair

3

67 68

Number of Electron-pair Arrangement Number of Electron-pair Arrangement
electron-pairs electron-pairs

5 6

69 70

108

In short, the basic shape with : Molecular shape of molecule or ion

• It is important to identify the electron-pair
about the central atom to determine the
molecular shape

• Central atom are surrounded by :
i. only bonding pair electrons or
ii. bonding pair and lone pair electrons

71 72

• If the central atom has : I) Molecular shape for central atom with only
bonding pair electrons (Basic Shape)
i. only bonding pair electrons - Assign the molecule
as ABm where A (central atom), B (terminal atom) 2 electron pairs in the valence shell of central atom:
and m is intergers . The shape of the molecule
adopt the electron-pair arrangement and called the Class of Number of Number of Shape
basic shape. molecules bonding lone pairs
pairs

180°

ii. bonding pair and lone pair electrons – the AB2 2 0
shape is different from the basic shape.

So, assign as ABmEn where A (central atom), Linear
B (terminal atom), E (lone pair), m and n are

intergers 73 74

3 electron pairs in the valence shell of central atom: 4 electron pairs in the valence shell of central atom:

Class of Number of Number of Shape Class of Number of Number of Shape
molecules bonding lone pairs molecules bonding lone pairs
pairs pairs

120° AB4 4 0

AB3 3 0 Tetrahedral

trigonal planar 76

75

109

5 electron pairs in the valence shell of central atom: 6 electron pairs in the valence shell of central atom:

Class of Number of Number of Shape Class of Number of Number of Shape
molecules bonding lone pairs molecules bonding lone pairs
pairs pairs

AB5 90° 120° AB6 90° 90°

50 60

Trigonal bipyramidal Octahedral
77
78

II) Molecular shape for central atom with lone pair 4 electron pairs in the valence shell of central atom:
and bonding pair electrons
Class of Number of Number of Shape
3 electron pairs in the valence shell of central atom: molecules bonding lone pairs
pairs
Number of
Class of bonding Number of Shape
molecules pairs lone pairs

AB3E 3 1

AB2E 2 1 Trigonal pyramidal

Bent / V-shaped 80

79

4 electron pairs in the valence shell of central atom: 5 electron pairs in the valence shell of central atom:

Class of Number of Number of Shape Class of Number of Number of Shape
molecules bonding lone pairs molecules bonding lone pairs
pairs pairs

AB2E2 2 2 AB4E 4 1

Bent / V-shaped Distorted tetrahedral
81 (see-saw)
82

110

5 electron pairs in the valence shell of central atom: 5 electron pairs in the valence shell of central atom:

Class of Number of Number of Shape Class of Number of Number of Shape
molecules bonding lone pairs molecules bonding lone pairs
pairs pairs

AB3E2 3 2 AB2E3 2 3

T-shaped Linear
83
84

6 electron pairs in the valence shell of central atom: 6 electron pairs in the valence shell of central atom:

Class of Number of Number of Shape Class of Number of Number of Shape
molecules bonding lone pairs molecules bonding lone pairs
pairs pairs

AB5E 5 1 AB4E2 4 2

Square pyramidal Square planar
85 86

As a conclusion : No. of Class of No. of bonding No. of Shape
electron molecule pair electron, lone pair,
Molecules of the same electron-pair linear
arrangement, but different repulsive force pair AB2 B E Trigonal planar
(due to the presence of lone pair electrons) will 2 AB3 2 0 Bent / V-shaped
have different bond angles thus have 3 AB2E 3 0
different molecular shape. 4 AB4 2 1 Tetrahedral
AB3E 4 0 Trigonal pyramidal
87 5 AB2E2 3 1
AB5 2 2 Bent / V-shaped
6 AB4E 5 0 Trigonal bipyramidal
4 1 See-saw /distorted
AB3E2 tetrahedral 88
AB2E3 3 2
2 3 T-shaped
AB6 6 0 linear
AB5E 5 1
AB4E2 4 2 octahedral
Square pyramidal

Square planar

111

Steps to predict a molecular shape :

1) Draw the correct Lewis structure Example
Predict the molecular shape of
2) Determine number of electron-pair i. CCl4
ii. SeF4
Each multiple bond in a molecule constitutes a single electron iii. XeF4
domain (single bond)

3) Apply the VSEPR theory

4) Identify electron-pair arrangement.

5) State the class of molecule.

6) Draw the shape and state the bond angle
based on VSEPR

89 90

Solution 5) VSEPR theory : Solution
i. CCl4 repulsion of bonding pair
electrons are equal ii. SeF4
3) Class of molecule : AB4E
1) Lewis structure : 1) Lewis structure :
4) Electron-pair arrangement :
Valence e- C = 4 Cl 6) The shape : Valence e- Se = 6 Trigonal bipyramidal
4Cl = 28 Cl C Cl
32 4F = 28

bonding - 8 Cl Cl 109.5o 34 5) VSEPR theory :
repulsion of bonding pair -lone pair
24 C bonding - 8 electrons > bonding pair- bonding
Cl pair electrons
2) No of electron pair: Cl 26

4 electron pairs and all are bonding Cl F -24
F Se F 2
pair electrons
6) The shape :
3) Class of molecule : AB4 tetrahedral F
Bond angle Cl-C-Cl: 109.5 ° FF
4) Electron-pair arrangement : Se see- saw shaped
Tetrahedral 2) No of electron pair: Bond angle F-Se-F :
5 electron pairs: FF
4 bonding pair electrons <120° , < 90°
91 1 lone pair electron
92

Solution Example:
Draw the Lewis structure and determine its shape for
iii. XeF4 3) Class of molecule : AB4E2 each of the following:

1) Lewis structure : 4) Electron-pair arrangement : 1. CH4 2. BeCl2 3. BF3 4. NH3
octahedral 5. H2O 6. SO2 7. PCl5 8. SF6
Valence e- Xe = 8 9. SF4 10. ClF3 11. BrF5 12. I3-
5) VSEPR theory : 13. NO2- 14. XeF2 15. IF5 16. CH2Cl2
4F = 28 repulsion of lone pair-lone pair
electron > bonding pair -lone pair 94
36 electrons > bonding pair- bonding
pair electrons
bonding - 8

26

F -24
F Xe F 4

F 6) The shape :
FF

2) No of electron pair: Xe Square planar shaped
6 electron pairs:
4 bonding pair electrons F Bond angle F-Xe-F : 90°
2 lone pair electron F

