Chemistry Reference
In-Depth Information
The modes give a further illustration of the selection rules derived at the beginning of
the chapter. In Section 6.3.1 it was shown that we expect only the
B
2u
and
B
3u
to be IR
active. The C
H bonds in 1,4-difluorobenzene will have local dipole moments with a
δ
+
charge on H and
charge on C: as a C H bond shortens, the local dipole will decrease;
and as the bond lengthens, the local dipole will increase. The motion of the
A
g
and
B
1g
modes does not give IR activity because the changes in local dipoles opposite one another
cancel out. For the
B
2u
and
B
3u
modes, the changes in local dipoles opposite to one another
reinforce and the net effect is a changing dipole moment in the
Y
-direction for
B
2u
and
in
X
for
B
3u
. This was predicted from the rightmost columns of the
D
2h
character table in
Section 6.3.1. The origin of Raman selectivity is more difficult to visualize, as it depends
on the properties of the polarization matrix.
δ
−
6.6.2 The Projection Operator and Degenerate Representations
The choice of the generating vector for a point group without degenerate states is quite
straightforward. In the examples so far we have simply taken the first basis vector in the
list and used this as the generating vector. In Problem 6.11 it was found that selecting any
of the other vectors in the basis would give equivalent results.
If the reduction formula has given degenerate representations then we expect two vibra-
tions for each
E
representation and three for each
T
representation. But the projection of a
single vector will give only a single mode, and so we must find an alternative generating
function for the others.
Ammonia, NH
3
,C
3
v
As an example we will look at the N H stretching modes in ammonia. The basis of
N H bonds in ammonia was used in Chapter 4 as part of the development of matrices
in symmetry. We can now use this basis (defined in Figure 4.7) to consider degenerate
vibrations of a molecule. The reducible representation and the application of the reduction
formula for the three-vector basis is shown in Table 6.12. Once the totals are divided by
the order of the group (
h
=
6in
C
3v
), we find
=
A
1
+
E
(6.40)
So the three basis vectors give rise to a vibration of
A
1
symmetry and two degenerate
vibrations conforming to the
E
irreducible representation. From the character table we
expect the
A
1
and
E
modes to be both IR and Raman active.
Table 6.12
Application of the reduction formula to the ammonia
N H bond basis defined in Figure 4.7.
C
3v
E
2
C
3
3
σ
h
=
6
v
3
0
1
C
g
c
χ
i
(
C
)
g
c
χ
i
(
C
)
χ
(
C
)
χ
(
C
)
A
1
3
3
6
A
2
3
−
3
0
E
6
0
6
