Chemistry Reference
In-Depth Information
z
x
(a)
C
3
C
3
C
3
1
z
y
x
y
z
x
(b)
C
3
C
3
C
3
1
z
y
x
y
Figure 5.20
The effect of a C
3
1
rotation in O
h
on (a)
d
x
2
−
y
2
and (b)
d
z
2
.
These transformations are illustrated in Figure 5.20, which shows that the corresponding
orbitals are orientated in a nonstandard way.
To find the characters for these orbitals we need to form linear combinations of the
original basis set that are equivalent to the transformed functions. The functions we seek
contain no products of axes; so, in a similar manner to the earlier
T
d
example, the
xy
,
xz
and
yz
functions can be disregarded. Taking the
z
2
−
x
2
case first:
a
√
3
(2
z
2
z
2
x
2
x
2
y
2
)
b
(
x
2
y
2
)
−
=
−
−
+
−
(5.48)
for which
b
is the amount of the original
x
2
y
2
remaining after the transformation, i.e.
the character. This appears in the coefficient of both
x
2
and
y
2
; however, these coefficients
will also contain the unknown coefficient
a
, so this must be obtained first. From the
z
2
coefficients we have
−
√
3
2
2
a
√
3
=
=
1
i.e.
a
(5.49)
Now, from the
y
2
coefficients:
a
√
3
−
a
√
3
1
2
0
=−
b
giving
b
=−
and so
b
=−
(5.50)
y
2
by the selected
C
3
1
operation is
i.e. the character for the transformation of
x
2
−
−
1
/
2.
y
2
case we start from the result of Equation (5.47) and form a new
linear combination with unknown coefficients:
For the 2
z
2
−
x
2
−
1
√
3
a
√
3
(2
y
2
−
z
2
−
x
2
)
=
(2
z
2
−
x
2
−
y
2
)
+
b
(
x
2
−
y
2
)
(5.51)
