Chapter 9
Molecular Geometry
Bonding Theories
Molecular Shape
• A bond angle is the
angle defined by lines
joining the centers of
two atoms to a third
atom to which they are
covalently bonded
• The molecular
geometry or shape is
defined by the lowest
energy arrangement of
its atoms in threedimensional space.
VSEPR
Valence-Shell Electron-Pair Repulsion Theory
The geometric arrangement of atoms bonded to a
given atom is determined principally by
minimizing electron pair repulsions of bonding and
non-bonding electrons.
Central Atoms without Lone Pairs
Steric number (SN) is
the number of volumes
of space occupied by
electrons surrounding
a central atom
Geometric Forms
Predicting a VSEPR Structure
1. Draw Lewis structure.
2. Determine the steric number of the
central atom.
3. Use the SN to determine the geometry
around the central atom.
4. The name for molecular structure is
determined by the number of volumes
of space occupied by bonding
electrons.
Examples
• What is the molecular geometry of BF3?
• What is the molecular geometry of CH4
Examples
• What is the molecular geometry of BF3?
Lewis Structure (exception to Law of Octaves)
F
B
F
F
• What is the molecular geometry of CH4
Examples
• What is the molecular geometry of BF3?
Lewis Structure (exception to Law of Octaves)
F
B
F
F
Bond Angles = 120°
Trigonal Planar
• What is the molecular geometry of CH4
Examples
• What is the molecular geometry of BF3?
Lewis Structure (exception to Law of Octaves)
F
B
F
F
Bond Angles = 120°
Trigonal Planar
• What is the molecular geometry of CH4
H
H
C
H
H
Examples
• What is the molecular geometry of BF3?
Lewis Structure (exception to Law of Octaves)
F
B
F
F
Bond Angles = 120°
Trigonal Planar
• What is the molecular geometry of CH4
H
H
C
H
Bond Angles = 120°
H
Tetrahedral
Central Atoms with Lone Pairs
• Electron-pair geometry describes the arrangement
of atoms and lone pairs of electrons about a
central atom.
 The electron-pair geometry will always be one of the
five geometries presented previously.
• The molecular geometry in these molecules
describes the shape of the atoms present (it
excludes the lone pairs).
Lone Pairs
• Lone pairs of electrons occupy more space
around a central atom than do bonding electrons.
• Lone pair-lone pair repulsion is the largest.
• Lone pair-bonding pair repulsion is the next
largest.
• Bonding pair-bonding pair repulsion is the smallest.
• In structures with lone pairs on the central atom,
the bond angles are a little smaller than predicted
based on the electron-pair geometry.
SN = 3, Electron-pair Geometry = Trigonal Planar
No. of Bonded
Atoms
No. of Lone
Pairs
Molecular
Geometry
Bond Angles
3
0
Trigonal Planar
120o
Bent
<120o
Like 119.6o
2
1
Non-bonding Electrons & Shape
H
H
B
B
H
H
H
H
Angles = 120
Angles < 120
Trigonal Planar
Bent
SN = 4, Electron-pair Geometry = Tetrahedral
No. of Bonded
Atoms
No. of Lone
Pairs
Molecular
Geometry
Bond Angles
4
0
Tetrahedral
109.5o
3
1
Trigonal
Pyramidal
<109.5o
2
2
Bent
<109.5o
Non-bonding Electrons & Shape
Non-bonding Electrons & Shape
Tetrahedral
Trigonal Pyramid
v-shape
SN = 5, Electron-pair Geometry = Trigonal
Bipyramidal
No. of Bonded
Atoms
No. of Lone
Pairs
Molecular
Geometry
Bond Angles
5
0
Trigonal
Bipyramidal
120o & 90o
4
1
Seesaw
<120o & 90o
3
2
T-shaped
<120o & 90o
2
3
Linear
180o
The lone pairs of electrons are always found in the trigonal
planar part of the structure to minimize repulsion.
SN = 6, Electron-pair Geometry = Octahedral
No. of Bonded
Atoms
No. of Lone
Pairs
Molecular
Geometry
Bond Angles
6
0
Octahedral
90o
5
1
Square
Pyramidal
<90o
4
2
Square Planar
90o
3
3
2
4
Although these arrangements
are possible, we will not
encounter any molecules with
these arrangements.
Polar Bonds and Polar Molecules
• Two covalently bonded atoms with different
electronegativities have partial electric charges
of opposite sign creating a bond dipole.
• A molecule is called a polar molecule when it
has polar bonds and a shape where the bond
dipoles don’t offset each other.
Examples
Measuring Polarity
The permanent dipole moment () is a measured
value that defines the extent of separation of positive
and negative charge centers in a covalently bonded
molecule.
Valence Bond Theory
• Hybridization is the mixing of atomic orbitals to
generate new sets of orbitals that are then
available to overlap and form covalent bonds with
other atoms.
