Chemistry Reference
In-Depth Information
Tab l e 5 . 4
The reducible representation for the F(p
z
) basis on BF
3
shown in Figure 5.6 and the application of the reduction formula to it.
D
3h
E
2
C
3
3C
2
σ
h
2
S
3
3
σ
v
h
=
12
3
0
-1
-3
0
1
C
g
c
g
c
χ
χ
χ
i
(
C
)
χ
(
C
)
i
A
1
3
-3
-3
3
0
A
2
3
3
-3
-3
0
E
6
0
-6
0
0
A
1
3
-3
3
-3
0
A
2
3
3
3
3
12
E
6
0
6
0
12
the 2
C
3
and 2
S
3
classes have character 0 in
, and so the terms for these columns in the
reduction formula must always be zero. The order of the
D
3h
point group is 12, and so
summation over classes shows that
1
A
2
1
E
=
+
(5.21)
This means that the three p
z
orbitals can be taken together in one pattern conforming to
A
2
and two degenerate patterns following
E
. In mathematical terms, the patterns are
symmetry-adapted linear combinations (SALCs) of the orbitals that conform to the irre-
ducible representations. The three orbitals taken together as drawn in Figure 5.6 would
give an SALC with a character of
1 for any operation that flips the molecule plane over
and +1 for all others. This is exactly the character set for
A
2
, and so we have found a pic-
torial representation of the SALC. Finding the two degenerate
E
SALCs is more complex
and is left to Chapter 7, where MOs are discussed in more detail.
−
5.6
A Complete Set of Vibrational Modes for H
2
O
In Chapter 1 it was noted that the number of vibrational modes of a molecule can be
calculated by counting the degrees of freedom of the atoms (three per atom for
X
,
Y
and
Z
movement) and subtracting the degrees of freedom for motion of the molecule as a whole,
three for its translation and (for nonlinear molecules) three for rotation. This was used in
Section 5.2 to arrive at a reducible representation for the basis of nine atomic degrees of
freedom for H
2
O, the classic
C
2v
molecule. The characters for this representation were
given in Table 5.1. We can now apply the reduction formula to identify the irreducible
representations for the three vibrations of H
2
O.
Table 5.5 gives the terms generated by the reduction formula for each of the irreducible
representations in
C
2v
. This shows that
=
3A
1
+
A
2
+
2B
1
+
3B
2
(5.22)
As expected, the nine basis vectors have produced nine irreducible representations, but
only 9
3 of these can correspond to molecular vibrations. The others are motions of
the molecule as a whole.
−
6
=
