CH 10: MOLECULAR GEOMETRY
& CHEMICAL BONDING THEORY
Vanessa N. Prasad-Permaul
Valencia Community College
CHM 1045
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Molecular Shapes: VSEPR
 Molecular Geometry: the general shape of a
molecule, as determined by the relative positions of
the atomic nuclei.
 Valence-Shell Electron-Pair Repulsion Theory:
predicts the shapes of molecules and ions by
assuming that the valence-shell electron pairs are
arranged about each atom so that the electron pairs
are kept as far away from one another as possible,
thus minimizing electron-pair repulsions.
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Molecular Shapes: VSEPR
 Two Electron Groups: Electron groups point
in opposite directions.
 LINEAR ARRANGEMENT
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Molecular Shapes: VSEPR
 Three Electron Groups: Electron groups lie in
the same plane and point to the corners of an
equilateral triangle.
 TRIGONAL PLANAR ARRANGEMENT
 Trigonal planar geometry or bent (angular)
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Molecular Shapes: VSEPR
 Four Electron Groups: Electron groups point to
the corners of a regular tetrahedron.
 TETRAHEDRAL ARRANGEMENT
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Molecular Shapes: VSEPR
 The approximate shape of molecules is given by
Valence-Shell Electron-Pair Repulsion (VSEPR).
 Step #1: Count the total electron groups.
 Step #2: Arrange electron groups to maximize
separation.
 Groups are collections of bond pairs between two
atoms or a lone pair.
 Groups do not compete equally for space:
Lone Pair > Triple Bond > Double Bond > Single Bond
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Molecular Shapes: VSEPR
Table of the Molecular Geometry
and Examples of Certain Compounds
Electron Groups
Lone Pairs
Bonds
Geometry
Examples
2
0
2
Linear
BeCl2
3
0
3
Trigonal planar
BF3
3
1
2
Bent
SO2
4
0
4
Tetrahedral
CH4
4
1
3
Trigonal pyramidal
NH3
4
2
2
Bent
H2O
5
0
5
Trigonal bipyramidal
PCl5
5
1
4
See-saw
SF4
5
2
3
T-Shaped
ClF3
5
3
2
linear
I3-
6
0
6
Octahedral
SF6
6
1
5
Square pyramidal
SbCl52-
6
2
4
Square planar
XeF4
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Molecular Shapes: VSEPR
EXAMPLE 10.1 : Predict the geometry of the following
Molecules or ions, using the VSEPR method:
A. BeCl2
B. NO2C. SiCl4
Be = 2 electrons Cl = 7 electrons each = 14
Total of 16 electrons
¨ ¨
:Cl:Be:Cl:
¨
¨
2 sets of electron pairs, 0 lone pairs  linear
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Molecular Shapes: VSEPR
B. NO2N = 5 electrons, O = 6 electrons each = 12
Total of 17 electrons plus 1 electrons = 18 e:O:N::O:
¨ ¨ ¨
¨
3 sets of electron pairs, 1 lone pair  trigonal planar
arrangement and a bent geometry
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Molecular Shapes: VSEPR
C. SiCl4
Si = 4 electrons, Cl = 7 electrons each = 28 electrons
Total of 32 electrons
4 electron pairs, 0 lone pairs tetrahedral
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Molecular Shapes: VSEPR
EXERCISE 10.1 : Use the VSEPR method to predict the
geometry of the following ion and molecules:
A. ClO3B. OF2
C. SiF4
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Molecular Shapes: VSEPR
Five Electron Groups: Electron groups point to the
corners of a trigonal bipyramid.
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Molecular Shapes: VSEPR
Six Electron Groups: Electron groups point to the
corners of a regular octahedron.
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Molecular Shapes: VSEPR
EXAMPLE 10.2 : What do you expect for the geometry of
tellurium tetrachloride?
