Covalent Bonding
Lone pairs, valence electrons
not involved in covalent bond

H





 H - Cl
  





Cl


Cl
H +

• Formation of hydrogen chloride:

Covalent bond, shared electrons
Structural Formula: H-Cl
(lone pairs are not drawn)
Lewis Structures
H2:
Cl2:
H
+




 

H  H H or H H


Cl +  Cl







Cl
Cl
 

Structural Formula: Cl-Cl
Double and Triple Bonds
• Atoms can share 4 electrons to form a double bond
or 6 electrons to form a triple bond.

N2:

O
=O


O2:



N N

• The number of shared electron pairs (covalent
bonds) that an atom can form is the
bonding capacity.
Drawing Lewis Structures
1. Arrange the element symbols.
•
Central atoms are generally those with the highest bonding capacity.
•
Carbon atoms are always central atoms
•
Hydrogen atoms are always peripheral atoms
1. Add up the number of valence electrons from all atoms.
•
Add one electron for each negative charge and subtract one for each
positive charge.
1. Draw a skeleton structure with atoms attached by single
bonds.
2. Complete the octets of peripheral atoms.
3. Place extra electrons on the central atom.
4. If the central atom doesn’t have an octet, try forming
multiple bonds by moving lone pairs.
Strategy for
Writing Lewis
Structures
Structural Formula
• From the Lewis structure, remove dots
representing lone pairs
• Replace bond dots with a dash
Draw Lewis structures and the structural formula for:

H 
F
HF:



 H
H O




H N H
H
H

 H
H C




H
Molecular
formula
For NF3
:
: F:
: F:
:
Sum of
valence e -
:
Atom
placement
N
N = 5eF = 7eTotal
X 3 = 21e26e-
:
: F:
Remaining
valence e -
H O
H

H
H
or H C H
H N H
H




 


CH4:


or

NH3:

H F


or
H2O:

or
Zero: NF3 is uncharged
Lewis
structure
SAMPLE PROBLEM 10.1 Writing Lewis Structures for Molecules with
One Central Atom
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
SOLUTION:
F
F
:
:
Cl
:
:
Steps 2-4:
C has 4 valence e-, Cl and F each have 7. The
sum is 4 + 4(7) = 32 valence e-.
Cl
Cl C
Cl C
:
:
Step 1: Carbon has the highest bonding capacity and is
the central atom.
The other atoms are placed around it.
:
:
:
F
:
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
F
:
:
PROBLEM:
SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with
More than One Central Atom
PROBLEM:
Write the Lewis structure for methanol (molecular formula
CH4O), an important industrial alcohol that is being used as a
gasoline alternative in car engines.
SOLUTION:
Hydrogen can have only one bond so C and O must be next
to each other with H filling in the bonds.
There are 4(1) + 4 + 6 = 14 valence e-.
C has 4 bonds and O has 2. O has 2 pair of nonbonding e-.
H
:
H
C
O
H
:
H
SAMPLE PROBLEM 10.3 Writing Lewis Structures for Molecules with
Multiple Bonds.
PROBLEM:
Write Lewis structures for the following:
(a) Ethylene (C2H4), the most important reactant in the
manufacture of polymers
(b) Nitrogen (N2), the most abundant atmospheric gas
PLAN:
For molecules with multiple bonds, there is a Step 5 which follows the
other steps in Lewis structure construction. If a central atom does not
have 8e-, an octet, then e- can be moved in to form a multiple bond.
SOLUTION:
(a) There are 2(4) + 4(1) = 12 valence e-. H can have only
one bond per atom.
H
:
H
C
C
H
H
H
H
H
C
C
H
:
:
:
:
.
.
:
:
(b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make
the octet around each N.
N
N
N
N
N
N
.
.
.
.
Polyatomic Ions
• Many compounds contain a combination of
covalent and ionic bonds.
• E.g. NaOH
• OH- is a polyatomic ion with a covalent
bond between O and H
• Ionic bond between Na+ and OH-
Coordinate Covalent Bonds
A covalent bond in which both of the shared electrons come
from the same atom.
E.g. NH3 (ammonia) and H+ (hydrogen ion) to form NH4
(ammonium)
Drawing Lewis Structures
HOCl
26 ve’s
14 ve’s










H O
Cl



 

 

O Cl O
CH3OH
14 ve’s







O
ClO3
24 ve’s
Cl C Cl
O
COCl2


 











H

H C O
H

H