• A hybrid atomic orbital is one of a set of
equivalent orbitals about an atom created when
specific atomic orbitals are mixed.
Valence-Bond Theory
• Valence-bond theory assumes that covalent
bonds form when orbitals on different atoms
overlap or occupy the same region of space.
• A sigma () bond is a covalent bond in which
the highest electron density lies between the
two atoms along the bond axis connecting
them.
Atomic Orbitals and Bonds
•
•
A tetrahedral molecule requires that four
orbitals of the central atom must overlap with
an orbital of an outer atom to form a bond.
The central atom would use its s orbital and
its three p orbitals, but these orbitals would
not yield the 109° bond angles observed in
the tetrahedral molecule.
Hybrid Orbitals
You may have noticed that the electron pairs in molecules
have different orientations in space compared to atomic
orbitals. Wave equations mathematically generated
volumes of space where electrons spend most of their time,
but what about molecules?
Hybrid Orbitals
You may have noticed that the electron pairs in molecules
have different orientations in space compared to atomic
orbitals. Wave equations mathematically generated
volumes of space where electrons spend most of their time,
but what about molecules? This brings us to the concept
of hybrid orbitals, combinations of atomic orbitals, or
molecular orbitals (from wave equations of electrons in
molecules)
Hybrid Orbitals
You may have noticed that the electron pairs in molecules
have different orientations in space compared to atomic
orbitals. Wave equations mathematically generated
volumes of space where electrons spend most of their time,
but what about molecules?
Hybrid Orbitals
You may have noticed that the electron pairs in molecules
have different orientations in space compared to atomic
orbitals. Wave equations mathematically generated
volumes of space where electrons spend most of their time,
but what about molecules? This brings us to the concept
of hybrid orbitals, combinations of atomic orbitals, or
molecular orbitals (from wave equations of electrons in
molecules)
Hybrid Orbitals
Hybridization is a concept you might be familiar
with. For example a grapefruit is a hybrid of
what two fruits?
Hybrid Orbitals
Hybridization is a concept you might be familiar
with. For example a grapefruit is a hybrid of
what two fruits?
Hybrid Orbitals
Hybridization is a concept you might be familiar
with. For example a grapefruit is a hybrid of
what two fruits? Lemon and orange
Hybrid Orbitals
How about a nectarine?
Hybrid Orbitals
How about a nectarine? Plumb and a peach.
Hybrid Orbitals
How about a nectarine? Plumb and a peach.
Broccoaflower? Broccoli and cauliflower
Hybrid Orbitals
How about a nectarine? Plumb and a peach.
Broccoaflower? Broccoli and cauliflower
And a Cocapoo?
Hybrid Orbitals
How about a nectarine? Plumb and a peach.
Broccoaflower? Broccoli and cauliflower
And a Cocapoo? Cocker spaniel and poodle
Hybrid Orbitals
On to Chemistry! How about an s-orbital and a porbital? Yes, sp orbital.
Hybrid Orbitals
On to Chemistry! How about an s-orbital and a porbital? Yes, sp orbital.
How about one s-orbital and two p-orbitals?
Hybrid Orbitals
On to Chemistry! How about an s-orbital and a porbital? Yes, sp orbital.
How about one s-orbital and two p-orbitals? Yes an
sp2 orbital.
Examples
Tetrahedral Geometry: sp3 Hybrid
Orbitals
A tetrahedral orientation of valence electrons is
achieved by forming four sp3 hybrid orbitals form one
s and three p atomic orbitals.
Other sp3 Hybrid Examples
sp2 Hybridization
• In a covalent pi () bond, electron density is
greatest above and below the bonding axis.
sp Hybridization
• Pi bonds will not exist between two atoms
unless a sigma bond forms first.
The Bonding in Carbon Dioxide
The carbon atom
is sp hybridized
and these orbitals
form the two
sigma bonds. The
 bonds are
rotated 90° from
one another.
d2sp3 Hybridization
dsp3 Hybridization
Delocalization of Electrons
Localized
Delocalized
• The electrons in the
 system with
alternating single
and double bonds
can be delocalized
over several atoms
or even an entire
molecule.
Hybrid Orbital Notation
In order to construct hybrid orbital notation, we
need to separate the central atom from the
surrounding electrons, usually the central atom is the
largest, the most electronegative, or the one that
there is one of.
Hybrid Orbital Notation
In order to construct hybrid orbital notation, we
need to separate the central atom from the
surrounding electrons, usually the central atom is the
largest, the most electronegative, or the one that
there is one of. When constructing a hybrid orbital
diagram, all of the valence electrons of the central
atom are used and only the single electrons of the
atoms attached to the central atom are use.
