CHAPTER 4 - CHEMICAL BONDING

CHAPTER 4
CHEMICAL BONDING

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY MISS DALINA BINTI DAUD

CHAPTER 4
CHEMICAL BONDING

4.1 Lewis Structure
4.2 Molecular Shape and Polarity
4.3 Orbital Overlap and Hybridization
4.4 Intermolecular Forces
4.5 Metallic Bond

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4.1 LEWIS STRUCTURE

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At the end of this lesson, students should be
able to :

(a) write the Lewis dot symbol for an atom.

(b) state the octet rule and describe how
atoms obtain the octet configuration.

(c) describe the formation of the following
bonds using Lewis dot symbol.
i. ionic or electrovalent bond
ii. covalent bond
iii. dative or coordinate bond

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(d) draw Lewis structures of covalent species with
single, double and triple bonds.

(e) compare the bond length between single, double
and triple bonds.

(f) determine the formal charge and the most plausible
Lewis structures.

(g) explain the exception to the octet rule : incomplete
octet, expanded octet and odd number electrons.

(h) Illustrate the concept of resonance using 5
appropriate examples.

 Chemical bond is the force that
holds two atoms together in a
molecule or compound

 Valence electrons play an important
role in the formation of chemical
bonds

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4.1.1 Lewis Symbol
A Lewis symbol consists of:

 the symbol of an element
 dots or cross to represent the valence

electrons in an atom of the element.

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 Elements in the same group have the
same valence electronic configurations
 similar Lewis symbols.

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Lewis symbol of elements in period 2

Group Number of valence Element Lewis
electrons symbol
1 1 Li
2 2 Be 9
13 B
3

14 4 C

15 5 N

16 6 O

17 7 F

18 8 Ne

4.1.2 Octet Rule

 Octet rule states that atoms tend to form
bonds to obtain 8 electrons in the valence
shell (as in atom of a noble gas)

 Atoms combine by gaining, losing or
sharing electrons to achieve stability as the
noble gas configuration.

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 Types of stabilities of cations and anions
electronic configurations are:

 a) noble gas configuration
 b) pseudo-noble gas configuration
 c) half-filled orbital configuration

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Electronic Configuration of Cations and Anions

a) Noble gas configuration
Group 1, 2 and 13 elements donate valence
electrons to form cations with noble gas
configurations

Example:
Na : 1s22s22p63s1
Na+ : 1s22s22p6 (isoelectronic with Ne)

Ca : 1s22s22p63s23p64s2
Ca2+ : 1s22s22p63s23p6 (isoelectronic with Ar1)2

 Group 15, 16 and 17 elements accept electrons to
form anions with noble gas configurations
Example:
O : 1s22s22p4
O2─: 1s22s22p6
(isoelectronic with neon)

Cl : 1s22s22p63s23p5
Cl─ : 1s22s22p63s23p6

(isoelectronic with Ar)

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b) Pseudo-noble gas configuration

 d block elements donate electrons from 4s
orbitals to form cations with pseudo-noble gas
configuration.

Example:
Zn : 1s22s22p63s23p63d104s2
Zn2+ : 1s22s22p63s23p63d10
(pseudo-noble gas configuration )

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c) Half-filled orbital configuration

 d block elements donate electrons to achieve
the stability of half-filled orbitals
Example:
Mn : 1s22s22p63s23p63d54s2
Mn2+ : 1s22s22p63s23p63d5
(stability of half-filled 3d orbital )

Fe : 1s22s22p63s23p63d64s2
Fe3+
: 1s22s22p63s23p63d5

(stability of half-filled 3d orbital)

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4.1.3 Formation of the bonds using
Lewis Symbols

a) Ionic bond / Electrovalent bond
b) Covalent bond
c) Dative bond / Coordinate covalent bond

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(a) Ionic bond (Electrovalent bond)

 Ionic bond (or electrovalent bond) is the
electrostatic attraction between
positively and negatively charged ions
in an ionic compound.

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

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Example 1: NaCl

Sodium, an electropositive metal, tends to remove
its valence electron to form Na+ ion with the stable
noble gas configuration.

Chlorine, an electronegative element (non-metal),
tend to accept electron from Na to obtain stable
noble gas configuration.

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 The electrostatic forces between Na+ and
Cl- produce ionic bond

 These two processes occur simultaneously

 The formation of ionic bond can be shown
using Lewis symbol:

+

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Example 2: CaCl2

++ 2

 Ca atom (metal) transfer 2 electrons, one to each
chlorine atom, it become a Ca2+ ion with the stable

electronic configuration of noble gas.

 At the same time each chlorine atom (non-metal) gained
one electron becomes a Cl- ion to achieve noble gas
configuration

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

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Exercises:

By using Lewis symbol, show how the ionic
bond is formed in the compounds below.

(a) MgO
(b) KF
(c) Na2O

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Answer:
(a) MgO

Mg + O Mg 2+ O 2-

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

 The 2 electrons are accepted by an oxygen atom
to achieve the octet configuration and formed the
oxide ,O2- ions.

• The electrostatic attraction between Mg2+ and O2-
ions created ionic bond.

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Answer :

(b) KF +-

K +F KF

 K atom remove 1 electron to achieve octet
configuration and become a K+ ion.

 The electron is accepted by a fluorine atom as
to achieve the octet configuration and formed
chloride,F- ion.

• The electrostatic attraction between K+ and F-

ions created ionic bond. 23

Answer :

(c) Na2O + 2-
Na + O + Na
2 Na O

 Each Na atom remove 1 electron to achieve octet
configuration and become a Na+ ion.

