 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2022/paper-1/a5/c/iii/solution

One quantum in an

E_{2 g}

E_{2 g}

vibrational pair. The electronic factor is

B_{2 u}

, hence

Γ_{final} = B_{2 u} \otimes E_{2 g} = E_{1 u} .

This follows by multiplying the characters of the two group representations. In particular, multiplication of

E_{2}

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.

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