Issue 17, 2016

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.

Graphical abstract: The stability of biradicaloid versus closed-shell [E(μ-XR)]2 (E = P, As; X = N, P, As) rings. Does aromaticity play a role?

Associated articles

Supplementary files

Article information

Article type
Paper
Submitted
25 Nov 2015
Accepted
05 Feb 2016
First published
25 Feb 2016

Phys. Chem. Chem. Phys., 2016,18, 11879-11884

The stability of biradicaloid versus closed-shell [E(μ-XR)]2 (E = P, As; X = N, P, As) rings. Does aromaticity play a role?

R. Grande-Aztatzi, J. M. Mercero and J. M. Ugalde, Phys. Chem. Chem. Phys., 2016, 18, 11879 DOI: 10.1039/C5CP07263H

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