93

112

Solution 5) VSEPR theory : Solution 5) VSEPR theory :
1. CH4 repulsion of bonding repulsion of bonding
pair-bonding pair 2. BeCl2 pair-bonding pair
electrons are equal. electrons are equal.
1) Lewis structure : 1) Lewis structure :

Valence e- C = 4 H Valence e- Be = 2

4H=4 6) The shape : 2Cl = 14 Cl Be Cl 6) The shape :
8 H
HC H 16

bonding - 4 H bonding -4

4 C 109.5 12 180o
2) No of electron pair:
2) No of electron pair: Cl Be
4 electrons pair and all are bonding HH H 2 electrons pair and all are bonding Cl
pair electrons
pair electrons

3) Class of molecule : AB4 tetrahedral 3) Class of molecule : AB2 linear

Bond angle H-C-H: 109.5° Bond angle Cl-Be-Cl: 180°

4) Electron-pair arrangement : 4) Electron-pair arrangement :
Tetrahedral linear

95 96

Solution 5) VSEPR theory : Solution 5) VSEPR theory :
repulsion of bonding repulsion of lone pair –
3. BF3 pair-bonding pair 4. NH3 bonding pair electrons
electrons are equal. >bonding pair-bonding pair
1) Lewis structure : 1) Lewis structure : electrons.
6) The shape :
Valence e- B = 3 FB F Valence e- N = 5 HN H 6) The shape :
3 F = 21 F 120o H N
3H=3
24 BF HH
bonding -6 F 8
F H 107.5o
18 Trigonal planar bonding -6
Trigonal pyramidal
Bond angle F-B-F: 120° 2
Bond angle H-N-H: 107.5°
2) No of electron pair: 97 2) No of electron pair:
3 electrons pair and all are bonding 4 electrons pair and 3 bonding pair 98
pair electrons electrons and 1 lone pair electron

3) Class of molecule : AB3 3) Class of molecule : AB3E

4) Electron-pair arrangement : 4) Electron-pair arrangement :
trigonal planar tetrahedral

Solution 5) VSEPR theory : Solution
repulsion of lone pair- lone
5. H2O pair electrons > lone pair – 6. SO2 5) VSEPR theory :
bonding pair electrons repulsion of lone pair –bonding
>bonding pair-bonding pair
1) Lewis structure : electrons. 1) Lewis structure : pair electrons > bonding pair-
Valence e- S = 6 bonding pair electrons
Valence e- O = 6 6) The shape :

2H=2 H O H 104.5o 2O = 12 O- S+
8 18 O
O
bonding -4 H bonding - 4 6) The shape :

4 H 14 O- S+
2) No of electron pair: <120o
2) No of electron pair: V-shaped O
4 electrons pair and 2 bonding pair 2 bonding pair electrons and 1
Bond angle H-O-H: 104.5° lone pair
electrons and 2 lone pair electron
99
3) Class of molecule : AB2E2 3) Class of molecule : AB2E V-shaped
Bond angle O-S-O: <120 °
4) Electron-pair arrangement : 4) Electron-pair arrangement :
tetrahedral Trigonal Planar 100

113

Solution 5) VSEPR theory : Solution 5) VSEPR theory :
7. PCl5 repulsion of bonding pair- 8. SF6 repulsion of bonding pair-
bonding pair electrons bonding pair electrons
1) Lewis structure : are equal. 1) Lewis structure : are equal.

Valence e- P = 5 Cl 6) The shape : Valence e- S = 6 F 6) The shape :

5 Cl = 35 Cl P Cl Cl 6 F = 42 FS F FF
40 Cl Cl 48 F FF F SF
Cl P Cl
bonding - 10 Cl120o Cl bonding - 12 FF

30 Trigonal bipyramidal 36 Octahedral
2) No of electron pair: 2) No of electron pair: Bond angle F-S-F: 90o
Bond angle Cl-P-Cl: 120°, 90o
5 electrons pair and all are bonding 6 electrons pair and all are bonding 102
101
pair electrons pair electrons

3) Class of molecule : AB5 3) Class of molecule : AB5

4) Electron-pair arrangement : 4) Electron-pair arrangement :
Trigonal bipyramidal octahedral

Solution 5) VSEPR theory : Solution 5) VSEPR theory :
9. SF4 repulsion of lone pair –bonding repulsion of lone pair- lone
pair electrons > bonding pair- 10. ClF3 pair electrons > lone pair –
1) Lewis structure : F bonding pair electrons F bonding pair electrons
Valence e- S = 6 1) Lewis structure : >bonding pair-bonding pair
6) The shape : Valence e- Cl = 7 electrons.
4 F = 28 FS F 3 F = 21 Cl F
34 F F SF 28 F 6) The shape :
FF bonding - 6
bonding - 8 22 F
See-saw 90o
26 octet terminal - 18
2) No of electron pair: Bond angle F-S-F: <120°, <90° 4 F Cl F

5 electrons pair there are 4 bonding

pair electrons and 1 lone pair 2) No of electron pair:
5 electrons pair , there are 3 bonding pair
electrons electrons and 2 lone pair electrons T-shaped
Bond angle Cl-P-Cl: ~90o
3) Class of molecule : AB4E 3) Class of molecule : AB3E2

4) Electron-pair arrangement : 4) Electron-pair arrangement :
trigonal bipyramidal Trigonal bipyramidal

103 104

Solution F 5) VSEPR theory : Solution 5) VSEPR theory :
11. BrF5 F Br F repulsion of lone pair –bonding repulsion of lone pair- lone
pair electrons > bonding pair- 12. I3- - pair electrons > lone pair –
bonding pair electrons bonding pair electrons
I I+ I >bonding pair-bonding pair
6) The shape : electrons.
1) Lewis structure : F F 1) Lewis structure :
Valence e- Br = 7 F Valence e- 3I = 21 6) The shape :
5 F = 35 FF charge = 1
42 22 180O
bonding - 10 Br <90o bonding - 4
32 18 I I+ I
FF linear
octet terminal - 30 octet terminal - 12
2 Square pyramidal 6 Bond angle I-I-I: 180o
Bond angle F-Br-F: <90o
2) No of electron pair: 2) No of electron pair: 106
6 electrons pair, there are 5 bonding 105 5 electrons pair, there are 2 bonding
pair electrons and 1 lone pair pair electrons and 3 lone pair
electrons electrons

