Molecular Shapes and Molecular
Polarity
• Recall that bond polarity is a measure of how
equally the electrons in a bond are shared
between two atoms. As the difference in
electronegativity between two atoms increases,
so does bond polarity.
• In molecules containing two atoms, a dipole
exists if there is a difference in electronegativity.
• For a molecule with more than two atoms, the
dipole moment depends on both the polarities of
the individual bonds and the geometry of the
molecule.
• Consider linear CO2
• Each C=O bond is polar, but since the bonds are
identical, the bond dipoles are equal in
magnitude. The overall dipole moment = 0.
• Bond dipoles and dipole moments are vector
quantities, which means they have magnitude
and direction. The overall dipole moment of a
molecule is the sum of its bond dipoles.
• Consider the bent molecule, water, with two
polar bonds.
• Both bonds are identical so the bond
dipoles are equal, but since the molecule
is bent, the bonds do not directly oppose
each other. Therefore, the bond dipoles do
not cancel each other out.
• The water molecule has a non zero dipole
moment so it is polar. The oxygen carries
a partial negative charge and the
hydrogen atoms each carry partial positive
charges.
Comparing Models
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
2. Add up the number of valence electrons from all atoms.
•
Add one electron for each negative charge and subtract one for each
positive charge.
3. Draw a skeleton structure with atoms attached by single
bonds.
4. Complete the octets of peripheral atoms. Hydrogen will
not have any lone pairs!
5. Place extra electrons on the central atom.
6. If the central atom doesn’t have an octet, try forming
multiple bonds by moving lone pairs.
Structural Formula
7. From the Lewis structure, remove dots
representing lone pairs
8. Replace bond dots with a dash
Water
Hydrogen (1) + Hydrogen (1) + Oxygen (6) = 8
SAMPLE PROBLEM A)
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
SOLUTION:
:
:
: Cl :
:Cl C
: F:
:
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
F
F
:
Steps 2-4:
C has 4 valence e-, Cl and F each have 7. The
sum is 4 + 4(7) = 32 valence e-.
Cl C
:
Step 1: Carbon has the highest bonding capacity and is
the central atom.
The other atoms are placed around it.
Cl
F:
:
PROBLEM:
Writing Lewis Structures for Molecules with
One Central Atom
Ammonia (NH3)
Nitrogen (5) and Hydrogen 3(1) = 8
Drawing Lewis Structures



14 ve’s


H O
Cl





HOCl
24 ve’s

COCl2


O


Cl
C
Cl


CH3OH
26 ve’s
14 ve’s

O
Cl
O




ClO3



O
H

H C O
H

H
SAMPLE PROBLEM B)
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
:
SOLUTION:
Write the Lewis structure for methanol (molecular formula
CH4O), an important industrial alcohol that is being used as a
gasoline alternative in car engines.
H
C
O
:
PROBLEM:
Writing Lewis Structure for Molecules with
More than One Central Atom
H
H
Multiple Bonds
• So far we have only looked at single bonds. The
sharing of a pair of electrons constitutes a single
bond.
• In many molecules atoms attain complete octets
by sharing more than one pair of electrons.
• When two electron pairs (4 electrons) are
shared, two lines are drawn to represent the
double bond.
• A triple bond corresponds to sharing of three
pairs of electrons (6 electrons).
SAMPLE PROBLEM C)
Writing Lewis Structures for Molecules with
Multiple Bonds.
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
PROBLEM:
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.
(a) There are 2(4) + 4(1) = 12 valence e-. H can have only
one bond per atom.
SOLUTION:
H
H
:
H
C
C
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)