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Chapter 11
Theories of Covalent Bonding
11-1
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Theories of Covalent Bonding
11.1 Valence Bond (VB) Theory and Orbital Hybridization
11.2 The Mode of Orbital Overlap and the Types of Covalent Bonds
11.3 Molecular Orbital (MO)Theory and Electron Delocalization
11-2
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The Central Themes of VB Theory
Basic Principle
A covalent bond forms when the orbitals of two atoms overlap
and the overlap region, which is between the nuclei, is
occupied by a pair of electrons.
The two wave functions are in phase so the amplitude increases
between the nuclei.
11-3
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The Central Themes of VB Theory
Themes
A set of overlapping orbitals has a maximum of two electrons
that must have opposite spins.
The greater the orbital overlap, the stronger (more stable) the
bond.
The valence atomic orbitals in a molecule are different from
those in isolated atoms.
There is a hybridization of atomic orbitals to form molecular
orbitals.
11-4
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Figure 11.1
Orbital overlap and spin pairing in three
diatomic molecules.
Hydrogen, H2
Hydrogen fluoride, HF
Fluorine, F2
11-5
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Hybrid Orbitals
Key Points
The number of hybrid orbitals obtained equals the number of
atomic orbitals mixed.
The type of hybrid orbitals obtained varies with the types of
atomic orbitals mixed.
Types of Hybrid Orbitals
sp
11-6
sp2
sp3
sp3d
sp3d2
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Figure 11.2
The sp hybrid orbitals in gaseous BeCl2.
atomic
orbitals
hybrid
orbitals
orbital box diagrams
11-7
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Figure 11.2
(continued)
The sp hybrid orbitals in gaseous BeCl2.
orbital box diagrams with orbital contours
11-8
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Figure 11.3
11-9
The sp2 hybrid orbitals in BF3.
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Figure 11.4
11-10
The sp3 hybrid orbitals in CH4.
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Figure 11.5
11-11
The sp3 hybrid orbitals in NH3.
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Figure 11.5
(continued)
11-12
The sp3 hybrid orbitals in H2O.
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Figure 11.6
11-13
The sp3d hybrid orbitals in PCl5.
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Figure 11.7
11-14
The sp3d2 hybrid orbitals in SF6.
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11-15
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Figure 11.8
The conceptual steps from molecular formula to the hybrid orbitals
used in bonding.
Step 1
Molecular
formula
Step 2
Lewis
structure
Figure 10.1
11-16
Step 3
Molecular shape
and e- group
arrangement
Figure 10.12
Table 11.1
Hybrid
orbitals
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SAMPLE PROBLEM 11.1
PROBLEM:
Postulating Hybrid Orbitals in a Molecule
Use partial orbital diagrams to describe mixing of the atomic
orbitals of the central atom leads to hybrid orbitals in each of
the following:
(a) Methanol, CH3OH
PLAN:
Use the Lewis structures to ascertain the arrangement of
groups and shape of each molecule. Postulate the hybrid
orbitals. Use partial orbital box diagrams to indicate the hybrid
for the central atoms.
SOLUTION:
(a) CH3OH
H
C O
H
H H
11-17
(b) Sulfur tetrafluoride, SF4
The groups around C are
arranged as a tetrahedron.
O also has a tetrahedral
arrangement with 2 nonbonding
e- pairs.
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SAMPLE PROBLEM 11.1
Postulating Hybrid Orbitals in a Molecule
continued
2p
2s
2p
sp3
single C atom
hybridized
C atom
2s
sp3
hybridized
O atom
single O atom
(b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs.
F
F S
F
F
3d
3d
3p
sp3d
3s
11-18
S atom
hybridized
S atom
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Figure 11.9
both C are sp3 hybridized
The s bonds in ethane(C2H6).
s-sp3 overlaps to s bonds
sp3-sp3 overlap to form a s bond
relatively even
distribution of electron
density over all s
bonds
11-19
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Figure 11.10
11-20
The s and p bonds in ethylene (C2H4).
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Figure 11.11
11-21
The s and p bonds in acetylene (C2H2).
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Figure 11.12
11-22
Electron density and bond order.
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SAMPLE PROBLEM 11.2
PROBLEM:
PLAN:
Describing the Bond in Molecules
Describe the types of bonds and orbitals in acetone, (CH3)2CO.
