Chapter 13
Molecular Shapes
VSEPR Theory
Molecular Shape
• Molecular shape or molecular geometry
is the three-dimensional arrangement of
the atoms in a molecule.
Molecular Shape
• Molecular shape determines several
properties of a substance including:
– reactivity
– polarity
– phase of matter
– color
– magnetism
– biological activity
Olfaction – sense of smell
• Lock-and-Key Theory: humans can
smell various odors because each threedimensional odor molecule fits into only
one type of receptor.
Gustation – the sense of taste
• Taste receptors are
located on the tongue
and are sensitive to four
major tastes: salty,
sweet, sour and bitter.
• Taste receptors respond
differentially to the
varying shapes of food
and liquid molecules.
H2
H
2(1) = 2e-
H
Any 2 atoms
linear
What about molecules consisting of
more than two atoms
For this we must use the VSEPR Theory
VSEPR Theory
• Valence Shell Electron Pair Repulsion Theory
• Predicts the molecular shape of a bonded
molecule containing a central atom(s).
• Electrons around the central atom arrange
themselves as far apart from each other as
possible
• So only electrons (lone pairs or bonds)
connected to the central atom are important.
Five Molecular Shapes
See the table in your notes
Steps in Determining Molecular Shape
1. Draw the Lewis Structure for the
Molecule.
2. Count the number of atoms attached to
the central atom.
3. Count the number of lone pairs attached
to the central atom.
4. Use your counts on steps 2 and 3 to
determine the shape of the molecule.
H2S
2(1)+6 = 8e-
• Dihydrogen monosulfide is commonly known as
sewer gas. This colorless, toxic and flammable
gas is responsible for the foul odor of rotten eggs.
H
••
S
••
H
2/2
angular
H2S
• Dihydrogen monosulfide is commonly known as
sewer gas. This colorless, toxic and flammable
gas is responsible for the foul odor of rotten eggs.
S
H
H
Methane (CH4)
4 + 4(1) = 8e-
Methane (CH4)
4 + 4(1) = 8e4 atoms attached to the
central atom.
0 lone pairs on the
central atom.
4/0
Tetrahedral shape
Methane (CH4)
4 + 4(1) = 8e4 atoms attached to the
central atom.
0 lone pairs on the
central atom.
4/0
Tetrahedral shape
HCN
H
••
C••
1+4+5 = 10e-
••
N••
••
HCN
H
••
C••
1+4+5 = 10e-
••
N••
••
HCN
H
••
C••
1+4+5 = 10e-
••
N••
••
HCN
H
••
C
1+4+5 = 10e-
••
N••
HCN
H
••
C
1+4+5 = 10e-
••
N••
HCN
H
C
2/0
linear
1+4+5 = 10e-
••
N
H2S
S
H
H
2/2
angular
SO2
••
••
O
••
••
S••
6+2(6) = 18e-
••
••
O
••
SO2
••
••
O
••
••
S••
6+2(6) = 18e-
••
••
O
••
SO2
••
••
O
••
••
S••
2/1
angular
6+2(6) = 18e-
••
••
O
••
CH4
4/0
tetrahedral
NO3••
O
••
••
N
••
••
O
••
••
••
••
5+3(6)+1 = 24e-
O
••
NO3••
O
••
••
N
••
••
O
••
••
••
••
5+3(6)+1 = 24e-
O
••
NO3••
O
••
••
N
••
••
O
••
••
••
••
5+3(6)+1 = 24e-
O
••
NO3-
O
••
N
••
]
••
••
O
••
••
[
••
••
5+3(6)+1 = 24e-
O
3/0
trigonal
planer
SOCl2
SOCl2
Draw the Lewis structure
for an acetate ion?
Draw the Lewis structure
for an acetate ion?
What’s the shape of acetate?
4/0
3/0
tetrahedral
trigonal
planer
Homework
 Worksheet Chapter 13.
 Study Guide Chapters 12 – 14 (due by the
end of the week).