Hybrid Orbital Example
Suppose we want to make a diagram of SF6
First we separate the central atom from the other atoms.
The central atom is A and the other atoms are called X’s
A
SF6
X’s
Hybrid Orbital Example
Suppose we want to make a diagram of SF6
First we separate the central atom from the other atoms.
The central atom is A and the other atoms are called X’s
A
SF6
X’s
Then we generate a set of degenerate hybrid orbitals to
house the valence electrons
Hybrid Orbital Example
Suppose we want to make a diagram of SF6
First we separate the central atom from the other atoms.
The central atom is A and the other atoms are called X’s
A
SF6
X’s
Then we generate a set of degenerate hybrid orbitals to
house the valence electrons F
F
F
F
F
F
F
Insert single electrons into the degenerate hybrid orbitals
Hybrid Orbital Example
Suppose we want to make a diagram of SF6
First we separate the central atom from the other atoms.
The central atom is A and the other atoms are called X’s
A
SF6
X’s
Then we generate a set of degenerate hybrid orbitals to
house the valence electrons F
F
F
F
F
F
F
Insert single electrons into the degenerate hybrid orbitals
Structure of Sulfurhexafluoride
F
F
F
s
F
F
Shape- Octahedral
Hybrid Orbitals- sp3d2
F Bond angles- 90°
Polarity- Nonpolar
Noble Gas Compounds
Critics of the hybrid orbital theory argued that the
hybrid orbital theory suggests that compounds of
Noble gases should exist or be made. In 1962 Neil
Bartlett created a compound of xenon, platinum and
fluorine. Today there are now several hundred Noble
gas compounds known.
University of British Columbia
Noble Gas Compounds
Practice
XeO2
KrF4
XeO2F22+
Problems with Bonding Theories
• Lewis structure and valence bond theory help
us understand the bonding capacities of
elements.
• VSEPR and valence bond theories account
for the observed molecular geometries.
• None of these models enables us to explain
why O2 is attracted to a magnetic field while
N2 is repelled slightly.
Molecular Orbital (MO) Theory
• The wave functions of atomic orbitals of
atoms are combined to create molecular
orbitals (MOs) in molecules.
 Each MO is associated with an entire
molecule, not just a single atom. MOs are
spread out, or delocalized over all the
atoms in a molecule.
Types of MOs
• Electrons in bonding orbitals serve to hold
atoms together in molecules by
increasing the electron density between
nuclear centers.
• Electrons in antibonding orbitals in a
molecule destabilize the molecule
because they do not increase the electron
density between nuclear centers.
MO Guidelines
1. The total number of MO formed equals the
number of atomic orbitals used in the mixing
process.
2. Orbitals with similar energy and shape mix more
effectively than do those that are different.
3. Orbitals of different principal quantum numbers
have different sizes and energies resulting in less
effective mixing.
4. A MO can accommodate two electrons with
opposite spin.
5. Electrons are placed in MO diagrams according to
Hund’s rule.
MOs for H2
• The two 1s orbitals may be added or subtracted to
yield two MOs.
Bond Order
Bond Order = 1/2 (# bonding electrons - # antibonding
electrons)
The bond order is zero in He2 and the molecule is not
stable.
BOND ORDER
Useful concept:
The net number of bonds existing after the cancellation
of bonds by antibonds.
In He2
the electronic configuration is….
(1s)2(1s*)2
the two bonding electrons were cancelled out by the two
antibonding electrons.
There is no BOND!
67
BOND ORDER = 0!!!!!
So……..
Bond Types
• A sigma, , bond is a covalent bond in
which the highest electron density lies
along the bond axis.
• A pi, , bond is formed by the mixing of
atomic orbitals that are not oriented
along the bonding axis in a molecule.
BOND ORDER
A measure of bond strength and molecular stability.
If # of bonding electrons > # of antibonding electrons
the molecule is predicted to be stable
Bond
order
69
=
BOND ORDER
A measure of bond strength and molecular stability.
If # of bonding electrons > # of antibonding electrons
the molecule is predicted to be stable
Bond
order
= 1/2{
= 1/2 (n
# of bonding
electrons(nb)
b
- na)
# of antibonding
electrons (na)
–
}
A high bond order indicates high bond energy and
short bond length.
70
Consider H2+,H2,He2+,He2……….