3) Class of molecule : AB5E 3) Class of molecule : AB2E3

4) Electron-pair arrangement : 4) Electron-pair arrangement :
octahedral trigonal bipyramidal

114

Solution 5) VSEPR theory : Solution 5) VSEPR theory :
13. NO2- repulsion of lone pair –bonding 14. XeF2 repulsion of lone pair- lone
pair electrons > bonding pair- pair electrons > lone pair –
1) Lewis structure : 1) Lewis structure : bonding pair electrons
- bonding pair electrons Valence e- Xe = 8 >bonding pair-bonding pair
Valence e- N = 5 2 F = 14 electrons.
22
2O = 12 O- N O bonding - 4 F Xe F 6) The shape :
charge = 1 18 180O

18 octet terminal - 12 F Xe F
6
bonding - 4 6) The shape : linear

14 O- N Bond angle F-Xe-F: 180o
<120o
octet terminal - 12 108

2) No of ele2ctron pair: O 2) No of electron pair:
5 electrons pair, there are 2 bonding
2 bonding pair electrons and 1 pair electrons and 3 lone pair
electrons
lone pair V-shaped
Bond angle O-N-O: <120 °
3) Class of molecule : AB2E 3) Class of molecule : AB2E3

4) Electron-pair arrangement : 107 4) Electron-pair arrangement :
Trigonal Planar trigonal bipyramidal

Solution F F 5) VSEPR theory : Solution 5) VSEPR theory :
15. IF5 FI repulsion of lone pair –bonding 16. CH2Cl2 repulsion of bonding pair
pair electrons > bonding pair- electrons are equal
1) Lewis structure : F F bonding pair electrons 1) Lewis structure :
Valence e- C = 4
Valence e- I = 7 6) The shape : 2Cl = 14 Cl
2H= 2
5 F = 35 F 20 Cl C H 6) The shape :
FF bonding - 8 H
42 12 Cl 109.5o
I <90o
bonding - 10
FF
32 H C
Square pyramidal Cl
octet terminal - 30 Bond angle F-I-F: <90o H

2 109 2) No of electron pair:
4 electron pairs and all are bonding
2) No of electron pair: pair electrons tetrahedral
Bond angle Cl-C-Cl: 109.5 °
6 electrons pair, there are 5 bonding

pair electrons and 1 lone pair 3) Class of molecule : AB4

electrons 4) Electron-pair arrangement :
Tetrahedral
3) Class of molecule : AB5E

4) Electron-pair arrangement : 110
octahedral

MOLECULAR SHAPE

Central atom WITHOUT lone Central atom WITH lone pair Bond angles in
pair (BASIC SHAPE) i. methane, CH4
• BENT / V-SHAPED ii. ammonia, NH3
• LINEAR • TRIGONAL PYRAMIDAL iii. water , H2O
• TRIGONAL PLANAR • DISTORTED TETRAHEDRAL
• TETRAHEDRAL • T-SHAPED
• TRIGONAL BIPYRAMIDAL • LINEAR
• OCTAHEDRAL • SQUARE PYRAMIDAL
• SQUARE PLANAR
112
111

115

109.5 H H NH
H107.3
H CH
H Ammonia, NH3
 Central atom N has 3 bonding pairs and 1 lone pair
Methane, CH4
 Central atom C has 4 bonding pair electrons electrons.
 According to VSPER theory, repulsion between lone pair-
 Repulsions between bonding pairs are equal
bonding pair > bonding pair-bonding pair.
 Every H-C-H bond angle is 109.5  Lone-pair repels the bonding-pair more strongly, the

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

114

OH Bond and Molecules polarity
H 104.5
• A covalent bond can be polar or non-polar.
Water, H2O • A non-polar bond will form a non-polar

 Central atom O has 2 bonding pairs and 2 lone pairs molecule however, molecules with polar
bond may form polar or non-polar
 Repulsion between lone pair-lone pair > lone pair- molecule.
bonding pair > bonding pair-bonding pair electrons
116
 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

115

Cl Cl Polar Covalent Bond
• Polar covalent bond forms when the bonding pair
Non-polar covalent bond
• Involves two atoms of same electronegativity electrons are not equally shared between two
atoms of different electronegativity .
(Cl-Cl). • The atom with greater electronegativity draws the
electron pair closer to it, thus forms a polar
• It is formed when bonding electrons are covalent bond.
shared equally between the two atoms.
118
117

116

Example (δ+) H Cl (δ-)
The polar bond of HCl
electron density
(δ+) H Cl (δ-)
• The shift of the electrons density is shown by a
cross arrow, ( ) from the δ+ to the δ- end,
which indicate the bond dipole.
• Bonding electrons are shared unequally between
two atoms of different electronegativity (H-Cl). (δ+) (δ-)

• Cl is more electronegative than H, thus attracts the H Cl

bonding electrons to it and bears a partial negative The bond dipole is a vector quantity which has
both magnitude and direction
charge (δ-).
120
• The hydrogen atom experiences electron deficient

and has a partial positive charge (δ+). 119

Polar Molecules and Dipole Moments • A polar molecule has a dipole moment which
• A quantitative measures of the polarity of a bond is (μ0).

is a dipole moment (μ) • A non-polar molecule has no dipole moment
μ = Qr which is (μ=0). It occurs when the bond dipole or
the resultant dipole counterbalance each other.
where ,
Q = the magnitude of the charges 122
r = the distance that separates the
charges.

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

121

• The dipole moment can be determine Polarity of molecules
qualitatively from the nett or resultant of bond • Molecules can exist as :
dipole.
a) diatomic ( only two atoms involved) or
Example : b) polyatomic (more than two atoms) molecules.

