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

The electronic transition starts from $A_{1g}$, so the electric-dipole test reduces to whether the final electronic irreducible representation occurs among $A_{2u}+E_{1u}$. The excited $E_{1u}$ state matches the $(x,y)$ dipole components and is therefore electric-dipole allowed.

The $B_{2u}$ state matches neither $z$ nor $(x,y)$ and is electric-dipole forbidden at the equilibrium geometry. Both excited states are singlets, so the difference is spatial symmetry rather than the spin selection rule. Inversion parity alone does not decide the result: both final states are ungerade, but only the matching vector representation is allowed.

The overall vibrational ground state is invariant under every molecular symmetry operation, so it transforms as $A_{1g}$. Combining it with the electronic ground state givesFor a degenerate harmonic normal mode, the zero-quantum wavefunction depends on the invariant sum $Q_x^2+Q_y^2$, so the ground vibrational symmetry remains totally symmetric even when excited vibrational states form multiplets.

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.

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.

A coordinate $Q_s$ of an $A_{1g}$ normal mode is unchanged by every molecular symmetry operation. Every harmonic-oscillator excitation $H_n(Q_s)e^{-Q_s^2/2}$ therefore also has symmetry $A_{1g}$, including odd values of $n$. Consequentlyfor any number of quanta in that totally symmetric coordinate. The same electric-dipole permission survives every $A_{1g}$ excitation.

This statement concerns symmetry. Actual progression intensities depend on vibrational inner products, equilibrium displacements and the coordinate dependence of the electric dipole transition moment; a permitted line need not have measurable intensity for every vibrational quantum number.