Chemistry Reference
In-Depth Information
5.3
Finding Reducible Representations
In the previous section we deduced irreducible representation labels for two of the vibra-
tional modes of H
2
O by inspecting the molecular geometry and thinking about the possible
movements of the atoms. This is easy enough for simple molecules but the representation
for a particular atomic motion may not be so straightforward for more complex cases.
Also, if there are many vibrational modes, it is unlikely that we would find them all by
inspection. What is required is a general method to find the representations for all possible
molecular vibrations. The approach we will use is first to identify all the irreducible rep-
resentations that are present for a given basis and then interpret each of them in terms of
combinations of the basis functions. The rest of this chapter is dedicated to the first part of
this process and the second part is the subject of Chapter 6.
Staying with the vibrational analysis problem, for a molecule with
N
atoms we would
expect 3
N
degrees of freedom, since each atom can move in three dimensions. For non-
linear molecules, such as H
2
O, the molecule as a whole will have six degrees of freedom:
three translations and three rotations. This means that we would expect 3
N
6 vibrations.
Each of these vibrations will be a collective motion, potentially involving all of the atoms
in the molecule.
However, we can derive the irreducible representations before the patterns of atomic
motion in the vibrational modes are actually identified. This is done by first imagining that
each individual atom is independent of the others. For example, Figure 5.2a shows a basis
of nine vectors for the atoms in the H
2
O molecule. This gives us the expected number of
degrees of freedom, since 3
N
−
9forH
2
O. If we were to set up a matrix representation
for, say, the
C
2
operation in the
C
2v
group for this basis, then we would have to use the
9
=
9 matrix shown in Figure 5.4a.
We saw in Figure 5.2 that because the H atoms are swapped over by the
C
2
rotation
so are their basis vectors. This shows up in the matrix representation as the two sets of
nonzero off-diagonal elements in the lines for H
1
and H
2
. These are off-diagonal elements
because the axes are moved to different atom positions by the rotation. Application of
the matrix to the nine basis vectors is shown in Figure 5.3b, confirming that it correctly
reproduces the operation.
The character for the
C
2
1
operation using this basis is given by trace of the matrix, i.e.
the sum of the diagonal terms, which is
×
1. This is the last time we will resort to the matrix
representation, because this example establishes the following simple rule:
−
If an operation moves an atom to a symmetry-equivalent position, then all basis vectors
to do with that atom will give rise to only off-diagonal elements in the transformation
matrix and so contribute zero to the character for the operation.
σ
v
(
XZ
) operation also swaps the H atoms, and so we need only consider the basis
vectors on the O atom to derive the total character for this operation. This plane causes the
transformation
The
