The stability of biradicaloid versus closed-shell [E(μ-XR)]2 (E = P, As; X = N, P, As) rings. Does aromaticity play a role?†
Abstract
High-level multiconfigurational self-consistent field calculations, supplemented with multiconfigurational quasi-degenerate perturbation theory ab initio calculations with the aug-cc-pVTZ basis set, demonstrate that the [E(μ-XH)]2 (E = P, As; X = N, P, As) compounds possess one planar and one butterfly-like isomer. The calculations predict that for X = N, planar isomers, which bear substantial biradicaloid character, are more stable than their butterfly-like counterpart isomers, which feature closed-shell electronic structures. This has been ascribed to the fact that the increased bond angle strain at E–N–E is not compensated by the E–E σ (deformed) bond formation in the butterfly-like isomers, yielding the planar structures, which hold wider E–N–E bond angles, as the most stable isomers. As N is substituted by heavier atoms, either P or As, the E–P(As)–E bond angle strain is released and, additionally, as the formed E–E σ-bond is less deformed, the butterfly isomer becomes the most stable isomer. Subsequent evaluation of the normalized Giambiagi multicenter electron delocalization indices revealed no sign of electron delocalization in the four-membered rings and consequently, it is concluded that aromaticity does not play any role in the stabilization of the planar isomers.
- This article is part of the themed collection: Electron delocalization and aromaticity: 150 years of the Kekulé benzene structure