First row diatomic molecules and ions
H2
E
1s*
1s
Magnetism
Bond order
Bond energy
(kJ/mol)
Bond length (pm)
71
H2+
He2+
He2
First row diatomic molecules and ions
H2
E
1s*
1s
Magnetism
Dia-
Bond order
1
Bond energy
(kJ/mol)
436
Bond length (pm)
74
72
H2+
He2+
He2
First row diatomic molecules and ions
H2
H2+
Magnetism
Dia-
Para-
Bond order
1
½
Bond energy
(kJ/mol)
436
225
Bond length (pm)
74
106
E
73
1s*
1s
He2+
He2
First row diatomic molecules and ions
H2
H2+
He2+
Magnetism
Dia-
Para-
Para-
Bond order
1
½
½
Bond energy
(kJ/mol)
436
225
251
Bond length (pm)
74
106
108
E
74
1s*
1s
He2
First row diatomic molecules and ions
H2
H2+
He2+
He2
Magnetism
Dia-
Para-
Para-
—
Bond order
1
½
½
0
Bond energy
(kJ/mol)
436
225
251
—
Bond length (pm)
74
106
108
—
E
75
1s*
1s
HOMONUCLEAR DIATOMICS
Now look at second period…..
First is Li2
Li : 1s22s1
Both the 1s and 2s overlap to produce  bonding and
anti-bonding orbitals.
This is the energy level diagram…..
76
DI-LITHIUM
 *
2s
2s
2s
2s
E
1s*
1s
77
Li2
Put the electrons in…….
1s
1s
ELECTRONS
FOR
DILITHIUM
2s*
Li2
2s
2s
2s
E
1s*
1s
78
Put the electrons in the MO’s
1s
1s
Electron configuration for DILITHIUM
2s*
Li2
(1s)2(1s*)2(2s)2
2s
2s
2s
E
Bond Order ??????
What do we need?
1s
1s
79
1s
Electron configuration for DILITHIUM
2s*
Li2
(1s)2(1s*)2(2s)2
2s
2s
na = 2
nb = 4
2s
E
Bond Order =
1/2 (nb - na)
= 1/2(4 - 2) =1
1s
80
1s
1s
A single bond.
Electron configuration for DILITHIUM
2s*
Li2
(1s)2(1s*)2(2s)2
2s
2s
na = 2
nb = 4
2s
E
Note:
The 1s and 1s* orbitals
cancel!
1s
81
1s
1s
So…….
Electron configuration for DILITHIUM
2s*
Li2
(1s)2(1s*)2(2s)2
2s
2s
2s
E
1s
82
1s
The 1s and 1s* orbitals can
be ignored when both are
FILLED!
1s
We often omit the
inner shell!
Li2
Only valence orbitals contribute to
molecular bonding
(2s)2
Li
Li2
Li
2s*
E
2s
2s
2s
(1s)2(1s*)2 assumed
83
The Li2 configuration….
But can be included.
Li2
Only valence orbitals contribute to
molecular bonding
(2s)2
Li
Li2
Li
2s*
E
2s
2s
2s
The complete configuration is: (1s)2(1s*)2 (2s)2
84
What is the bond order????
Li2
Only valence orbitals contribute to
molecular bonding
(2s)2
Li
E
Li2
Li
2s*
Ignoring the filled
(1s)2(1s*)2
2s
2s
nb = 2
2s
Bond Order =
85
A single
bond.
na = 0
1/2(nb - na) = 1/2(2 - 0) =1
Is the molecule stable or unstable???
Li2
Only valence orbitals contribute to
molecular bonding
(2s)2
Li
E
Li2
Li
2s*
Ignoring the filled
(1s)2(1s*)2
2s
2s
nb = 2
2s
Bond Order =
86
A single
bond.
na = 0
1/2(nb - na) = 1/2(2 - 0) =1
STABLE!
Now Be2…….
VALENCE ELECTRONS FOR DIBERYLLIUM
Be
Be2
Be
2s*
E
2s
2s
2s
Put the electrons in the MO’s...
87
Be2
Electron configuration for DIBERYLLIUM
Be
Be2
Be
Be2
2s*
E
2s
2s
2s
Configuration:
88
(2s)2(2s*)2
Bond order?
Electron configuration for DIBERYLLIUM
Be2
Be
Be2
Be
(2s)2(2s*)2
2s*
E
nb = 2
2s
2s
na = 2
2s
Bond Order =
No89 bond!!!
1/2(nb - na) = 1/2(2 - 2) =0
We conclude???????
Electron configuration for DIBERYLLIUM
Be2
Be
Be2
Be
(2s)2(2s*)2
2s*
E
nb = 2
2s
2s
na = 2
2s
Bond Order =
No90 bond!!!
1/2(nb - na) = 1/2(2 - 2) =0
The molecule is not stable!
Now B2...
B2
The Boron atomic configuration is
1s22s22p1
So we expect B to use 2p orbitals to
form molecular orbitals.
How do we do that???
Combine them by ???
Addition and subtraction….
91
This is what they look like…….
 molecular orbitals
SUBTRACT
* antibonding
2p
ADD
92
-
+
+
2p bonding
+
PHH picture
-molecular orbitals
PHH picture
-
93
-
+
-
+
+
- MO’s
-
The  molecular orbitals.