Use the Lewis structures to ascertain the arrangement of groups and
shape at each central atom. Postulate the hybrid orbitals taking note of
the multiple bonds and their orbital overlaps.
sp2
SOLUTION:
sp2
sp3 hybridized
O
sp3
hybridized H
C
sp
H
C
C
H
H H
H
sp2 hybridized
H
2
sp3
H sp
2
sp2 C sp
C
3
sp H
sp3
H sp3
sp3
sp3
H
3
sp 3 H
sp
sbonds
11-23
O
2
O
C
C
H3 C
CH3
pbond
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The Central Themes of MO Theory
A molecule is viewed on a quantum mechanical level as a
collection of nuclei surrounded by delocalized molecular orbitals.
Atomic wave functions are summed to obtain molecular wave
functions.
If wave functions reinforce each other, a bonding MO is formed
(region of high electron density exists between the nuclei).
If wave functions cancel each other, an antibonding MO is formed
(a node of zero electron density occurs between the nuclei).
11-24
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Figure 11.13
An analogy between light waves and atomic wave functions.
Amplitudes of wave
functions added
Amplitudes of
wave functions
subtracted.
11-25
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Figure 11.14
Contours and energies of the bonding and antibonding
molecular orbitals (MOs) in H2.
The bonding MO is lower in energy and the antibonding MO is higher in
energy than the AOs that combined to form them.
11-26
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Figure 11.15
The MO diagram for H2.
Filling molecular orbitals with electrons follows the
same concept as filling atomic orbitals.
Energy
s*1s
1s
1s
H2 bond order
= 1/2(2-0) = 1
s1s
AO
of H
11-27
MO
of H2
AO
of H
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Figure 11.16
MO diagram for He2+ and He2.
s*1s
1s
1s
Energy
Energy
s*1s
1s
1s
s1s
AO of
He
MO of
He+
s1s
AO of
He+
He2+ bond order = 1/2
11-28
AO of
He
MO of
He2
AO of
He
He2 bond order = 0
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SAMPLE PROBLEM 11.3 Predicting Stability of Species Using MO Diagrams
PROBLEM:
PLAN:
Use MO diagrams to predict whether H2+ and H2- exist.
Determine their bond orders and electron configurations.
Use H2 as a model and accommodate the number of electrons in
bonding and antibonding orbitals. Find the bond order.
SOLUTION:
bond order
= 1/2(1-0)
= 1/2
s
1s
1s
AO of H
AO of H
s
MO of H2
H2- does exist
H2 does exist
1s
11-29
s
+
bond order
= 1/2(2-1)
= 1/2
configuration is
+
1s
AO of H-
AO of H
(s1s)1
s
MO of H2-
configuration is
(s1s)2(s2s)1
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Energy
Figure 11.17
Bonding in s-block homonuclear diatomic molecules.
s*2s
s*2s
2s
2s
s2s
2s
2s
s2s
Be2
Li2
Li2 bond order = 1
11-30
Be2 bond order = 0
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Figure 11.18
11-31
Contours and energies of s and p MOs through
combinations of 2p atomic orbitals.
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Figure 11.19
Relative MO energy levels for Period 2 homonuclear
diatomic molecules.
without 2s-2p
mixing
with 2s-2p
mixing
MO energy levels
for O2, F2, and Ne2
MO energy levels
for B2, C2, and N2
11-32
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Figure 11.20
MO occupancy
and molecular
properties for B2
through Ne2
11-33
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Figure 11.21
The paramagnetic
properties of O2
11-34
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SAMPLE PROBLEM 11.4
PROBLEM:
Using MO Theory to Explain Bond Properties
As the following data show, removing an electron from N2 forms
an ion with a weaker, longer bond than in the parent molecules,
whereas the ion formed from O2 has a stronger, shorter bond:
N2
N2+
O2
O 2+
Bond energy (kJ/mol)
945
841
498
623
Bond length (pm)
110
112
121
112
Explain these facts with diagrams that show the sequence and occupancy of MOs.
PLAN:
Find the number of valence electrons for each species, draw the MO
diagrams, calculate bond orders, and then compare the results.
SOLUTION:
N2 has 10 valence electrons, so N2+ has 9.
O2 has 12 valence electrons, so O2+ has 11.
11-35
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SAMPLE PROBLEM 11.4
Using MO Theory to Explain Bond Properties
continued
N2+
N2
bonding e- lost
1/2(8-2)=3
11-36
O2 +
O2
s2p
s2p
p2p
p2p
s2p
s2p
p2p
p2p
s2s
s2s
s2s
s2s
1/2(7-2)=2.5
bond
orders
1/2(8-4)=2
antibonding
e- lost
1/2(8-3)=2.5