Chemistry Reference
In-Depth Information
Tab l e 5 . 3
The terms required for the application of the reduction formula to the x, y, z basis
on a central atom in a
D
4h
complex.
D
4h
2
C
2
2
C
2
E
2
C
4
C
2
i
2
S
4
σ
h
2
σ
v
2
σ
d
h
=
16
3
1
−
1
−
1
−
1
−
3
−
1111
C
g
c
χ
i
(
C
)
g
c
χ
i
(
C
)
χ
(
C
)
χ
(
C
)
A
1g
3
2
−
1
−
2
−
2
−
3
−
2122 0
A
2g
3
2
−
1
2
2
−
3
−
21
−
2
−
2
0
B
1g
3
−
2
−
1
−
2
2
−
3
2
1
2
−
2
0
B
2g
3
−
2
−
1
2
−
2
−
3
2
1
−
2
2
0
E
g
6
0
2
0
0
−
6
0
−
2
0
0
0
A
1u
3
2
−
1
−
2
−
232
−
1
−
2
−
2
0
A
2u
3
2
−
1
2
2
3
2
−
1
2
2
16
B
1u
3
−
2
−
1
−
2
2
3
−
2
−
1
−
2
2
0
B
2u
2 0
E
u
6020060200 16
3
−
2
−
1
2
−
23
−
2
−
1
2
−
except
A
2u
and
E
u
, for which the summation gives 16. The order of the group is also 16,
and so Equation (5.19) shows that
=
1
A
2u
+
1
E
u
(5.20)
i.e. the reducible representation
contains 1
A
2u
and 1
E
u
irreducible representations, as we
found using matrices in Chapter 4. The representation
was constructed from three basis
functions and we have found irreducible representations which are for a single object (
A
2u
)
and a degenerate pair of objects (
E
u
).
For any application of the reduction formula we will always find that the number of
objects in the irreducible set of representations is equal to the number used in the
definition of the reducible representation, i.e. the number of basis functions.
Problem 5.5:
BF
3
is a molecule in the
D
3h
point group. Show that the basis of F(p
z
)-
orbitals shown in Figure 5.6 has the reducible representation given in Table 5.4.
p
z
3
F
3
B
F
1
p
z
1
F
2
p
z
2
Figure 5.6
A basis of F(p
z
) orbitals in BF
3
.
Table 5.4 also gives another illustration of the application of the reduction formula to the
basis of three F(p
z
)-orbitals for BF
3
, shown in Figure 5.6. In the reducible representation,
