Chemistry Reference
In-Depth Information
(a)
H
1
H
3
H
2
1
3
(
b
1
+
b
2
+
b
3
)
A
1
:
(b)
(c)
H
1
H
1
H
3
H
3
H
2
H
2
E
mode from projection of
b
1
:
E
mode from projection of
b
2
−
b
3
:
1
6
1
2
(2
b
1
-
b
2
-
b
3
)
(
b
2
-
b
3
)
Figure 6.20
The N H stretching modes of ammonia shown as SALCs of the three-bond
vector basis. (a) The A
1
symmetric stretch mode; (b) and (c) the degenerate pair of modes
for the E irreducible representation.
and we have assumed
b
1
,
b
2
and
b
3
are orthogonal to one another and normalized (Equa-
tions (6.29) and (6.30)). This is just the same as the
A
1
mode in Table 6.13, and so is not a
good candidate as a generating vector.
The second orthogonal function is
b
2
2
b
3
2
b
2
−
b
3
because
(
b
2
−
b
3
)(2
b
1
−
b
2
−
b
3
)
=−
+
=
0
(6.44)
Problem 6.13:
Show that
b
2
−
b
3
is also orthogonal to
b
1
+
b
2
+
b
3
.
The projection operator for the
E
representation of this new function is shown in
Table 6.14. The function is unaltered by the process other than a scaling factor; after
normalization we obtain
1
√
2
φ
2
(
E
)
=
(
b
2
−
b
3
)
(6.45)
The form of this vibration is shown in Figure 6.20c. It should be remembered that
φ
1
(
E
)
and
φ
2
(
E
) form a degenerate pair which are found in this form because of our choice of
b
1
as the initial projection operator. They look like very different motions, but symmetry tells
us that they belong to the same degenerate irreducible representation and so must have the
same vibrational frequency.
