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CHEMISTRY 161
Chapter 10
Chemical Bonding II
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1
MOLECULAR ORBITAL THEORY
electrons occupy orbitals each of which spans
the entire molecule
molecular orbitals each hold up to two electrons
and obey Hund’s rule, just like atomic orbitals
2
H2 molecule:
1s orbital on Atom A
1s orbital on Atom B
the H2 molecule’s molecular orbitals can be
constructed from the two 1s atomic orbitals
1sA + 1sB = MO1
constructive interference
1sA – 1sB = MO2
destructive interference
3
 1s
 1 
 R ( r )  2

 a0 
3/2
e
- r/a
0
4
ADDITION OF ORBITALS
builds up electron density in overlap region
1sA + 1sB = MO1
A
B
combine them by addition
5
ADDITION OF ORBITALS
builds up electron density in overlap region.
1sA + 1sB = MO1
A
B
what do we notice?
electron density between atoms
6
SUBTRACTION OF ORBITALS
results in low electron density in overlap region..
1sA – 1sB = MO2
A
B
subtract
7
SUBTRACTION OF ORBITALS
results in low electron density in overlap region..
1sA – 1sB = MO2
A
B
what do we notice?
no electron density between atoms
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COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei
9
COMBINATION OF ORBITALS
1sA – 1sB = MO2 ANTI-BONDING
results in low electron density between nuclei
1sA + 1sB = MO1
BONDING
builds up electron density between nuclei
10
11
THE MO’s FORMED BY TWO
1s ORBITALS
12
1sA – 1sB = MO2
s1s*
sigma anti-bonding = s1s*
s1s
1sA + 1sB = MO1
sigma bonding = s1s
13
COMBINING TWO 1s ORBITALS
E
Energy of a
1s orbital in
a free atom
A
B
Energy of
a 1s orbital
in a free
atom
14
E
Energy of a
1s orbital in
a free atom
A
B
s1s
Energy of
a 1s orbital
in a free
atom
1sA+1sB
MO
15
1sA-1sB
MO
E
Energy of a
1s orbital in
a free atom
A
s1s*
s1s
B
Energy of
a 1s orbital
in a free
atom
1sA+1sB
MO
16
COMBINING TWO 1s ORBITALS
s1s*
E
1sA
A
B
1sB
s1s
17
bonding in H2
H
H2
H
s1s*
E
1s
1s
s1s
18
H
H2
H
s1s*
E
1s
1s
s1s
the electrons are placed in the s1s molecular orbitals
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H
H2
H
s1s*
E
1s
1s
s1s
H2: (s1s)2
20
He2
atomic configuration of He
He
He2
1s2
He
s1s*
E
1s
1s
s1s
21
He2: (s1s)2(s1s*)2
He
He2
He
s1s*
E
1s
1s
s1s
bonding effect of the (s1s)2 is cancelled by the
antibonding effect of (s1s*)2
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BOND ORDER
net number of bonds existing after the
cancellation of bonds by antibonds
He2
the electronic configuration is….
(s1s)2(s1s*)2
the two bonding electrons were cancelled out
by the two antibonding electrons
BOND ORDER = 0
23
BOND ORDER
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
the molecule is predicted to be stable
Bond
order
=
24
BOND ORDER
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
the molecule is predicted to be stable
Bond
order
# of bonding
–
electrons(nb)
= 1/2{
= 1/2 (n
b
# of antibonding
electrons (na)
}
- na)
high bond order indicates high bond
energy and short bond length
H2+,H2,He2+
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H2
H2+
He2+
He2
s1s*
E
s1s
Magnetism
Bond order
Bond energy
(kJ/mol)
Bond length
(pm)
26
H2
H2+
He2+
He2
s1s*
E
s1s
Magnetism
Dia-
Bond order
1
Bond energy
(kJ/mol)
436
Bond length
(pm)
74
27
H2
H2+
Magnetism
Dia-
Para-
Bond order
1
½
Bond energy
(kJ/mol)
436
225
Bond length
(pm)
74
106
He2+
He2
s1s*
E
s1s
28
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
He2
s1s*
E
s1s
29
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
—
s1s*
E
s1s
30
second period
HOMONUCLEAR DIATOMICS
Li2
Li : 1s22s1
both the 1s and 2s overlap to produce s
bonding and anti-bonding orbitals
31
ENERGY LEVEL DIAGRAM FOR DILITHIUM
s2s*
Li2
2s
2s
s2s
E
s1s*
1s
1s
s1s
32
ELECTRONS FOR DILITHIUM
s2s*
2s
Li2
2s
s2s
E
s1s*
1s
1s
s1s
33
Electron configuration for DILITHIUM
s2s*
Li2
(s1s)2(s1s*)2(s2s)2
2s
2s
s2s
E
Bond Order ?
1s
1s
s1s
34
Electron configuration for DILITHIUM
s2s*
Li2
(s1s)2(s1s*)2(s2s)2
2s
2s
nb = 4
s2s
E
1s
Bond Order = 1
1s
s1s
na = 2
single bond.
35
Electron configuration for DILITHIUM
s2s*
Li2
(s1s)2(s1s*)2(s2s)2
2s
2s
s2s
E
1s
the s1s and s1s* orbitals
can be ignored when
both are FILLED!
