Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A5 / c / iii / Solution

One quantum in an $E_{2g}$ normal mode gives an $E_{2g}$ vibrational pair. The electronic factor is $B_{2u}$, henceThis follows by multiplying the characters of the two group representations. In particular, multiplication of $E_2$ by $B_2$ produces $E_1$ in the hexagonal group, and the inversion parities multiply as $u\times g=u$. The resulting vibronic state is twofold degenerate.