SUBTRACT
ADD
94
+
-
-
+
* antibonding
2p
+
-
PHH...
2p bonding
The  molecular orbitals.
+
-
+
+
-
95
ENERGY
LEVELS?
ENERGY LEVEL DIAGRAM
When we form MO’s we get orbitals of different
energies
2s
Example the
E
2s*
From 2s
2s*
2s
2s
2s
The p- MO’s……..
96
The M.O.’s formed by p orbitals
2p*
2p*
2p
2p
2p
E
2p
The  do not split as much because of weaker
overlap. Combine this with the s-orbitals…..
97
Modified Molecular Orbital Diagram
It should be noted that both sigma 1s and sigma 2p
orbitals have similar shape and energy, thus mixing
occurs lowering the sigma 1s orbital and elevating the
sigma 2p orbital above the bonding pi molecular
orbitals.
This is illustrated on the next slide.
Expected orbital splitting:
2p*
The  do not split as
much because of
weaker overlap.
2p*
2p
2p
2p
E
2s*
2s
99
2p
But the s and p along
the internuclear axis
interact…….
2s
2s
This pushes the 2p up..
MODIFIED ENERGY LEVEL DIAGRAM
2p*
2p*
2p
E
2p
2p
2s*
2s
100
Shows additional 
interaction.
2p
Notice that the 2p and 2p
have changed places!!!!
2s
2s
Now look at B2...
Electron configuration for B2
2p*
B is [He] 2s22p1
2p*
2p
E
2p
2p
2s*
2s
101
2p
Place electrons from 2s into
2s and 2s*
2s
2s
Electron configuration for B2:
2p*
2p*
2p
E
2p
2p
2s*
2s
102
2p
Place electrons from 2p into
2p and 2p
2s
2s
Remember HUND’s RULE
Electron configuration for B2:
ELECTRONS
ARE
UNPAIRED
2p*
Abbreviated configuration
 *
2p
(2s)2(2s*)2(2p)2
2p
E
2p
2p
2s*
2s
103
2p
Complete configuration
(1s)2(1s*)2(2s)2(2s*)2(2p)2
2s
2s
Bond order????
Electron configuration for B2:
2p*
(2s)2(2s*)2(2p)2
2p*
na = 2
2p
E
2p
2p
2p
Bond order
1/2(nb - na)
2s*
= 1/2(4 - 2) =1
2s
2s Molecule is
predicted to be
stable and
paramagnetic.
2s
104
nb = 4
HOMONUCLEAR DIATOMICS
Li2
B2
C2
N2
O2
F2
Which energy level diagram???
The one with s and p interaction or the one without?
We find……….
105
Li2
SECOND ROW DIATOMICS
B2
C2
N2
O2
USE 2p*
E
2p*
2p
2p
2p
F2
2p*
USE
2p*
2p
2p
2p
2p
2p
2s*
2s*
2s
106
2s
2s
2s
2s
2s
Second row diatomic molecules
B2
2p*
2p*
E
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
107 length(pm)
Bond
C2
N2
O2
F2
Second row diatomic molecules
B2
2p*
2p*
E
2p
2p
2s*
2s
Magnetism
Para-
Bond order
1
Bond E. (kJ/mol)
290
108 length(pm)
Bond
159
C2
N2
O2
F2
Second row diatomic molecules
B2
C2
Magnetism
Para-
Dia-
Bond order
1
2
Bond E. (kJ/mol)
290
620
159
131
2p*
2p*
E
2p
2p
2s*
2s
109 length(pm)
Bond
N2
O2
F2
Second row diatomic molecules
B2
C2
N2
Magnetism
Para-
Dia-
Dia-
Bond order
1
2
3
Bond E. (kJ/mol)
290
620
942
159
131
110
2p*
2p*
E
2p
2p
2s*
2s
110 length(pm)
Bond
O2
F2
Second row diatomic molecules
NOTE SWITCH
OF LABELS
B2
C2
N2
O2
Magnetism
Para-
Dia-
Dia-
Para-
Bond order
1
2
3
2
Bond E. (kJ/mol)
290
620
942
495
111
Bond
length(pm)
159
131
110
121
2p*
2p*
E
2p
2p
2s*
2s
F2
Second row diatomic molecules
NOTE SWITCH
OF LABELS
B2
C2
N2
O2
F2
Magnetism
Para-
Dia-
Dia-
Para-
Dia-
Bond order
1
2
3
2
1
Bond E. (kJ/mol)
290
620
942
495
154
112
Bond
length(pm)
159
131
110
121
143
2p*
2p*
E
2p
2p
2s*
2s
QUESTION
The molecule X+2 that has the molecular orbital configuration
(1s)2(1s*)2(2s)2(2s*)2(2p)4 (2p)2 (2p*)1 is
1
B2+
2
C2+
3
N2+
4
O2+
5
F2+
ANSWER…...