 2 bond dipoles, a and b in the water molecule • For a diatomic molecule, the atoms concerned
produces a net or resultant bond dipole upwards. could be of same or different electronegativity.

 therefore, μ0 so, water molecule is polar. 124

a b a
resultant bond dipole

b

123

117

a) i. Diatomic molecules of the same element : ii. Diatomic molecules of different elements
 do not have bond dipole ( different electronegativity) :
 form non-polar molecules
 have bond dipoles (polar bonds)
e.g : H2, N2 and Cl2  have resultant bond dipole
 thus, dipole moments (μ0)
Cl2 : Cl  Cl  so, the molecules are polar .
 = 0,  non-polar bond
 non-polar molecule • e.g: HF, HCl, NO and CO

125 μ0
126

b) For a polyatomic molecule, Example

the dipole moment is determine by the polarity Determine the polarity of the following molecules
of the bonds and the molecular geometry. i. carbon dioxide, CO2
ii. carbon tetrachloride, CCl4
even if the molecule has polar bonds, the iii. chloromethane, CH3Cl
molecules will not necessarily have a dipole iv. ammonia, NH3
moment.
128
127

Solution ii) CCl4

Cl resultant dipole bond

i. CO2 bond dipole C
Cl Cl
• The CO2 molecular geometry is linear.
• Oxygen is more electronegative than carbon so Cl

there is a bond dipole in each carbon-oxygen • Carbon tetrachloride, CCl4 is a symmetrical
bond. molecule with tetrahedral shaped.

• The bond dipoles are equal in magnitude and • Cl is more electronegative than C, therefore each of
they pointed at the opposite direction, leading to C-Cl bond is polar.
a cancellation of their effects.
• Due to its symmetrical structure the resultant bond
dipoles cancelled each other.

• The dipole moment is zero (μ=0), CO2 is 129 • Therefore, the molecule has no dipole moment 130
nonpolar. (μ=0); CCl4 is a nonpolar molecule.

118

iii) CH3Cl Cl iv) NH3

C H

H
H

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

• Has resultant bond dipole.

• Dipole moment (μ0)

• Therefore, CH3Cl is a polar molecule. 131 132

Steps to determine the polarity of a molecule: Example
Determine the polarity of the following molecules :
Draw the correct Lewis structure find out
HBr ; SO3 ; ClF3 ; CF4 ; H2O ; SO;2
the class of molecule draw the correct XeF4 ; NF3 ; Cis-C2H2Cl2 ; trans-C2H2Cl2

molecular shape draw the respective 134

bond dipoles find out the resultant bond

dipole state the dipole moment, 

state the polarity

133

ANSWER ANSWER O-
i) HBr ii) SO3 S
O- O
H Br
• Molecular shape : trigonal planar
• Molecular shape : linear
• Br is more electronegative than H • O is more electronegative than S
• Form resultant bond dipole
• Dipole moment μ0 • Due to its symmetrical structure the resultant
• HBr is a polar molecule. bond dipoles cancelled each other.

• Dipole moment μ=0

135 • SO3 is a nonpolar molecule. 136

119

iv) CF4

iii) ClF3 F resultant dipole bond

F bond dipole C
F Cl F FF

F

• Molecular shape: T-shaped • Carbon tetrafluoride, CF4 is a symmetrical
• F is more electronegative than Cl molecule with tetrahedral shaped.
• Has resultant bond dipole.
• Dipole moment (μ0) • F is more electronegative than C, therefore each of
• Therefore, ClF3 is a polar molecule. C-F bond is polar.

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

• Therefore, the molecule has no dipole moment
137 (μ=0); CF4 is a nonpolar molecule. 138

v) H2O vi) SO2

O S+
H O- O

H

• Molecular shape: V-shaped 139 • Molecular shape: V-shaped 140
• O is more electronegative than H • O is more electronegative than S
• Has resultant bond dipole. • Has resultant bond dipole.
• Dipole moment (μ0) • Dipole moment (μ0)
• Therefore, H2O is a polar molecule. • Therefore, SO2 is a polar molecule.

vii) XeF4 viii) NF3

XeF F2-

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

• F is more electronegative than Xe, therefore each • Molecular shape : trigonal pyramidal 142
of Xe-F bond is polar. • F is more electronegative than N
• Form resultant bond dipole
• Due to its unsymmetrical structure the resultant • Dipole moment μ0
bond dipoles cannot cancelled each other. • NF3 is a polar molecule.

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

141

120

ix) Cis-C2H2Cl2 x) trans-C2H2Cl2

Cl Cl

C Cl C H
H C H C

H Cl

• Molecular shape : trigonal planar • Molecular shape : trigonal planar

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

• Form resultant bond dipole • The resultant bond dipoles cancelled each other.

• Dipole moment μ0 • Dipole moment μ=0

• Cis-C2H2Cl2 is a polar molecule. 143 • Trans-C2H2Cl2 is a nonpolar molecule. 144

BOND

Conclusion POLAR BOND 2 type of bonds NON-POLAR BOND

Factors that affect the polarity of molecules may formed NON-POLAR MOLECULES
are:
NON-POLAR MOLECULES POLAR MOLECULES
• molecular shape. Non-symetrical molecules
• electronegativity of the bonded atoms. Symetrical molecules (has the i. basic shape with different
same terminal atom)
145 terminal atom
i. basic shape ii. molecules with lone pairs
ii. molecules with lone pairs :
* linear (from trigonal 146

bipyramidal)
* square planar

121

4.3 ORBITAL OVERLAP LEARNING OUTCOME
AND HYBRIDISATION At the end of this lesson, students should be able to

147 a) illustrate the formation of sigma () and pi ()
bonds from overlapping orbitals.

b) describe the formation of hybrid orbitals of central
atom : sp, sp2 ,sp3, sp3d and sp3d2

c) illustrate the hybridisation of the central atom and
the overlapping of orbitals in molecules.

148

Covalent Bond and Orbital Overlap • Orbitals overlapping may occur between:
a) orbitals with unpaired electrons
• Valence Bond Theory explains the formation of b) an orbital with paired electrons and another
covalent bonds and the molecular shape outlined empty orbitals (dative bond)
by the VSEPR.
150
• The theory states that a covalent bond is formed
when the neighboring atomic orbitals overlap.

149

Example: The s-orbital of the Types of covalent bond
Hydrogen atom • Overlapping of orbitals produces two major
HH
types:
Change in electron density as a)  bond (sigma bond)
two hydrogen atoms b)  bond (pi bond)
approach each other.
152
High electron density as the
orbitals overlap

(covalent bond formed)

151

122

a) bond Overlap of two s orbitals
 bond
• formed when orbitals overlap along its
internuclear axis (end to end overlapping) + or

• Atoms involved have free rotation about the s orbitals
axis
Example:
i. overlapping s orbitals

s+ s ss In  bond, the electron density is concentrated
between the nuclei of the bonding atoms
 bond
154
153

 bond Overlap of s and p orbitals  bond
ii. Overlapping of s and p orbitals + or

s+ Px orbital x
x

 bond

 bond

155 156

 bond Overlap of two p orbitals  bond
or
iii. Overlapping of end to end p orbitals +

x+ x x

 bond

 bond

157 158

123

b)  bond  bond

• Formed when two p-orbitals of the same + or
orientation overlap sideways
In  bond, the electron density is concentrated
yy y  bond above and below the plane of the nuclei of the
bonding atoms
+
160
159

Formation of bonds in a molecule Overlapping of pure orbitals

• Covalent bonds may form by: Example :
a) overlapping of pure orbitals Draw the orbital overlapping for the
b) overlapping of hybrid orbitals following molecules.