Tellurium has 6 electron pairs in its valence shell
Cl = 7 electrons each = 28 electrons
Total of 34 electrons
¨
There are 4 bonding pairs and 1 lone pair of electrons 
Trigonal bipyramidal arrangement and a seesaw geometry14
Molecular Shapes: VSEPR
EXERCISE 10.2 : According to the VSEPR model, what
molecular geometry would you predict for iodine
trichloride?
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Dipole Moment & Molecular Geometry
Dipole Moment: a quantitative measure of the degree
of charge separation in a molecule.
+
-
H
Cl
 If a molecule has a dipole moment it is polar
 A compound could contain polar bonds but the
molecule could be non-polar because there is no
dipole moment
 Bond dipoles can cancel each other out
O
C
O
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Dipole moment
 Based on electronegativity differences between
atoms in a molecule
 The most electronegative atom is partially negative
 The less electronegative atom is partially positive
 The dipole moment is the average of all dipoles in
the molecule
 Exception (C-H bonds do not have a dipole)
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Molecular Shapes: VSEPR
EXAMPLE 10.3 : Each of the following molecules has a
non-zero dipole moment. Select the molecular geometry
that is consistent with this information. Explain.
SO2: linear or bent
BENT because in linear geometry, the dipole moment
would be zero. There are 2 electron pairs and 1 lone pair
which is a trigonal planar arrangement giving a bent
geometry.
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Molecular Shapes: VSEPR
PH3: trigonal planar or trigonal pyramidal
In a trigonal planar geometry, the dipole moment is zero,
therefore PH3 has a trigonal pyramidal geometry.
Tetrahedral arrangement; 3 electron pairs and one lone
pair.
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Molecular Shapes: VSEPR
EXERCISE 10.3 : Bromine trifluoride BrF3, has a nonzero
dipole moment. Indicate the correct arrangement and
molecular geometry is consistent with this information.
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Molecular Shapes: VSEPR
EXERCISE 10.4 : Which of the following would be
expected to have a dipole moment of zero on the basis of
symmetry. Explain.
A. SOCl2
B. SiF4
C. OF2
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Valence Bond Theory
1. Covalent bonds are formed by overlapping of
atomic orbitals, each of which contains one
electron of opposite spin.
2. Each of the bonded atoms maintains its own
atomic orbitals, but the electron pair in the
overlapping orbitals is shared by both atoms.
3. The greater the amount of orbital overlap, the
stronger the bond.
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Valence Bond Theory
According to Valence Bond Theory:
 An orbital on one atom comes to occupy a
portion of the same region of space as an orbital
on the other atom. The two orbitals are said to
overlap.
 The total number of electrons in both orbitals is
no more than two.
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Valence Bond Theory
Combined
hybrid
orbitals
Hybrid orbitals
# of
orbitals
VSEPR
geometry
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Valence Bond Theory
• sp hybrid:
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Valence Bond Theory
• sp2 hybrid:
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Valence Bond Theory
• sp2 hybrid (π bond):
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Valence Bond Theory
• sp3 hybrid:
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Valence Bond Theory
sp3d hybrid:
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Valence Bond Theory
• sp3d2 hybrid:
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Valence Bond Theory
Steps to obtain the bonding description about an atom in
a molecule:
1. Write the Lewis electron-dot formula
2. From the Lewis formula, use the VSEPR model to
obtain arrangement
3. From the geometry, s what type of hybrid orbitals are
required
4. Assign valence electrons to the hybrid oritals of this
atom one at a time, pairing them only when necessary
5. Form the bonds by overlapping singly occupied orbitals
of other atoms
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Valence Bond Theory
EXAMPLE 10.4: Describe the bonding in H2O according to
valence bond theory.
Using the VESPR theory, notice there are 3 sets of electron
pairs, and two lone pairs giving a tetrahedral arrangement
and a bent geometry
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Valence Bond Theory
EXERCISE 10.5: Using hybrid orbitals, describe the
bonding in NH3 according to valence bond theory.
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Valence Bond Theory
EXAMPLE 10.5: Describe the bonding in XeF4 using hybrid
orbitals.