 The two electrons are accepted by a fluorine atom
as to achieve the octet configuration and formed
oxide,O2- ion.

• The electrostatic attraction between Na+ and O2-
ions created ionic bond.

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Properties of Ionic compounds

Ionic bond is very strong, therefore ionic
compounds:

a) Have very high melting and boiling points

b) Solid at room temperature

c) Hard and brittle

d) Soluble in water

e) Can conduct electricity when they are in

molten form or aqueous solution because of

the moving ions 25

b) Covalent Bond

• A chemical bond formed by the sharing of one or
more electron pairs between two atoms.

• Involve non-metal atoms (atoms of the same or
similar electronegativity)

Why should two atoms share electrons?
To gain stability by having noble gas
configuration

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The formation of covalent bonds can be represented
using Lewis symbol.

Example 1: H H or H H

H +H

By sharing the electron, each hydrogen atom achieve
the stable noble gas electronic configuration of
helium (eight electrons in its valence shell)

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Example 2: F F or F F

F +F

• By sharing the electrons, each chlorine atom has
eight electrons in its valence shell.

• A bonding pair electrons can be represented by
two dots or a single line.

• lone pairs electrons are electron pairs that are not
involved in bonding

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c) Coordinate Covalent Bond (Dative Bond)

 Dative bond is a bond in which the pair of
shared electrons is supplied by one of
the two bonded atoms

 Involve overlapping of a full orbital and an
empty orbital

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Requirement for dative bonds:
i. Donor atoms should have at least one

lone pair electrons
ii. The acceptor atoms should have empty

orbitals in the valence shell.

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Example 1: H3O+

H+ + O H +

H OH

HH

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

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Example 2:

Use the Lewis symbol to show the formation of NH4+

Solution

H+ + H H

NH +

H H NH
H

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Exercise :

Use the Lewis symbol to show the formation of the
following compound:

i) F3BNH3 FH F H +
FB + NH FB NH
H
FH F

ii) Al2Cl6 Cl Cl Cl

Cl Cl Al Al
Cl Al + Al Cl
Cl Cl Cl Cl
Cl
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4.1.4 Lewis Structure of Covalent Species

Steps in Writing Lewis Structures

1. Count total number of valence electron of atoms involved.
- Add one electron for each negative charge.
- Subtract one electron for each positive charge.

2. Draw skeletal structure of the compound.
- the least electronegative atom occupies the central atom.
- H atom always occupies terminal atom
- draw a single covalent bond between the central atom and each of the
surrounding atoms.

3. Complete an 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.

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Example

Draw the Lewis structure for each of the

following compounds:

i. HF ii. CH4

HF H
HCH

H

iii. CHCl3 iv. NH3
Cl
HNH
H CH H

H 35

Solution vi. HCN

v. H2O HC N

OH
H

vii. PO43 3- viii. NH4+

O- H+
O- P O H N+ H

O- H

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Solution

ix. C2H4

HH
HC CH

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 Compounds may have these covalent bonds:

i. Single bond. Example : ClCl
ii. Double bond (two atoms share two

pairs of electrons). Example : OO
iii. Triple bond (two atoms share three pairs

of electrons). Example : NN

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4.1.5 Bond Length

The distance between the nuclei of two bonded
atoms is called bond length.

C C C C CC

1.54 Å 1.34 Å 1.20 Å

• As the number of bonds between the carbon
increase, the bond length decreases because C
are held more closely and tightly together.

• As the number of bonds between two atoms
increases, the bond grows shorter and stronger

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Example: Bond length, Å 40
1.43
C─N 1.38
C=N 1.16
CN 1.47
N─N 1.24
N=N 1.10
NN 1.43
C─O 1.23
C=O 1.13
CO

Formal Charge and the most
plausible Lewis structure

• Formal charge is the charge on a certain
atom in a Lewis Structure

Formal = Number of ― Total number of ― Number of
Charge valence non-bonding bonds

electrons electrons involved

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• Formal charge is used to find the most
stable/most plausible Lewis structure

• The most stable/most plausible Lewis Structure
may has:

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.

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EXAMPLE 1:
a) Draw all the possible Lewis structure of COCl2.
b) Predict the most plausible structure.

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Answer: Structure 2
00 0
Structure 1

-1 0 +1

00

The most plausible structure is structure 2
because it has zero formal charge.

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 The sum of the formal charge in a Lewis Structure must :
i. Equal zero for a neutral molecule
Example : PCl3 (formal charge = 0)
ii. Equal the magnitude of the charge for a polyatomic ion.
Example: NH4+ (formal charge = +1)

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EXAMPLE 2:

a) Draw all the possible Lewis structure of SCN.
b) Predict the most plausible structure.

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ANSWER: - Structure 3 -

a) Structure 1 Structure 2 (0) (-1)

(-1) (0) (+1) (-2) S C N-

S- C N S C N+ 2- (0)

(0) (0) -

b) The most plausible structure is structure 3
because formal charge of each atoms closest
to zero.

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Exception to the Octet Rule
Three conditions:
a) Incomplete octet
b) Expanded octet
c) Odd number electrons

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a) Incomplete octet

 Occurs when central atom has less than 8
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

electrons. 49

Example :
i. Draw the Lewis structures for these compounds:
a) BeCl2
b) BF3
c) AlBr3

ii. Choose the most plausible Lewis structures
for the respective compounds. Explain

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