Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A5 / a / ii

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 ${\Gamma }_H={\Gamma }_C$ and the complete benzene displacement group representation is ${\Gamma }_{3N}=2{\Gamma }_C$, of dimension $36$.

Subtract translations $A_{2u}+E_{1u}$ and rotations $A_{2g}+E_{1g}$. The full set of normal modes isThe 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 ${\mathrm{C}}_6{\mathrm{H}}_6$. The hydrogen contribution is required for the actual molecule.