Chemistry Reference
In-Depth Information
into
1 times the original function, or changed into a different product altogether. This
makes the assignment of a character for these three functions straightforward. For exam-
ple,
xy
becomes
−
yz
under
C
3
1
, and so because the transformation has taken
xy
into another
member of the d-orbital basis set we assign a character of 0.
In most of the entries of Table 5.11 the functional form for the d
z
2
and d
x
2
−
y
2
functions
can be seen as simply unchanged or multiplied by
−
−
1 and so character assignment can be
made accordingly.
However, for d
z
2
and d
x
2
−
y
2
under the
C
3
1
rotation, the character assignment is a little
less clear-cut. In this case, the
x
2
−
y
2
function follows
x
2
z
2
(5.36)
The result indicates that the orbital has changed, but it is not instantly recognizable as
another member of the basis set. This transformation is shown in Figure 5.17a, and com-
parison with Figure 5.13 shows that one of the orbitals used to create the d
z
2
orbital has
been produced. So, we may expect some part of the 2
z
2
−
y
2
→
y
2
−
−
x
2
−
y
2
function to now be present
at the expense of the
x
2
−
y
2
.
C
3
C
3
(a)
Z
Z
C
3
1
Y
Y
X
X
C
3
C
3
(b)
Z
Z
C
3
1
Y
Y
X
X
Figure 5.17
The C
3
1
operation in T
d
showing the result for (a) the
d
x
2
−
y
2
and (b) the
d
z
2
orbitals.
The new function contains
y
and
z
squared, but no products such as
xy
,
yz
or
xz
, and so its
composition in terms of the d-orbital basis will only contain contributions from 2
z
2
−
x
2
−
y
2
and the original
x
2
−
y
2
function. So we can write
a
√
3
y
2
−
z
2
=
(2
z
2
−
x
2
−
y
2
)
+
b
(
x
2
−
y
2
)
(5.37)
