1 chemistry 161 chapter 10 chemical bonding ii
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1
CHEMISTRY 161
Chapter 10
Chemical Bonding II
www.chem.hawaii.edu/Bil301/welcome.html
2
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
3
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
1sA – 1sB = MO2
constructive interference
destructive interference
4
0r/a-3/2
01 e
12)(
a
rRs
5
ADDITION OF ORBITALSbuilds up electron density in overlap region
1sA + 1sB = MO1
combine them by addition
A B
6
ADDITION OF ORBITALSbuilds up electron density in overlap region.
1sA + 1sB = MO1
A Bwhat do we notice?
electron density between atoms
7
SUBTRACTION OF ORBITALSresults in low electron density in overlap region..
1sA – 1sB = MO2
A B
subtract
8
SUBTRACTION OF ORBITALSresults in low electron density in overlap region..
1sA – 1sB = MO2
A Bwhat do we notice?
no electron density between atoms
9
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei
10
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei
1sA – 1sB = MO2
results in low electron density between nuclei
BONDING
ANTI-BONDING
11
12
THE MO’s FORMED BY TWO 1s ORBITALS
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1sA + 1sB = MO1
1sA – 1sB = MO2
sigma anti-bonding = 1s*
sigma bonding = 1s
1s
1s*
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E
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
COMBINING TWO 1s ORBITALS
15
E
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
1sA+1sB
MO
1s
16
E
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
1sA-1sB
MO
1sA+1sB
MO
1s
1s*
17
E 1sAA B
1s
1s*
1sB
COMBINING TWO 1s ORBITALS
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E1s
1s*
1s
1s
H HH2
bonding in H2
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E1s
1s*
1s
1s
H HH2
the electrons are placed in the 1s molecular orbitals
20
E1s
1s*
1s
1s
H2: (1s)2
H HH2
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E1s
1s*
1s
1s
He2
He HeHe2
atomic configuration of He 1s2
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E1s
1s*
1s
1s
He2: (1s)2(1s*)2
He HeHe2
bonding effect of the (1s)2 is cancelled by the
antibonding effect of (1s*)2
23
BOND ORDER
net number of bonds existing after the cancellation of bonds by antibonds
the two bonding electrons were cancelled out by the two antibonding electrons
He2
(1s)2(1s*)2
the electronic configuration is….
BOND ORDER = 0
24
BOND ORDER
=
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
Bondorder
the molecule is predicted to be stable
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BOND ORDER
= {
high bond order indicates high bond energy and short bond length
# of bonding electrons(nb)
# of antibonding electrons (na)
– 1/2 }
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
Bondorder
the molecule is predicted to be stable
H2+,H2,He2
+
= 1/2 (nb - na)
26
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
E
He2+ He2H2
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1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
E
He2+ He2H2
Dia-
1
436
74
28
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
Para-
½
225
106
E
He2+ He2H2
Dia-
1
436
74
29
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
Para-
½
225
106
E
He2+
Para-
½
251
108
He2H2
Dia-
1
436
74
30
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
First row diatomic molecules and ions
H2+
Para-
½
225
106
E
He2+
Para-
½
251
108
He2
—
0
—
—
H2
Dia-
1
436
74
31
HOMONUCLEAR DIATOMICS
Li2 Li : 1s22s1
both the 1s and 2s overlap to produce bonding and anti-bonding orbitals
second period
32
E
1s
1s*
1s
1s
2s
2s*
2s
2s
ENERGY LEVEL DIAGRAM FOR DILITHIUM
Li2
33
E
1s
1s*
1s
1s
2s
2s*
2s
2s
Li2
ELECTRONS FOR DILITHIUM
34
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
Li2
Bond Order ?
35
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
Li2
nb = 4 na = 2
Bond Order = 1
single bond.
36
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
the 1s and 1s* orbitals can be ignored when
both are FILLED!
Li2
omit the inner shell
37
E2s
2s*
2s
2s
Li LiLi2
The complete configuration is: (1s)2(1s*)2 (2s)2
Li2 (2s)2 only valence orbitals contribute to molecular bonding
38
E2s
2s*
2s
2s
Be BeBe2 Be2
39
E2s
2s*
2s
2s
Be2Be BeBe2
Electron configuration for DIBERYLLIUM
Configuration: (2s)2(2s*)2 Bond order = 0
40
E2s
2s*
2s
2s
(2s)2(2s*)2Be BeBe2
Be2
Electron configuration for DIBERYLLIUM
nb = 2
na = 2
Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0
No bond!!! The molecule is not stable! Now B2...
41
B2
the Boron atomic configuration is
1s22s22p1
form molecular orbitals
we expect B to use 2p orbitals to
addition and subtraction
42
-molecular orbitals
43
molecular orbitals
44
ENERGY LEVEL DIAGRAM
E
2s
2s*
2s
2s
45
2p*
2p
2p
2p*
E2p 2p
46
E
expected orbital splitting
2s
2s*
2s
2s
2p
2p*
2p
2p
2p
2p*
This pushes the 2p up
47
E
MODIFIED ENERGY LEVEL DIAGRAM
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
Notice that the 2p and 2p
have changed places!!!!
48
E
2s
2s*
2s
2s
Electron configuration for B2
2p
2p*
2p2p
2p
2p*
Place electrons from 2s into 2s and 2s*
B is [He] 2s22p1
49
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
Place electrons from 2p into 2p and 2p
Remember HUND’s RULE
50
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*(2s)2(2s*)2(2p)2
Abbreviated configuration
Complete configuration
(1s)2(1s*)2(2s)2(2s*)2(2p)2
ELECTRONS ARE UNPAIRED
51
E
2s
2s*
2s
2s
Electron configuration for B2:
Bond order
2p
2p*
2p2p
2p
2p*(2s)2(2s*)2(2p)2
Molecule is predicted to be stable and paramagnetic.
na = 2
nb = 4
1/2(nb - na)
= 1/2(4 - 2) =1
52
A SUMMARY OF THE MO’s
Emphasizing nodal planes
53
ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR DIATOMICS
B2 C2 N2 O2 F2Li2
54
B2 C2 N2
O2 F2
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
2s
2s*
2s
2s
2p*
2p
2p
2p
2p*
2p
Li2
55
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2 C2 N2 O2 F2
E
56
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2 N2 O2 F2
E
57
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2 O2 F2
E
58
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2 F2
E
59
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2
Para-
2
495
121
F2
E
NOTE SWITCH OF LABELS
60
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2
Para-
2
495
121
F2
Dia-
1
154
143
E
NOTE SWITCH OF LABELS
61
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
–
O2 :
O2+ :
O2– :
O22-:
O22-
62
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
63
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
64
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
65
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
66
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
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
67
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
O2 : B.O. = 2
O2+ : B.O. = 2.5
O2– : B.O. = 1.5
O22- : B.O. = 1
O2+ >O2 >O2
– > O22-
BOND ENERGY ORDER
68
O O
OXYGEN
How does the Lewis dot picture correspond to MOT?
2p*
2p*
2p
2p
2s*
2s
E
12 valence electrons
BO = 2 but PARAMAGNETIC
69
Homework
Chapter 10
pages 397-409, problem sets