Chemistry Reference
In-Depth Information
.
Resonance between all VB structures
The complete VB problem, arising from the mixing of all five VB
structures, is rather tedious since it would involve solution of a fifth-
order determinantal equation. We can, however, simplify the problem
using symmetry arguments, if we are only interested in the ground-state
energy of the system, as we are. In fact, symmetry suggests that
Y ¼ðc
1
þc
2
Þ
c
K
þðc
3
þc
4
þc
5
Þ
c
D
¼ Y
K
c
K
þY
D
c
D
ð
9
:
121
Þ
where
Y
K
and
Y
D
are the un-normalized combinations of equivalent
Kekul
e and Dewar structures respectively. In this way, we reduce the
full-VB problem to the solution of a simple quadratic secular equation.
We have
¼
0
H
KK
ES
KK
H
KD
ES
KD
H
KD
ES
KD
H
DD
ES
DD
ð
9
:
122
Þ
where, from (9.108)-(9.110):
8
<
5
2
ð
Q
E
Þþ
6K
;
9
2
ð
Q
E
Þþ
9K
H
KK
ES
KK
¼
H
DD
ES
DD
¼
:
H
KD
ES
KD
¼
3
ð
Q
E
Þþ
9K
ð
9
:
123
Þ
Therefore, the (2
2) secular equation is
5
2
x
þ
6
3x
þ
9
¼
0
ð
9
:
124
Þ
9
2
x
þ
9
3x
þ
9
giving upon expansion
p
p
13
;
x
¼
1
x
2
þ
2x
12
¼
0
)
x
¼
1
þ
13
ð
9
:
125
Þ
Taking the lowest root, we obtain the strongly bonding ground state
E
¼
Q
þ
2
:
6055K
ð
9
:
126
Þ