161 i. H2
ii. HF
iii.F2
iv. O2
v. N2

162

Answer • Orbital overlap:  bond
i. H2
Valence orbital diagram: + 164
163
H : 1s
124
Lewis structure : H―H

ii. HF

Valence orbital diagram; • Orbital overlap:

H: 2p  bond

1s 165

F:

2s

Lewis structure : H F

166

iii. F2

Valence orbital diagram; • Orbital overlap:
 bond
F: 2 electrons shared and form
 bond
2s 2p

F:

2s 2p

Lewis structure : F F

 bond 167 168

iv. O2 Orbital Overlap:
Valence orbital diagram;



O: two unpaired electrons to be used in Orbital overlapping occurs
bonding; 1 for π bond , 1 for  bond O p side ways between the p-
2s 2p pO σ orbitals of each atom form

O: p p a  bond

2s 2p

Lewis structure : O  O



169 170

125

v. N2 Orbital overlap :
Valence orbital diagram;
 • 2 p-orbitals of each N
 atom overlapped side

N:  ways to form 2  bonds

2s 2p three unpaired electrons to be pN Np
used in bonding; 2 for π bond , 1
N: for  bond p p • 1 p-orbitals of each N
p p atom overlapped end to
2s 2p
end and form 1σ bonds
Lewis structure :
N  N




171 172

Hybrid Orbitals • Hybridization is the mixing of different type
of atomic orbitals to form a set of equivalent
• Valence bond theory is able to explain the orbitals
formation of bond pair electron and • 5 types of orbital hybridizations are:
observed molecular shape sp
sp2
• Equal repulsion of bonding pair electron sp3
around the central atom suggests that the sp3d
orbitals are equal sp3d2

• Therefore, atomic orbitals of the central 174
atom are assumed to mix and form new
orbitals called hybrid orbitals

173

How do I predict the hybridization of the central atom? sp3 hybridization

Count the number of lone pairs AND the number of • sp3 hybridization is the mixing of 1 s orbital
atoms bonded to the central atom and 3 p orbitals to form 4 equivalent sp3
orbitals.
No of Lone Pairs Hybridization Examples
+ sp • The shape of the four hybrid orbitals is
sp2 BeCl2 tetrahedral with the angle of 109.5o
No of Bonded Atoms sp3
2 BF3 176
sp3d
3 sp3d2 CH4, NH3, H2O

4 PCl5 175
SF6
5

6

126

s px sp3 sp3 • simplified drawing of sp3 orbitals:

py pz sp3 sp3 sp3
one s + three p orbitals four sp3 orbitals
sp3
177
sp3 Shown together (large
sp3 lobes only)

178

1) CH4 Example: sp3-Hybridized C atom in CH4
H

• Lewis structure : H CH
H

• Valence orbital diagram ; sp3

H : 1s Excitation: to form 4 unpaired electrons 1s sp3 sp3 1s
C ground state : 2p

2s sp3

C excited : 2s 2p

1s

C hybrid : sp3 hybrid 179 180

Example:

2) NH3 .. Orbital Overlap:
• Lewis structure :
H NH
H

• Valence orbital diagram; sp 3

H ground state : 1s  N sp 3
N ground state : 2s
N excited state : 2s H sp 3 H

1s  sp 3 1s

2p H
2p
sp3 1s

N hybrid : Molecular Geometry : Trigonal pyramidal

181 182

127

3) H2O H Answer
 Lewis structure : HO
Orbital Overlap:

 Valence orbital diagram; sp 3
O ground state :
 O
2s 2p sp 3
H sp 3
O excited state :
1s  sp 3
2s 2p
H
O hybrid : sp3
1s

H ground state : 1s Molecular Geometry : V- shaped
183 184

sp2 hybridization s sp2

• sp2 hybridization is the mixing of 1 s orbital px py sp2 sp2
and 2 p orbitals to form 3 equivalent sp2 one s orbital + two p orbitals three sp2 orbital1s86
orbitals.

• The shape of the three hybrid orbitals is
trigonal planar with the angle of 120o

185

• simplified drawing of sp2 orbitals: 1) BF3 Example: F F
• Lewis structure B
sp2 :

Shown together (large F
lobes )
• Valence orbital diagram;
sp2 sp2
F ground state : 2s 2p
187
B ground state : 2p
2s

B excited : 2s 2p

B hybrid : 188
sp2

128

p p 2) C2H4 Example:
• Lewis structure :
 189 HH
HC C H
sp2 sp2
• Valence orbital diagram;

C ground state :

Shape: trigonal planar sp2 C excited : 2s 2p
2s 2p

sp2 2p
C hybrid : 1s

H ground state :

p 190

Orbital Overlap: Example:

C2H4 

H  H 1s  bonds

1s sp 2  bond
sp 2
C
C  sp2 

sp2 sp2 H 1s

 sp2

H

1s

191 192

Example: H

3) C6H6 H C C H
• Lewis structure : C

HCCCH

H

• Valence orbital diagram;

C ground state : 2s 2p
2s 2p
C excited :

C hybrid : sp2 2p
1s
H ground state : 193 194

129

195 196

sp hybridization s px sp sp
198
• sp hybridization is the mixing of 1 s orbital
and 1 p orbitals to form 2 equivalent sp
orbitals.

