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Chemical Bonding II:
Molecular Geometry and
Hybridization of Atomic Orbitals
Chapter 10
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Molecular Geometry
___ arrangement of atoms affects _________&_________
properties: MP, BP, density, types of reactions, etc.
The geometry must be determined __________________,
but we can use the valence-shell electron-pair repulsion
model (VSEPR) to predict the geometry of a molecule with
much success.
Remember that the valence shell is the outermost shell that
has electrons.
Basic assumption: electron pairs in the valence shell
______________________ each other and thus will remain
_______________________ each other as possible.
Two general rules:
1. Double bonds can be treated as single bonds
(with regard to __________________________).
2. VSEPR can be applied to any resonance
structure (if the molecule has resonance structures).
The valence-shell electron-pair repulsion model
accounts for the _________________________ of
electron pairs around a central atom in terms of the
___________________between pairs of electrons.
So let’s look at some examples that can serve as
patterns for you to remember.
Valence-shell electron-pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsions between the electron (bonding and nonbonding) pairs.
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
_________
________
B
B
10.1
0 lone pairs on central atom
Cl
Be
Cl
2 atoms bonded to central atom
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
________
________
AB3
3
0
10.1
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
________
AB3
3
0
trigonal
planar
AB4
4
0
_________
10.1
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
_________
_________
10.1
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
AB6
6
0
_________
_________
10.1
10.1
______________
______________
<
______________
______________
<
______________
______________
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB3
3
0
AB2E
2
1
Arrangement of
electron pairs
Molecular
Geometry
________
________
________
________
________
________
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
AB3E
3
1
Arrangement of
electron pairs
Molecular
Geometry
_________
________
_________
________
________
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
AB3E
3
1
tetrahedral
trigonal
pyramidal
AB2E2
2
2
_________
_________
O
H
H
10.1
VSEPR
Class
AB5
AB4E
# of atoms
bonded to
central atom
5
4
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
0
_________
_________
_________
_________
1
_________
_________
_________
_________
10.1
VSEPR
Class
AB5
# of atoms
bonded to
central atom
5
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
_________
_________
distorted
tetrahedron
_________
F
F
Cl
F
10.1
VSEPR
Class
AB5
# of atoms
bonded to
central atom
5
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
AB2E3
2
3
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
distorted
tetrahedron
_________
_________
T-shaped
_________
I
I
I
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
_________
_________
AB5E
5
1
_________
________
_________
F
F
F
Arrangement of
electron pairs
Molecular
Geometry
Br
F
F
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
AB4E2
4
2
_________
square
pyramidal
_________
_________
Arrangement of
electron pairs
Molecular
Geometry
F
F
Xe
F
F
10.1
Predicting Molecular Geometry
1. Draw Lewis structure for molecule.
2. Count number of lone pairs on the central atom and
number of atoms bonded to the central atom.
3. Use VSEPR to predict the geometry of the molecule.
What are the molecular geometries of SO2 and SF4?
O
S
F
O
AB2E
___________
F
S
F
AB4E
F
_________
____________
10.1
Predicting Molecular Geometry
The VSEPR model has some theoretical
problems—but chemists use it because it is
simple and generally leads to useful (reliable)
predictions of molecular geometry.
10.1
Dipole Moments and Polar Molecules
electron poor
region
electron rich
region
H
F
d+
d-
m=Qxr
Q is the ________________
r is the ___________________________
1 D = 3.36 x 10-30 C m
Why are the units of D?
What does the “D” stand for?
10.2
___________ molecules are affected by an electric field (p.302)
10.2
10.2
Which of the following molecules have a dipole moment?
H2O, CO2, SO2, and CH4
S
O
H
O
C
O
H
C
H
H
10.2
Dipole Moments of Some Polar Molecules
10.2
Dipoles (polar molecules) and Microwaves
10.2
Read p.304
Lewis theory does not explain why
bonds exist or why bonds are of
different lengths and different energies.
The Valence Bond (VB) Theory and the
Molecular Orbital (MO) Theory improve
our understanding.
How does Lewis theory explain the bonds in H2 and F2?
Sharing of two electrons between the two atoms.
Overlap Of
Bond Dissociation Energy
Bond Length
H2
436.4 kJ/mole
74 pm
2 1s
F2
150.6 kJ/mole
142 pm
2 2p
__________________ – bonds are formed by
sharing of e- from overlapping atomic orbitals.
10.3
10.4
Change in electron
density as two hydrogen
atoms approach each
other.
10.3
____________________________ posits that reacting atoms
form a stable molecule when/because the potential energy of
the system decreases to a minimum (p.306).
How does this apply to polyatomic molecules?
(p.307)
___________________________________ – atomic orbitals
obtained when two or more nonequivalent orbitals of the same
atom combine to form a covalent bond
_______________________________ – the mixing of atomic
orbitals in an atom (central) to produce a set of hybrid orbitals
We’ll repeat this in a minute. Let’s look at an example….
Valence Bond Theory and NH3
N – 1s22s22p3
3 H – 1s1
If the bonds form from overlap of three 2p orbitals on
nitrogen with the 1s orbital on each hydrogen atom,
what would the molecular geometry of NH3 be?
