Abstract
By virtue of their large frequency change and equilibrium displacement, the vibronically active out‐of‐plane bending modes are argued to be important accepting modes for radiationless transitions of aromatic molecules with nonbonding electrons. The conclusion is shown to be consistent with the position dependence of the deuterium isotope effect in T1(ππ*) → S0 radiationless transitions of nitrogen‐heterocyclic and aromatic carbonyl compounds.