• The shape of the two hybrid orbitals is linear
with the angle of 180o

197

Example:

• simplified drawing of sp orbitals: 1) BeCl2
• Lewis structure :
sp sp Shown together
(large lobes) • Valence orbital diagram;

199 Cl ground state : 3s 3p
Be ground state : 2s 2p

Be excited : 2p

2s

Be hybrid : 200

sp

130

2) C2H2 Example:

• Lewis structure dia:grHam; C CH
• Valence orbital

 C ground state :

p sp sp p 2s 2p

C :excited 2s 2p

C hybrid : sp 2p

H ground state : 202
1s

201

Orbital Overlap: Example:



 C  sp C 

1s sp sp sp 1s

molecular geometry: linear

203 204

Example: 3) CO2
3) CO2  Lewis structure : O C O
• Lewis structure :
 Valence orbital diagram;
• Valence orbital diagram;
Oground state : C ground state : O ground state :
C ground state :
C excited : 2s 2p 2s 2p
C hybrid : 2s 2p
C excited : O excited :
• Orbital overlap:
2s 2p O hybrid : sp2 2p
sp 2p
C hybrid :

205 206

131

Orbital Overlap: O CO sp3d hybridization
• sp3d hybridization is the mixing of 1 s orbital, 3
 
p orbitals and 1 d orbital to form 5 equivalent
sp 2 sp 2 sp3d orbitals
• The shape of the five hybrid orbitals is trigonal
 O bipyramidal with the angle of 120o and 90o
sp C sp
O sp2 sp 2 sp 2 208
p
sp 2

p pp

molecular shape: linear

207

Example:

• simplified drawing of sp3d orbitals: 1) PCl5 Cl
• Lewis structure : Cl P Cl

• Valence orbital diagram; Cl Cl

sp 3d P ground state : 3d
3d
sp 3d P excited : 3s 3p
3s 3p
sp 3d
sp3d
sp 3 d P hybrid : 3s 3p

sp 3d

Cl ground state :

209 210

Orbital overlap: p Example:
p
Cl 2) ClF3
• Lewis structure :

p
sp 3 d  • Valence orbital diagram;
Cl
P sp 3d Cl ground state :

Cl sp3d

sp 3 d sp3d  Cl excited :
Cl

p

 Cl hybrid :
Cl

p

molecular geometry: trigonal bipyramidal 211 F ground state : 212

132

Example: F Orbital overlap: p
F Cl F p
2) ClF3 F
• Lewis structure : 

• Valence orbital diagram; sp 3d sp 3 d
sp 3 d
Cl ground state :  sp 3d Cl
Cl excited : F
sp 3d
3s 3p 3d
3s 3p 3d 
F
Cl hybrid :
p
sp3d
F ground state : 2s 2p molecular geometry: T- shaped

213 214

sp3d2 hybridization • Simplified drawing of sp3d2 orbitals:

• sp3d2 hybridization is the mixing of 1 s sp3d2
orbital, 3 p orbitals and 2 d orbitals to form 6
equivalent orbitals sp3d2 sp3d2

• The shape of the six hybrid orbitals is sp3d2 sp3d2
octahedral with the angle 90o sp3d2

215 216

Example: 1) SF6 FF
1) SF6  Lewis structure : F SF
• Lewis structure :
 Valence orbital diagram; F F
• Valence orbital diagram;
F ground state : S ground state : 3s 3p 3d
S ground state : 3d
S excited : S excited : 3s 3p
S hybrid :
S hybrid : sp3d2
• Orbital overlap:
F ground state : 218

217 2s 2p

133

Orbital overlap: Example:
2) ICl5
p • Lewis structure :

F • Valence orbital diagram;
I ground state:
p   F p I excited :
I hybrid :
F sp 3 d 2 Clground state :

sp 3 d 2 sp 3 d 2  • Orbital overlap:

 sp3d2 Cl

sp 3d 2

F sp 3 d 2 F

p p



F

p

molecular geom etry: octahedral

219 220

2) ICl5 Cl Orbital overlap:
 Lewis structure : Cl I Cl
p
Cl Cl
Cl
 Valence orbital diagram;
p Cl  
I ground state:  Cl p
sp 3 d 2

5s 5p 5d sp 3 d 2 sp 3 d 2
5s 5p 5d
I excited :  sp3d2 I 

sp 3d 2

I hybrid : Cl sp 3 d 2 Cl

p p

sp3d2

Cl ground state : Molecular Shape : Square pyramidal

3s 3p 221 222

Exercise: a) XeF2 F Xe F
• Lewis structure :
• For each of the following, draw the orbital
overlap to show the formation of covalent • Valence orbital diagram;
bond
a) XeF2 Xe ground state : 5s 5p 5d
b) O3 5p 5d
c) ICl4 Xe excited : 5s
d) OF2
Xe hybrid : sp3d 224
223 F ground state : 2s 2p

134

Orbital overlap: b)O3
• Lewis structure :
p O O+ O-
• Valence orbital diagram; 1 2
F
 O+ ground state : 2p
2p
sp 3d sp 3d sp3d O+ excited : 2s
sp 3 d 2s 2p
Xe
O+ hybrid :
sp 3d

 sp2
F

p
Molecular Shape : Linear

225 226

O O+ O- Orbital Overlap:
1 2
• Valence orbital diagram; 
2p
O1 ground state : 2p

O1 excited : 2s 2p sp 2  O+sp2
2s 2p
O 2- sp2 sp2 sp2 

O1 hybrid : sp 2 2p
2p

sp2 O-

O-2 ground state : 2p

2s Molecular Shape : V - shaped

227 228

C) ICl-4 -
 Lewis structure :
Cl Orbital overlap:

Cl +

 Valence orbital diagram; I Cl p Cl   Cl p
Cl
I- ground state: sp 3 d 2
I- excited : 5d
I- hybrid : 5s 5p sp 3 d 2 I- sp3d2 
5s 5p 5d sp 3d 2
 sp3d2
sp3d2
Cl sp 3 d 2 Cl

p p

Cl ground state : molecular geometry: square planar

3s 3p 229 230

135

3) OF2 Answer Orbital Overlap:
 Lewis structure : Molecular Shape : V- shaped
F
FO

 Valence orbital diagram; sp 3

O ground state :  O

2s 2p F2p - sp 3 sp 3

O excited state :  sp3
O hybrid :
2s 2p F-
F ground state : sp3
2p
2s 2p
231 232

136

Learning Outcomes:

a) Describe intermolecular forces : van der Waals
forces, dipole-dipole interactions or permanent
dipole, London forces or dispersion forces,
hydrogen bonding.