Overlap of
three 2p orbitals
predict _____0
Actual H-N-H
bond angle is
__________0 !
10.4
____________ – mixing of two or more atomic
orbitals to form a new set of hybrid orbitals.
1. Mix at least __________________ atomic orbitals
(e.g. s and p). Hybrid orbitals have very different
shape from original atomic orbitals.
2. Number of hybrid orbitals is equal to number of
_____________ used in the hybridization process.
3. Covalent bonds are formed by:
a. Overlap of ______ orbitals with _______ orbitals
b. Overlap of ______orbitals with _____________
orbitals
10.4
10.4
10.4
Predict correct
bond angle
10.4
Carefully consider
and study the summary
of hybridization on the
bottom of page 310 in
the text.
Formation of sp Hybrid Orbitals
10.4
Formation of sp2 Hybrid Orbitals
10.4
How do I predict the hybridization of the central atom?
Count the number of ___________________ AND the
number of ___________________________________
Compare top of p.312
# of Lone Pairs
+
# of Bonded Atoms
Hybridization
Examples
2
sp
BeCl2
3
sp2
BF3
4
sp3
CH4, NH3, H2O
5
sp3d
PCl5
6
sp3d2
SF6
10.4
p.316
10.5
10.5
Sigma bond (s) – electron density between the 2 atoms
10.5
Pi bond (p) – electron density above and below plane of nuclei of the bonding atoms
10.5
p.317
10.5
10.5
Sigma (s) and Pi (p) Bonds
1 _________ bond
Single bond
Double bond
1 ______ bond & 1 ___ bond
Triple bond
1 _____ bond & 2 ____ bonds
How many s and how many p bonds are in the
acetic acid molecule CH3COOH?
H
C
O
H
C
O
H
s bonds = 6 + 1 = 7
p bonds = 1
H
10.5
O
O
No unpaired eShould be diamagnetic
Experiments show O2 is paramagnetic
Molecular orbital theory – bonds are formed
from interaction of atomic orbitals to form
molecular orbitals that are associated with
the entire molecule.
p.318
10.6
This suggests a fundamental deficiency
in Valence Bond Theory—which brings
us to the Molecular Orbital Theory.
MO describes covalent bonds in terms of molecular orbitals,
that result from interaction of the atomic orbitals of the
bonding atoms and are associated with the entire molecule.
MO theory posits the formation of
two molecular orbitals:
one bonding molecular orbital and
one antibonding molecular orbital.
p.318
Energy levels of bonding and antibonding molecular
orbitals in hydrogen (H2).
A _________________ orbital has lower energy and greater
stability than the atomic orbitals from which it was formed.
An _________________________ orbital has higher energy
and lower stability than the atomic orbitals from which it was
formed.
10.6
p.319
10.6
Two possible interactions between two equivalent
p orbitals and the corresponding molecular orbitals
10.6
Energy levels of bonding and antibonding molecular
orbitals in hydrogen (H2).
A _________________ orbital has lower energy and greater
stability than the atomic orbitals from which it was formed.
An _________________________ orbital has higher energy
and lower stability than the atomic orbitals from which it was
formed.
10.6
A ___________ bond is almost
always a sigma bond and a pi bond.
A __________ bond is almost
always a sigma bond and two pi
bonds.
Now, carefully consider the
Rules Governing Molecular Electron
Configuration and Stability
p.322
Molecular Orbital (MO) Configurations
1. The number of molecular orbitals (MOs) formed is always
equal to the number of atomic orbitals combined.
2. The more stable the bonding MO, the less stable the
corresponding antibonding MO.
3. The filling of MOs proceeds from low to high energies.
4. Each MO can accommodate up to two electrons.
5. Use Hund’s rule when adding electrons to MOs of the
same energy.
6. The number of electrons in the MOs is equal to the sum of
all the electrons on the bonding atoms.
10.7
Note: this ignores the s2s and the s2s orbitals (pp.324f)
10.6
s1s < s1s < s2s < s2s < p2py
=
p2px < s2px < p2py
=
p2pz < s2px
p.324
1
bond order =
2
p.322
(
Number of
electrons in
bonding
MOs
-
Number of
electrons in
antibonding
MOs
)
Bond order can be a _____________.
A bond order of _____ or a _________
bond order means the bond has no
stability…and does not exist.
Bond order can be used only
_____________________ for
comparative purposes.
p.323
1
_________ =
2
bond
order
½
(
Number of
electrons in
bonding
MOs
1
-
½
Number of
electrons in
antibonding
MOs
)
0
10.7
Properties of Homonuclear Diatomic Molecules of the Second-Period Elements
p.325
10.7
MO theory describes bonding in terms of the combination
and rearrangement of atomic orbitals to form orbitals
associated with the whole molecule.
Bonding molecular orbitals increase electron density
between the nuclei and are lower in energy than the atomic
orbitals from which they are formed.
Antibonding molecular orbitals have a region of zero
electron density between the nuclei, and an energy level
higher than the atomic orbitals from which they are formed.
Molecules are stable if the number of electrons in bonding
molecular orbitals is greater than the number of electrons in
antibonding molecular orbitals.
p.326
______________________________
are not confined between two
adjacent bonding atoms, but actually
extend over three or more atoms.
10.8
__________________________ above and below the
plane of the _________________________________.
10.8