Applications in Vibrational Spectroscopy - Molecular Symmetry - page 207

Table 6.16 The application of the reduction formula to the reducible representation for the
vibrational modes of benzene (Table 6.15). The classes with zero in are omitted as they will

not contribute to the summation. However, the order of the group h is set by the total number
of operations in the complete D 6h character table.

3 C 2 D 6h E σ h 3 σ v h = 24

36 − 4 12 4

C g c χ i ( C )

g c χ i ( C ) χ ( C ) χ ( C ) n i

A 1g 1 × 36 × 13 ×− 4 × 1 1 × 12 × 13 × 4 × 1 48 2

A 2g 1 × 36 × 13 ×− 4 ×− 11 × 12 × 13 × 4 ×− 1 48 2

B 1g 1 × 36 × 13 ×− 4 × 1 1 × 12 ×− 13 × 4 ×− 1 0 0

B 2g 1 × 36 × 13 ×− 4 ×− 11 × 12 ×− 13 × 4 × 1 48 2

E 1g 1 × 36 × 23 ×− 4 × 0 1 × 12 ×− 23 × 4 × 0 48 2

E 2g 1 × 36 × 23 ×− 4 × 0 1 × 12 × 23 × 4 × 0 96 4

A 1u 1 × 36 × 13 ×− 4 × 1 1 × 12 ×− 13 × 4 ×− 1 0 0

A 2u 1 × 36 × 13 ×− 4 ×− 11 × 12 ×− 13 × 4 × 1 48 2

B 1u 1 × 36 × 13 ×− 4 × 1 1 × 12 × 13 × 4 × 1 48 2

B 2u 1 × 36 × 13 ×− 4 ×− 11 × 12 × 13 × 4 ×− 1 48 2

E 1u 1 × 36 × 23 ×− 4 × 0 1 × 12 × 23 × 4 × 0 96 4

E 2u 1 × 36 × 23 ×− 4 × 0 1 × 12 ×− 23 × 4 × 0 48 2

To find the irreducible representations present requires the use of the reduction formula.
It is worth noting that several of the classes have given a 0 in the character set for , and

so will not contribute to the sum in the reduction formula (Equation (6.17)). In this case,
only the E ,3 C 2 ,

σ v classes need be considered. The application of the reduction
formula is summarized in Table 6.16; even though only a few classes contribute, the order
h

σ h and 3

24 is still the total number of operations in the full D 6h point group. From the final
column in the table we find that

=

=

2 A 1g + 2 A 2g + 2 B 2g + 2 E 1g + 4 E 2g + 2 A 2u + 2 B 1u + 2 B 2u + 4 E 1u + 2 E 2u (6.59)

This means we have 12 singly degenerate vibrations (labelled A or B ) and 12 doubly degen-
erate vibrations (labelled E ), giving a total of 36, as required by the use of 36 basis vectors
to describe the atomic motions. Of these, six will be simple movements or rotations of
the molecule as a whole. We can find out which these are from the character table: the
symbols x and y occur with the irreducible representation E 1u , z with A 2u and R x , R y and R z
with E 1g and A 2g , so the corresponding number of each irreducible representation needs to
be removed from , remembering that E states are doubly degenerate. This leaves the 30

modes

( ν ib) = 2 A 1g + A 2g + 2 B 2g + E 1g + 4 E 2g + A 2u + 2 B 1u + 2 B 2u + 3 E 1u + 2 E 2u (6.60)

The character table indicates that IR-active modes will have symmetry A 2u and E 1u . These
are present in 3 E 1u , i.e. there are seven vibrational modes that are IR active,

but we only expect four bands in the IR spectrum, since six form degenerate pairs. Way
back in Figure 1.5 we saw that the experimentally observed IR spectrum of benzene does
indeed have only four bands. The products of axes in the character table follow A 1g , E 1g

( ν ib) as A 2u +