113
QUESTION
The molecule X+2 that has the molecular orbital configuration
(1s)2(1s*)2(2s)2(2s*)2(2p)4 (2p)2 (2p*)1 is
+
How?
The molecule X2 has 16 electrons
1
B2
2
C2+
Therefore Z = 8
3
N2+
Therefore O2+
4
O2+
5
F2+
PARAMAGNETIC?
BOND ORDER
ANSWER…...
114
YES!
2.5
EXAMPLE…..
Example: Give the electron configuration and bond order for O2, O2+
, O2- & O22-. Place them in order of bond strength and describe their
magnetic properties.
Step 1:Determine the number of valence electrons in each:
O2 :
6 + 6 = 12
O2+ :
6 + 6 - 1 = 11
O2– :
6 + 6 + 1 = 13
O22- : 6 + 6 + 2 = 14
115
Step 2:
Determine the valence electrons configurations:
O2
2p*
E
2p*
2p
2p
2s*
2s
O2 :
O2+ :
116
O2– :
O22-:
O2+
O2–
O22-
Step 2:
Determine the valence electrons configurations:
O2
O2+
O2–
2p*
2p*
2p
2p
2s*
E
2s
O2 :
O2+ :
117
O2– :
(2s)2(2s*)2 (2p)2(2p)4 (2p*)2
O22-
Step 2:
Determine the valence electrons configurations:
O2
O2+
O2–
2p*
2p*
2p
2p
2s*
E
2s
118
O2 :
(2s)2(2s*)2 (2p)2 (2p)4(2p*)2
O2+ :
(2s)2(2s*)2 (2p)2 (2p)4(2p*)1
O2– :
O22-
Step 2:
Determine the valence electrons configurations:
O2
O2+
O2–
2p*
2p*
2p
2p
2s*
E
2s
119
O2 :
(2s)2(2s*)2 (2p)2 (2p)4(2p*)2
O2+ :
(2s)2(2s*)2 (2p)2 (2p)4(2p*)1
O2– :
(2s)2(2s*)2 (2p)2 (2p)4(2p*)3
O22-
What about O22- ???:
Step 2:
O2
O2+
O2–
O22-
2p*
E
2p*
2p
2p
2s*
2s
120
O2 :
O2+ :
(2s)2(2s*)2 (2p)2 (2p)4(2p*)2
(2s)2(2s*)2 (2p)2 (2p)4(2p*)1
O2– :
(2s)2(2s*)2 (2p)2 (2p)4(2p*)3
O22- : (2s)2(2s*)2 (2p)2 (2p)4(2p*)4
Step 3:
Determine the bond orders of each species:
O2
2p*
E
2p*
2p
2p
2s*
2s
O2 :
B.O. = (8 - 4)/2 = 2
O2+ :
B.O. = (8 - 3)/2 = 2.5
O2– :
B.O. = (8 - 5)/2 = 1.5
O
121 2
2- :
B.O. = (8 - 6)/2 = 1
O2+
O2–
O22-
Step 4: Use the bond orders to place the species in order of
bond strength:
O2
O2+
O2–
O22-
2p*
E
2p*
2p
2p
2s*
2s
O2 :
B.O. = 2
O2+ :
B.O. = 2.5
O2– :
B.O. = 1.5
O22B.O. = 1
122:
BOND ENERGY ORDER
O2+ >O2 >O2– > O22HETERONUCLEAR DIATOMICS…...
SECOND ROW DIATOMICS
2p*
E
Can be used for
2p*
2p*
2p
2p
2p
2p*
2p
2p
heteronuclear diatomics
123
2s
2s
2p
2p
2s*
2s*
2s
2p
Which????
2s
2s
2s
NITRIC OXIDE (NO)
Number of valence electrons: 5 + 6 = 11
use the mo diagram for homonuclear diatomic
molecules with s-p interaction as an approximation
2p*
2p*
E
2p
2p
2s*
2s
124
Put the electrons in…..
NITRIC OXIDE (NO)
Number of valence electrons: 5 + 6 = 11
use the mo diagram for homonuclear diatomic
molecules with s-p interaction as an approximation
2p*
E
2p*
2p
2p
2s*
2s
125
NITRIC OXIDE (NO)
Number of valence electrons: 5 + 6 = 11
use the mo diagram for homonuclear diatomic
molecules with s-p interaction as an approximation
2p*
E
126
2p*
2p
2p
2s*
2s
Bond order
83

 2.5
2
NITRIC OXIDE (NO)
Number of valence electrons: 5 + 6 = 11
use the mo diagram for homonuclear diatomic
molecules with s-p interaction as an approximation
2p*
2p*
E
2p
2p
2s*
127
2s
Bond order
83

 2.5
2
Molecule is stable and
paramagnetic.