4 single bonds, 2 lone pairs octahedral arrangement and
square planar geometry.
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Valence Bond Theory
EXERCISE 10.6: Describe the bonding in PCl5 using hybrid
orbitals.
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Molecular Orbital Theory
 Additive and subtractive combination of p
orbitals leads to the formation of both sigma
and pi orbitals.
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Valence Bond Theory
EXAMPLE 10.6: Describe the bonding on a given N atom
in dinitrogen difluoride, N2F2, using valence bond theory.

¨
:F¨ N
¨
¨
N

¨
F:
¨
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Valence Bond Theory
EXERCISE 10.7: Describe the bonding on the carbon atom
in carbon dioxide using valence bond theory.
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Valence Bond Theory
EXERCISE 10.8: Dinitrogen difluoride exists as cis and
trans isomers. Write structural formulas for these isomers
and explain using valence bond theory why they exist.
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Molecular Orbital Theory
 The molecular orbital (MO) model provides a
better explanation of chemical and physical
properties than the valence bond (VB) model.
 Atomic Orbital: Probability of finding the electron
within a given region of space in an atom.
 Molecular Orbital: Probability of finding the
electron within a given region of space in a
molecule.
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Molecular Orbital Theory
 Additive combination of orbitals () is lower
in energy than two isolated 1s orbitals and is
called a bonding molecular orbital.
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Molecular Orbital Theory
 Subtractive combination of orbitals (*) is
higher in energy than two isolated 1s orbitals
and is called an antibonding molecular orbital.
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Molecular Orbital Theory
 Bond Order is the number of electron pairs
shared between atoms.
 Bond Order is obtained by subtracting the
number of antibonding electrons from the
number of bonding electrons and dividing by 2.
BO = Bonding electrons – antibonding electrons
2
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B2
*2p
B has 5 electrons
*2p
2p
2p
So B2 has 10 elec
Core electrons don’t
count toward BO
*2s
2s
*1s
1s
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C2
*2p
*2p
2p
Carbon has 6
electrons so C2 has
12 electrons
2p
*2s
2s
*1s
1s
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Molecular Orbital Theory
 Molecular Orbital Diagram for H2:
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Molecular Orbital Theory
 Molecular Orbital Diagrams for H2– and He2:
47
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Molecular Orbital Theory
 Second-Row MO Energy Level Diagrams:
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Valence Bond Theory
EXAMPLE 10.7: Give the orbital diagram of the molecule.
Is the molecular substance diamagnetic or paramagnetic?
What is the electron configuration? What is the bond
order?
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Valence Bond Theory
EXERCISE 10.9: The C2 molecule exists in the vapor phase
over carbon at high temperature. Describe the molecular
orbital structure of this molecule (orbital diagram and
electron configuration). Is the substance paramagnetic or
diamagnetic? What is the bond order?
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Valence Bond Theory
EXAMPLE 10.8: Write the orbital diagram for dinitrogen
monoxide (nitric oxide). What is the bond order for NO?
52
Valence Bond Theory
EXERCISE 10.10: Give the orbital diagram and electron
configuration for the carbon monoxide molecule. What is
the bond order and is this molecule paramagnetic or
diamagnetic?
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Molecular Orbital Theory
 MO Diagrams Can Predict Magnetic Properties:
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Example 1: VSEPR
 Draw the Lewis electron-dot structure and predict
the shapes of the following molecules or ions:
O3
H3O+
XeF2
PF6–
XeOF4
AlH4–
BF4–
SiCl4
ICl4–
AlCl3
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Example 2: Molecular Orbital Theory
 The B2 and C2 molecules have MO diagrams
similar to N2. What MOs are occupied in B2
and C2, and what is the bond order in each?
Would any of these be paramagnetic?
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Example 3: Dipole moment
 Draw the dipole moment for the following
molecules, are they polar?
HCl
NH3
CHCl3
H2O
SF6
CCl4
CO2
CH2Cl2
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Hybridization Easy Way
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