For a displacement group representation, an atom contributes only if the operation leaves its site fixed. Its three Cartesian displacements then contribute the trace of the polar-vector transformation.

Take

z

perpendicular to the ring and an in-plane

x

axis through opposite carbon atoms. To avoid confusing the principal twofold rotation with the in-plane axes, write

C_{2} (z)

for the former,

C_{2 a}

for an in-plane axis through atoms and

C_{2 b}

for one through opposite bond midpoints. The carbon displacement characters in the D6h molecular point group are

C χ_{C} i C χ_{C} (i C) E 18 i 0 2 C_{6} 0 2 S_{3} 0 2 C_{3} 0 2 S_{6} 0 C_{2} (z) 0 σ_{h} 6 3 C_{2 a} - 2 3 σ_{d} 0 3 C_{2 b} 0 3 σ_{v} 2

(1)

A through-atom half-turn fixes two carbon sites and has Cartesian trace

- 1

, giving

- 2

. The molecular-plane reflection fixes all six carbons with trace

+ 1

, giving

6

. A through-atom vertical mirror fixes two sites with trace

+ 1

, giving

2

Using molecular character reduction, the 18-dimensional carbon-only result is

Γ_{C} = A_{1 g} + A_{2 g} + B_{2 g} + E_{1 g} + 2 E_{2 g} + A_{2 u} + B_{1 u} + B_{2 u} + 2 E_{1 u} + E_{2 u} .

(2)

The six one-dimensional terms contribute six dimensions and the six copies of two-dimensional representations contribute twelve, totaling eighteen. The distinction between carbon-only and whole-molecule displacements is needed in part (ii).

ii

 0  0

Solution

 0  0

Each hydrogen lies on the radial line through its attached carbon. Consequently every molecular symmetry operation fixes or permutes the hydrogen sites exactly as it does the carbon sites, and the Cartesian trace is the same. Thus

Γ_{H} = Γ_{C}

and the complete benzene displacement group representation is $Γ_{3 N} = 2 Γ_{C}$ , of dimension

36

Subtract translations

A_{2 u} + E_{1 u}

and rotations

A_{2 g} + E_{1 g}

. The full set of normal modes is

Γ_{vib} = 2 A_{1 g} + A_{2 g} + 2 B_{2 g} + E_{1 g} + 4 E_{2 g} + A_{2 u} + 2 B_{1 u} + 2 B_{2 u} + 3 E_{1 u} + 2 E_{2 u} .

(1)

The one-dimensional components contribute

2 + 1 + 2 + 1 + 2 + 2 = 10

coordinates. The two-dimensional components contribute

2 (1 + 4 + 3 + 2) = 20

, giving $30 = 3 (12) - 6$ vibrational degrees of freedom. There are twenty distinct symmetry-labelled fundamentals before accidental coincidences, because each doubly degenerate pair has one frequency.

Subtracting translations and rotations directly from the carbon-only result would instead give twelve carbon-skeleton coordinates. That is a six-site mechanical model, not all the normal modes of