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

The initial vibronic state is $A_{1g}$ and the final one is $E_{1u}$, so an in-plane dipole component satisfiesThe product contains the totally symmetric representation. The combined electronic–vibrational transition is therefore symmetry-allowed, with an in-plane electric dipole transition moment.

A nonzero intensity still needs vibronic coupling. In a coordinate-independent separation, the forbidden electronic matrix element would remain zero. Its first derivative with respect to the $E_{2g}$ coordinate can instead be nonzero, and the vibrational factor $⟨1|Q|0⟩$ supplies the required one-quantum transition. This is vibrational activation of an electronic transition and explains why the band can be weak.