Experimental data agrees.
NO+ and CN-
Localized/MO Combination
Recalling the molecular orbital diagram for carbon we
notice that the sigma and sigma antibonding fill before
the pi bonds. Since this does not contribute to the bond
order it makes sense that we can use localized bonding
(line) to describe a sigma bond and use molecular
orbitals on just the pi bonding electrons.
Localized/MO Combination
Consider ozone, O3, we can use lines to connect the
three oxygen atoms together and do a molecular orbital
diagram on just the unhybridized p-orbitals. Then we
arrange the three combinations of p-orbitals in
increasing potential energy order by counting the nodes.
Potential energy is inversely proportional to the number
of nodes.
Localized/MO Combination
A review of adding orbitals
Constructive Interference
+
+
-
+
Destructive Interference
+
+
-
+
Makes bonging orbitals
+
-
-
Makes bonging orbitals
Localized/MO Combination
O
O
O
P=2 , which means two pi bonding electrons
Ozone Molecular Orbital Diagram
two nodes
Pi antibonding
one node
non bonding
zero Node
Pi bonding
Note: Odd electron systems will always have a
nonbonding molecular orbital, which is equal in
potential energy of the p-orbital they were
derived from.
Localized/MO Combination
O
O
O
P=2 , which means two pi bonding electrons
Ozone Molecular Orbital Diagram
two node
Pi antibonding
one node
non bonding
zero Node
Pi bonding
Now fill Pi MO with electrons
Localized/MO Combination
O
O
O
P=2 , which means two pi bonding electrons
Ozone Molecular Orbital Diagram
one node
Pi antibonding
one node
non bonding
zero Node
Pi bonding
Now fill Pi MO with electrons
Localized/MO Combination
O
O
O
P=2 , which means two pi bonding electrons
Ozone Molecular Orbital Diagram
one node
Pi antibonding ( π*)
one node
non bonding
zero Node
Pi bonding ( π)
Now fill Pi MO with electrons
Localized/MO Combination
O
O
O
P=2 , which means two pi bonding electrons
Ozone Molecular Orbital Diagram
one node
Pi antibonding ( π*)
one node
non bonding
zero Node
Pi bonding ( π)
Now fill Pi MO with electrons
The MO diagram shows one pi
bond. And the diagram above shows
two sigma bonds.
Localized/MO Combination
O
O
O
P=2 , which means two pi bonding electrons
Ozone Molecular Orbital Diagram
one node
Pi antibonding ( π*)
one node
non bonding
zero Node
Pi bonding ( π)
Now fill Pi MO with electrons
The MO diagram shows one pi
bond. And the diagram above shows
two sigma bonds.
Therefore the B.O. = 3-0/2 =1.5
Stable and diamagnetic.
Comparison of Theories
• MO theory may provide the most complete
picture of covalent bonding, but it is also the
most difficult to apply to large molecules
and it does not account for molecular
shape.
The End
ChemTour: Partial Charges and
Bond Dipoles
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Students learn that covalent bonds often include unequal
distribution of electrons leading to partial charges on atoms,
bond dipole moments, and molecule polarity. Interactive
Practice Exercises ask students to calculate dipole
moments of polar molecule.
ChemTour: Greenhouse Effect
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This unit explores how excess carbon dioxide and CFCs in
the atmosphere contribute to global warming.
ChemTour: Vibrational Modes
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This tutorial illustrates the three vibrational modes: bending,
symmetric stretching, and asymmetric stretching. Students
learn that molecules can absorb specific wavelengths of
infrared radiation by converting this energy into molecular
vibrations.
ChemTour: Hybridization
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This tutorial animates the formation of hybrid orbitals from
individual s and p orbitals, shows examples of their
geometry, and describes how they can produce single,
double, and triple bonds. Includes Practice Exercises.
ChemTour: Chemistry of the
Upper Atmosphere
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This ChemTour examines how particles of the upper
atmosphere absorb and emit electromagnetic radiation.
ChemTour: Molecular Orbitals
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This animated tutorial offers a patient explanation of
molecular orbital theory, an alternative to the bonding theory
depicted by Lewis dot structures. Includes Practice
Exercises.
Ethylene, which has the molecular formula C2H4, is a
rigid molecule in which all 6 atoms lie in a plane.
Which of the following molecules also has a rigid planar
structure?
A) H2C=C=CH2
Planar Hydrocarbons
B) H2C=C=C=CH2
C) Neither
Consider the following arguments for each answer and
vote again:
A. A combination of 3 carbons and 4 hydrogens can
form the rigid planar molecule H2C=C=CH2.
B. The orientations of the π bonds in H2C=C=C=CH2
alternate in such a way as to create a planar structure.
