Molecular Geometry and
Polarity
http://www.scl.ameslab.gov/MacMolPlt/Surface.JPG
Bond Angles in Carbon Compounds
electron configuration = 1s22s22p2
2p orbitals with one
electron in each.
Can p orbitals with one
electron in each find
the place where the
3rd p orbital should
be?
If they can, the bond angles
should be 90o.
Orbitals with one
electron in each will
overlap to form single
bonds.
But…the bond angles are
109.5o!
It’s All in the Shape…
• So what’s going on?
• Think back to the lab…
• What is the primary reason molecules
form the geometry we find?
• Electron Pair Repulsion
VSEPR Theory
• Electron groups around the central atom
will be most stable when they are as far
apart as possible – we call this valence
shell electron pair repulsion theory
– because electrons are negatively charged,
they should be most stable when they are
separated as much as possible
• The resulting geometric arrangement will
allow us to predict the shapes and bond
angles in the molecule
Electron-group repulsions and the five basic molecular shapes.
linear
trigonal bipyramidal
tetrahedral
trigonal planar
octahedral
Two electron pairs on central atom
Examples:
CS2, HCN, BeF2
Electron vs Molecular Geometry
• The geometry of electron pairs around a
central atom is called the electron
geometry.
• The arrangement of bonded nuclei around
a central atom forms the molecular
geometry.
• Lone pair electrons on a central atom will
repel other pairs but will not be visible in
the molecular geometry (no nuclei)
• If there are lone pairs on the central atom
the electron geometry and the molecular
geometry will differ.
Three electron
pairs on central
atom
Examples:
SO3, BF3, NO3-, CO32-
Examples:
SO2, O3, PbCl2, SnBr2
Four electron pairs on central atom
Examples:
CH4, SiCl4,
SO42-, ClO4-
Examples: NH3, PF3, ClO3. H3O+
Examples: H2O, OF2, SCl2
Five electron pairs on central atom
Six electron pairs on central atom
Representing 3-Dimensional
Shapes on a 2-Dimensional Surface
• One of the problems with drawing molecules is
trying to show their dimensionality
• By convention, the central atom is put in the
plane of the paper
• Put as many other atoms as possible in the same
plane and indicate with a straight line
• For atoms in front of the plane, use a solid
wedge
• For atoms behind the plane, use a hashed
wedge
The steps in determining a molecular shape
Molecular
formula
Step
1
Lewis
structure
Step
2
Electron-group
arrangement
(electron
geometry)
Count all e- pairs around central
atom
Step
3
Bond
angles
Note lone pairs and
double bonds
Step
4
Consider bonding
e- pairs only
Molecular
geometry
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles when the atoms
attached to the central atom are the same and when all electrons
are bonding electrons of the same order.
Effect of Double Bonds
ideal
1200
H
1200
C
H
O
1220
larger
EN
H
1160
greater
electron
density
H
real
Effect of Nonbonding(Lone) Pairs
Lone pairs repel bonding pairs
more strongly than bonding pairs
repel each other.
C
Sn
Cl
Cl
95
0
O
Predicting Molecular Shapes with Two, Three, or Four Electron Groups
PROBLEM:
Draw the molecular shape and predict the bond angles
(relative to the ideal bond angles) of (a) PF3 and (b) COCl2.
SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding
pair
The shape is based upon the tetrahedral arrangement.
F P F
The F-P-F bond angles should be <109.50 due
F
to the repulsion of the nonbonding electron
P
F pair.
F
F
The final shape is trigonal pyramidal.
0
<109.5
Predicting Molecular Shapes with Two, Three, or Four Electron Groups
(b) For COCl2, C has the lowest EN and will be the center atom.
There are 24 valence e-, 3 atoms attached to the center atom.
Cl
C
O
Cl
The shape for an atom with three atom
attachments and no nonbonding pairs on the
central atom is trigonal planar.
O
O
C
Cl
C does not have an octet; a pair of nonbonding
electrons will move in from the O to make a
double bond.
Cl
The Cl-C-Cl bond angle
will be less than 1200 due
to the electron density of
the C=O.
124.50
C
Cl
1110
Cl
Predicting Molecular Shapes with Five or Six Electron Groups
PROBLEM:
Determine the molecular shape and predict the bond angles
(relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central
atom will be in bonding pairs; shape is trigonal
bipyramidal.
F
F
F
F
Sb
F
F
F
Sb
F
F
F
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on
central atom. Shape is square pyramidal.
F
F
F
Br
F
F
Predicting Molecular Shapes with More Than One Central Atom
PROBLEM:
Determine the shape around each of the central atoms in
acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the
Lewis structure.
SOLUTION:
tetrahedral
H
H C
H
O
H
C C H
tetrahedral
H
trigonal planar
O
>1200
H
C
H
H C
C
H
HH
<1200
Molecular Polarity
• Just like bonds can be polar because of
even electron distribution, molecules
can be polar because of net electrical
imbalances.
• These imbalances are not the same as
ion formation.
• How do we know when a molecule is
polar?
The orientation of polar molecules in an electric field.
Electric field OFF
Electric field ON
Polarity of Molecules
• For a molecule to be polar it must
1.
have polar bonds


2.
electronegativity difference - theory
bond dipole moments - measured
have an unsymmetrical shape

vector addition
• Nonbonding pairs affect molecular polarity,
strong pull in their direction
Molecule Polarity
The H─Cl bond is polar. The bonding electrons are
pulled toward the Cl end of the molecule. The net result
is a polar molecule.
27
Molecule Polarity
The O─C bond is polar. The bonding electrons are
pulled equally toward both O ends of the molecule. The
net result is a nonpolar molecule.
28
Molecule Polarity
The H─O bond is polar. Both sets of bonding
electrons are pulled toward the O end of the
molecule. The net result is a polar molecule.
29
Predicting the Polarity of Molecules
PROBLEM:
From electronegativity (EN) values (button) and their
periodic trends, predict whether each of the following
molecules is polar and show the direction of bond dipoles
and the overall molecular dipole when applicable:
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
Draw the shape, find the EN values and combine the concepts to
determine the polarity.
SOLUTION: (a) NH3
The dipoles reinforce each
other, so the overall
molecule is definitely polar.
ENN = 3.0
H
ENH = 2.1
N
H
H
H
N
H
H
bond dipoles
H
N
H
H
molecular
dipole
Predicting the Polarity of Molecules
(b) BF3 has 24 valence e- and all electrons around the B will be involved
in bonds. The shape is AX3, trigonal planar.
F
B
F
1200
F
F (EN 4.0) is more electronegative
than B (EN 2.0) and all of the
dipoles will be directed from B to F.
Because all are at the same angle
and of the same magnitude, the
molecule is nonpolar.
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is
quite polar(DEN) so the molecule is polar overall.
S
C
O
More Molecular Polarity…
• http://academic.pgcc.edu/~ssinex/polar
ity/polarity.htm