= Solution
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 \b[electric-dipole allowed].
The $B_{2u}$ state matches neither $z$ nor $(x,y)$ and is \b[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.
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