4.4 Intermolecular forces b) Explain factors that influence van der Waals forces
and Hydrogen bond
233
c) Relate the effects of hydrogen bonding on the
physical properties:

i) boiling point

ii) solubility

iii) density of water compared to ice 234

Intermolecular Forces Classification of intermolecular forces:
 Intermolecular forces are attractive
Intermolecular Forces
forces between molecules.
van der Waal Forces Hydrogen Bond
• Intermolecular forces are responsible
for the physical properties such as boiling Dipole-dipole London forces
point, melting point and solubilities of interactions
molecular compounds.
236
235

A. Van der Waal Forces (i) Dipole-dipole interactions
(or permanent dipole forces)
 Forces that act between covalent molecules.
 Exist between polar covalent molecules
 Two types of van der Waals forces:  Polar molecules have permanent dipole due to
(i) Dipole-dipole interactions
- The intermolecular attraction between the uneven electron distributions
oppositely charged poles of nearby polar
molecules. Example: - + -

+

H Cl H Cl

(ii) London forces Chlorine is more Dipole-dipole forces; the partial
- The intermolecular attraction between electronegative,
molecules as a result of instantaneous thus it has higher positive end of one molecule attracts
polarisation of their electron clouds. electron density
the partial negative end of the other

237 molecule 238

137

• The electrostatic forces between the permanent (ii) London Forces (or dispersion forces)
+ pole of a molecule and permanent  pole of  Exist between non-polar molecules
adjacent molecules produce dipole-dipole
forces  Result from the temporary (instantaneous)
polarization of molecules
• Molecules with higher polarity;
- stronger dipole-dipole forces  The temporary dipole molecules will be attracted
to each other and these attractions is known as
• Polar molecules experience dipole-dipole forces the London forces or dispersion forces
as well as London dispersion forces
240
239

The formation of London forces Example:
London forces in Br2
 Electrons move randomly. At any instant,
electron distributions in one molecule may be Electrons in a molecule
unsymmetrical.
Br Br move randomly about the
 The end having higher electron density is nucleus
partially negative (–) and the other is partially
positive (+). At any instant, the - The temporary dipole
electron density might Br
 An instant dipole moment that exists in a be higher on one side molecule induce the
molecule induces the neighbouring molecule to be + - neighbouring atom to
polar. +
Br Br be partially polar
 The attraction that exists between the Temporary Br
instantaneous dipole and induced dipole is dipole molecule
known as London forces.
London forces
241
242

Factors that influence the strength of the Van 2. The polarity of molecules
der Waals forces.

1. The molecular size For molecules that have similar molecular
Molecules with larger molecular size have mass, the dipole-dipole interaction will be
stronger van der Waals forces as they tend to more dominant. Polar molecules have higher
have more electrons involved in the London boiling point due to the existence of stronger
forces. dipole-dipole attraction.

Example: Example: Molecular Polarity Boiling Point
Mass (oC)
Molecular Boiling Point F2 Non polar
Mass (oC) HCl 38.0 Polar 85.01

N2 28 77.3 36.5 188.11
Cl2 71 239.1
243 244

138

B. Hydrogen bond Other examples: H2O
Hydrogen bond
 Dipole-dipole interaction that acts between NH3 liquid
molecules that have a hydrogen atom that is OO
covalently bonded to a highly electronegative ..
atom; F, O ,N in one molecule and F,O or N of O
another molecule. N
O Hydrogen bond
Example: + - + - .. Hydrogen bond
HF HF
N

Covalent bond

Hydrogen bond 245 246

Effect of Hydrogen Bonding on Physical Example:
Properties
Explain the trend of boiling point given by the
(a) Boiling point graph below:

• Boiling point of compounds having hydrogen T/oC
bonds are relatively higher than compounds HF
having dipole-dipole interaction or London forces.
HBr HI
• It is due to Hydrogen bond is the strongest HCl
attraction force compared to the dipole-dipole
interaction or the London forces. Molecular mass

247 248

Answer:  The boiling points of compounds with hydrogen
bonds are affected by:
 HF form hydrogen bonds between molecules
while HCl, HBr and HI have van der Waals forces a) the number of hydrogen bonds per
acting between molecules. molecule
Hydrogen bond is stronger that the van der
Waals forces. More energy is required to break b) the strength of hydrogen bond which
the Hydrogen bond. HF has the highest boiling directly depends on the polarity of the
point. hydrogen bond

 Boiling point increases from HCl to HI. The 250
strength of van der Waals forces increases with
molecular size. Since molecular size increases
from HCl to HI, thus the boiling point will also
increase in the same pattern.

249

139

Example: Answer:
The order of the increase in boiling point is:
H2O > HF > NH3 > CH4  by looking at the polarity of the bond, we have
Explain the trend of boiling points. (order of polarity: HF > H2O > NH3)
but H2O has the highest boiling point.
251 For H2O, the number of hydrogen bonds per molecule
affects the boiling point.

 Each water molecule form more hydrogen bonds
with other water molecules. More energy is
required to break the Hydrogen bonds.

 HF has higher boiling point than NH3 because F is
more electronegative than Nitrogen. Therefore the
hydrogen bond in HF is stronger.

 CH4 is the lowest - it is a non polar compound and
has weak van der Waals forces acting between
molecules.
252

Hydrogen bonds between HF molecules and (b) Solubility
H2O molecules
H2O Hydrogen bond  Molecules which form hydrogen bond with
water molecules are soluble in water.
Examples: NH3, HF, CH3OH

+ - + - O O
HF HF
O Hydrogen bond

O .. ..

Hydrogen bond N N

O Hydrogen bond

253  NH3 dissolves in water because it form 254
hydrogen bond with water.

• Organic compound that soluble in water include: R OH H..
i. Amine (eg: methylamine, CH3NH2)
ii. Alcohol (eg: ethanol, CH3CH2OH) Hydrophobic O.. H
iii. Carboxylic acid (eg: ethanoic acid, CH3COOH) group
Hydrogen
• As the relative mass increase, the non-polar (water-hating) bonding
hydrocarbon portion (hydrophobic) become
larger while polar group (hydrophilic) represents 256
an increasingly smaller portion of the molecule.

•Therefore, the solubility decreases.

255

140

Example

Name Formula Solubility Solubility of butanol and hexanol in water:
(mol/100g of
Methanol CH3OH butanol: CH3CH2CH2CH2 OH .. H
Ethanol C2H5OH water) ..
Propanol C3H7OH miscible hexanol: CH3CH2CH2CH2CH2CH2 OH O.. H
Butanol C4H9OH miscible
Pentanol C5H11OH miscible H
Hexanol C6H13OH
Heptanol C7H15OH 0.11 O.. H
0.030
0.0058 Larger hydrophobic group,
0.0008 less soluble in water

257 258

(c) Density Ice (solid H2O) has lower density compared to its liquid.