1s
s1s
omit the inner shell
36
Li2
only valence orbitals contribute to
molecular bonding
(s2s)2
Li
Li2
Li
s2s*
E
2s
2s
s2s
The complete configuration is: (s1s)2(s1s*)2 (s2s)2
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Be
Be2
Be
s2s*
Be2
E
2s
2s
s2s
38
Electron configuration for DIBERYLLIUM
Be2
Be
Be2
Be
s2s*
E
2s
2s
s2s
Configuration:
(s2s)2(s2s*)2
Bond order = 0
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Electron configuration for DIBERYLLIUM
Be2
Be
Be2
Be
(s2s)2(s2s*)2
s2s*
nb = 2
E
2s
2s
na = 2
s2s
Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0
No bond!!! The molecule is not stable!
Now B2...
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B2
the Boron atomic configuration is
1s22s22p1
we expect B to use 2p orbitals to
form molecular orbitals
addition and subtraction
41
s-molecular orbitals
42
p molecular orbitals
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ENERGY LEVEL DIAGRAM
E
s2s*
2s
2s
s2s
44
s2p*
p2p*
2p
E
2p
p2p
s2p
45
expected orbital splitting
s2p*
p2p*
2p
p2p
2p
s2p
E
s2s*
2s
2s
s2s
This pushes the s2p up
46
MODIFIED ENERGY LEVEL DIAGRAM
s2p*
p2p*
2p
E
s2p
p2p
s2s*
2s
2p
Notice that the s2p and p2p
have changed places!!!!
2s
s2s
47
Electron configuration for B2
s2p*
B is [He] 2s22p1
p2p*
2p
E
s2p
p2p
s2s*
2s
2p
Place electrons from 2s
into s2s and s2s*
2s
s2s
48
s2p*
p2p*
2p
E
s2p
p2p
s2s*
2s
2p
Place electrons from 2p
into p2p and p2p
2s
s2s
Remember HUND’s RULE
49
ELECTRONS
ARE
UNPAIRED
s2p*
Abbreviated configuration
p *
2p
(s2s)2(s2s*)2(p2p)2
2p
E
s2p
p2p
s2s*
2s
2p
Complete configuration
(s1s)2(s1s*)2(s2s)2(s2s*)2(p2p)2
2s
s2s
50
Electron configuration for B2:
s2p*
p2p*
(s2s)2(s2s*)2(p2p)2
na = 2
2p
E
s2p
p2p
s2s*
2s
s2s
nb = 4
2p
Bond order 1/2(nb - na)
= 1/2(4 - 2) =1
2s Molecule is
predicted to be
stable and
51
paramagnetic.
A SUMMARY OF THE MO’s
Emphasizing nodal planes
52
ELECTRONIC CONFIGURATION OF THE
HOMONUCLEAR DIATOMICS
Li2
B2
C2
N2
O2
F2
53
Li2
E
2p
B2
C2
O2
N2
s2p*
s2p*
p2p*
s2p
p2p
F2
p2p*
2p
2p
p2p
2p
s2p
s2s*
s2s*
2s
2s
s2s
2s
2s
s2s
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Second row diatomic molecules
B2
C2
N2
O2
F2
s2p*
p2p*
E
s2p
p2p
s2s*
s2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
55
Second row diatomic molecules
B2
C2
N2
O2
F2
s2p*
p2p*
E
s2p
p2p
s2s*
s2s
Magnetism
Para-
Bond order
1
Bond E. (kJ/mol)
290
Bond length(pm)
159
56
Second row diatomic molecules
B2
C2
Magnetism
Para-
Dia-
Bond order
1
2
Bond E. (kJ/mol)
290
620
Bond length(pm)
159
131
N2
O2
F2
s2p*
p2p*
E
s2p
p2p
s2s*
s2s
57
Second row diatomic molecules
B2
C2
N2
Magnetism
Para-
Dia-
Dia-
Bond order
1
2
3
Bond E. (kJ/mol)
290
620
942
Bond length(pm)
159
131
110
O2
F2
s2p*
p2p*
E
s2p
p2p
s2s*
s2s
58
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
Bond length(pm)
159
131
110
121
F2
s2p*
p2p*
E
p2p
s2p
s2s*
s2s
59
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
Bond length(pm)
159
131
110
121
143
60
s2p*
p2p*
E
p2p
s2p
s2s*
s2s
O2
O 2+
O 2–
O22-
s2p*
E
p2p*
p2p
s2p
s2s*
s2s
O2 :
O 2+ :
O 2– :
O22-:
61
O2
O 2+
O 2–
O22-
s2p*
E
p2p*
p2p
s2p
s2s*
s2s
62
O2
O 2+
O 2–
O22-
s2p*
E
p2p*
p2p
s2p
s2s*
s2s
63
O2
O 2+
O 2–
O22-
s2p*
E
p2p*
p2p
s2p
s2s*
s2s
64
O2
O 2+
O 2–
O22-
s2p*
E
p2p*
p2p
s2p
s2s*
s2s
65
O2
O 2+
O 2–
O22-
s2p*
E
p2p*
p2p
s2p
s2s*
s2s
O2 :
B.O. = (8 - 4)/2 = 2
O2+ : B.O. = (8 - 3)/2 = 2.5
O2– : B.O. = (8 - 5)/2 = 1.5
O22- : B.O. = (8 - 6)/2 = 1
66
O2
O 2+
O 2–
O22-
s2p*
E
p2p*
p2p
s2p
s2s*
s2s
O2 :
B.O. = 2
O2+ : B.O. = 2.5
BOND ENERGY ORDER
O2+ >O2 >O2– > O22-
O2– : B.O. = 1.5
O22- : B.O. = 1
67
OXYGEN
O
O
How does the Lewis dot picture correspond to MOT?
12 valence electrons
s2p*
p2p*
E
s2p
BO = 2 but PARAMAGNETIC
p2p
s2s*
s2s
68
Homework
Chapter 10
pages 397-409, problem sets
69