256
H Solutions to Problems
Note that the two-electron states are symmetrized, except the A 2 combination.
The symmetrized states will combine with singlet spin states, while the A 2 state
will be a triplet. One thus has:
1 A 1 =
1
√
2
x(1)x(2) + y(1)y(2)
1
√
2
α(1)β(2) − β(1)α(2)
=
1
√
2
(xα)(xβ)
+
(yα)(yβ)
1 B 1 =
1
√
2
(xα)(yβ)
+
(yα)(xβ)
1 B 2 =
1
√
2
−
(xα)(xβ)
+
(yα)(yβ)
3 A 2 =
(xα)(yα)
The 1 A 1 and 1 B 2 states are the zwitterionic states, while the 1 B 1 and 3 A 2 states
are called the diradical states. It is clear from the expressions that in both cases
the two radical carbon sites are neutral. The zwitterionic states are easily polarizable though.
6.6 The carbon atoms form two orbits. The p z orbital on the central atom is in
the center of the symmetry group and transforms as a ′′
2 . The three methylene
orbitals are in C 2v sites, transforming as the b 2 irrep of the site group, i.e., they
are antisymmetric with respect to ˆ
σ h and symmetric with respect to ˆ
σ v .T h e
induced representation is
b 2 C 2v ↑ D 3h = a
′′
2 + e
′′
(11)
The SALCs are entirely similar to the hydrogen SALCs in the case of ammonia; this implies, for instance, that the component labeled x is symmetric
under the vertical symmetry plane through atom A. It will be antisymmetric
for the twofold-axis going through atom A since the relevant orbital is of p z
type:
|Ψ a =
1
√
3
|p A +|p B +|p C
|Ψ x =
1
√
6
2|p A −|p B −|p C
|Ψ y =
1
√
2
|p B −|p C
The a ′′
2 orbitals interact to yield bonding and antibonding combinations at
E = α ±
√
3β. Since the graph is bipartite, the remaining e ′′ orbitals are neces-