 The density of water is relatively high • The hydrogen bonds in ice (aorpraenngsetrtuhcetuHre2O) molecules
compared to other molecules with similar in open hexagonal crystal
molar mass.
Hydrogen bond takes
Reason: one of the tetrahedral
Hydrogen bonds are stronger than the orientation and occupy
dipole-dipole or the London forces. Thus the some space
water molecules are drawn closer to one
another and occupy a smaller volume. 260

259

• This arrangement leaves a relatively large • When ice melts, H2O molecules have higher
amount of empty space between them. kinetic energy and can overcome the hydrogen
bond.
• As a result, ice has larger volume
compared to the liquid water. • The hydrogen bonds are broken and the open
structure collapses. V-shaped water molecules
• Thus, the density of ice is less than liquid slide between each other.
water.
• This causes the volume occupied by water to
261 decrease causing a corresponding increase in
the density

262

141

263 264
265 266

267 268

142

4.5 Metallic Bond

269 270

Learning Outcomes: Metallic bond

a) Explain the formation of metallic bond by using  An electrostatic force between positive charge
electron sea model. metallic ions and the sea of delocalised valence
electrons.
b) Relate metallic bond to the properties of metal:
malleability, ductility, electrical conductivity and  The metallic bond can be imagined as an array
thermal conductivity. of positively charged ions immersed in a sea of
delocalized valence electron.
c) Explain electrical conductivity of metal by using
band theory.  In the solid state, metal atoms are closely
packed in a regular arrangement.
d) Explain the factors that affect the strength of
metallic bond.  The solid lattice held together by the strong
attractive forces between the delocalised
e) Relate boiling/melting point to the molecular valence electrons and positively charge ions.
structure, types of bonding and intermolecular
forces for elements of: period 3, group 1 and group 272
17. 271

Example : Na Physical properties of metals
Metal have the following physical properties:
(a) Malleability
(b) Ductility
(c) Electrical conductivity
(d) Thermal conductivity

273 274

143

Malleability and Ductility Electrical and Thermal Conductivity
• Metals have high electrical and thermal
Malleable – can be pressed into different shapes.
Ductile – can be pulled into wire conductivity because the sea of delocalised
electrons are free to move from one end to
• Metal atoms are arranged closely packed another end when there are differences in
• When sufficient force is applied to the metal, one electrical potential or heat.

layer of atoms can slide over another without 276
disrupting the metallic bonding.
• As a result, metals are malleable and can be
drawn into wires(ductile)

275

BAND THEORY OF ELECTRICAL
CONDUCTIVITY

Band theory of solids
• The delocalized electrons move freely through

“band” formed by overlapping molecular orbitals.

• The electronic structure of a bulk solid is referred
to as a band structure.

• When valence band (lower energy) and

conduction band (higher energy) overlapping,

allowing electrons to flow through the metals with

minimal applied voltage. 277 278

BAND THEORY BAND THEORY

• Band: An array of closely spaced molecular orbitals band theory of solids:
occupying a continuous range of energy • electrons jump from valence band to conduction
band even at ordinary temperature and if this
• Band gap: The energy gap between a fully occupied happens then the solid conducts electricity.
valence band and an empty conduction band. • conductivity depends on the gap between the
valence band and conduction band.
• Conduction band: A band of unoccupied molecular
orbitals lying higher in energy than the occupied 280
valence band.

• Valence band: a band closely spaced bonding

molecular orbitals that is essentially fully occupied
279
by electron.

144

Band structure of conductors, semiconductors Factors that affect
and insulators the strength of the metallic bond
 The strength of metallic bonding depends on:
281 i. Size of positive metal ions
ii. Number of valence electrons

 The strength of metallic bonds is proportional to
the number of valence electrons and inversely

proportional to the size of the atom.

 The smaller size of positive metal ions and the

more number of valence electrons will exert

stronger attraction of metallic bond. 282

Effect of the strength of metallic bond on Example:
boiling point
Explain the difference in the boiling point of the two
 The boiling point of a metal depends on the metals given:
strength of metallic bond.
Boiling Point (oC)
 The stronger the metallic bond, the higher the Mg 1120
boiling point because high energy is required to Al 2450
overcome these strong electrostatic forces
between the positive ions and the delocalised 284
valence electron sea in the metallic bond

283

Boiling point of Al is higher than that of Mg:
 The strength of metallic bond in Aluminium is
stronger than Magnesium because Al has three
valence electrons and smaller cationic radius
while Mg has only two valence electrons and
larger cationic radius.
 The strength of metallic bond is directly
proportional to the boiling point.
 The stronger metallic bond, the higher the
boiling point.

285

145

CHAPTER 4: CHEMICAL
BONDING

QUESTIONS

TUTORIAL.MEKA.KUMBE

lecture notes & questions

146

CHEMISTRY
TUTORIAL TOPIC 4: CHEMICAL BONDING

4.1: Lewis Structure

1. Redraw and complete the table below:
Element Li Be B C N O F Ne
No. of
Valence
Electron
Lewis

dot symbol

2. a) Define octet rule.

b) Atoms of an element tend to lose or gain electrons in order to form bond. For

each of the following elements, write the electronic configuration for its most

stable ion and state the type of stability:
i) O : 1s2 2s2 2p4
ii) Mg : 1s2 2s2 2p6 3s2
iii) Mn : 1s2 2s2 2p6 3s2 3p6 4s2 3d5
iv) Zn : 1s2 2s2 2p6 3s2 3p6 4s2 3d10

3. a) i. Define ionic bond or electrovalent bond.
ii. Describe the formation of ionic bond in Na2O by using Lewis dot
symbol.

b) i. Define covalent bond.
ii. Describe the formation of covalent bond in tetraflouromethane, CF4 by
using Lewis dot symbols.

c) i. Explain the difference between covalent bond and dative bond in term
of its formation.

ii. Describe the formation of dative bond in NH4+ by using Lewis dot
symbol.

4. Draw the Lewis structure for each of the following species. Identify those that do not

obey the octet rule and state the type of exception for each case.

a) BeCl2 b) CO2 c) BF3

d) SO2 e) NO2 f) CH3Cl

g) PCl5 h) SF4 i) SF6
j) XeF4 k) CO32− l) NO2+

5. a) Draw the Lewis structure for C2H6 and C2H4.
b) Compare the bond length of C-C bond in both structures.

147