C. The hybridization of the atomic orbitals on the
carbons prevents the retention of a planar structure in
molecules longer than C2H4.
Planar Hydrocarbons
What is the bond order of the N-O bond in nitrate, NO3-?
A) 1
Bond Order of Nitrate
B) 11/3
C) 2
Consider the following arguments for each answer and
vote again:
A. The majority of the bonds in NO3- are single bonds,
so the bond order is 1.
B. The N-O bond is twice as likely to be a single bond as
it is to be a double bond, so the bond order should be
11/3.
C. The bond order is dictated by the strongest bond,
which in NO3- is a double bond.
Bond Order of Nitrate
Which of the following species is not
paramagnetic in its ground state?
A) NO+
Bond Order of Nitrate
B) NO
C) NO-
Consider the following arguments for each answer
and vote again:
A. NO+ is isoelectronic with N2, which has no unpaired
electrons and hence is not paramagnetic.
B. NO has no electrical charge and thus cannot be
paramagnetic.
C. By pairing an additional electron with the one
unpaired electron in NO, a diamagnetic anion, NO-, is
formed.
Bond Order of Nitrate
According to Valence Shell Electron Pair
Repulsion (VSEPR) theory, 4 objects
around a central atom will have the
tetrahedral arrangement shown to the left
with bond angles of ~109.5º. Which of the
following compounds has a bond angle of
~109.5º?
A) SF2
B) SF3-
Molecular Geometry of SF , SF -, and SF
C) SF4
Please consider the following arguments for each answer and
vote again:
A. SF2 consists of a sulfur atom surrounded by 2 lone electron
pairs and bonded to 2 fluorine atoms, therefore, it has an
approximately tetrahedral bond angle.
B. The tetrahedral VSEPR arrangement of SF3- is formed by a
sulfur atom surrounded by 3 fluorine atoms and by the
additional electron (from the negative charge).
C. Sulfur tetrafluoride is the only molecule with a central atom
(sulfur) surrounded by 4 additional atoms (4 fluorines) and so
is the only molecule with a bond angle of ~109.5º.
Molecular Geometry of SF , SF -, and SF
Which of the following is true of the
bond angle (θ1) in BrF2+ compared to
the bond angle (θ2) in ICl2-?
A) θ1 = θ2
B) θ1 > θ2
Bond Angles of BrF2 and ICl2
C) θ1 < θ2
Please consider the following arguments for each
answer and vote again:
A. Both BrF2+ and ICl2- consist of a central halogen
atom bonded to two halogen atoms, and therefore
should have the same arrangement of atoms.
B. ICl2- has 1 more lone pair of electrons than BrF2+,
which forces the chlorine atoms closer together.
C. ICl2-, with 3 lone pairs, is linear whereas BrF2+, with
2 lone pairs, is bent.
Bond Angles of BrF2 and ICl2
Boron trifluoride (BF3), which has the
structure shown to the left, is capable of
reacting with an unknown compound to
form a new compound without breaking
any bonds. Which of the following
could be the unknown compound?
A) BF3
Reaction of Boron Trifluoride
B) CH4
C) NH3
Please consider the following arguments for each
answer and vote again:
A. BF3 can dimerize to BF3-BF3 by forming a boronboron single bond.
B. By forming a boron-carbon bond, the carbon atom in
CH4 will increase its steric number to 5, thus
expanding its octet to compensate for boron's
incomplete octet.
C. The nitrogen lone electron pair can form a nitrogenboron bond yielding BF3-NH3, isoelectronic with
CH3-CH3.
Reaction of Boron Trifluoride
Pictured to the left is the planar
molecule ethylene, C2H4, which does
not have a permanent electric dipole
moment.
If chlorine atoms were substituted for two hydrogen
atoms, how many of the possible structures would also
not possess a dipole moment?
A) 0
B) 1
Dipole Moments of Dichloroethylene
C) 2
Consider the following arguments for each answer
and vote again:
A. Chlorine atoms always draw electron density away
from carbon atoms, so all possible structures will
possess a dipole moment.
B. Only if the chlorine atoms are diagonally opposite
will the two carbon-chlorine dipole moments cancel
each other.
C. So long as the two chlorine atoms are on different
carbon atoms, no permanent dipole moment will
form.
Dipole Moments of Dichloroethylene
For which central atom "X" does the
anion pictured to the left have a square
planar geometry?
A) C
Molecular Geometry of XF
2-
B) S
C) Xe
Please consider the following arguments for each
answer and vote again:
A. CF42- forms a structure in which the 4 fluorine
atoms form a square plane with one negative charge
on either side of the plane.
B. With 2 lone electron pairs on the sulfur in SF42-, its
steric number is 6.
C. To maximize fluorine-fluorine distances, the 4
fluorine atoms in XeF42- will lie in a plane.
Molecular Geometry of XF
2-