Why does the reaction of the dihydrogen molecule with [p2n2]Zr(μ- η2-N2)Zr[p2n2] produce [p2n2]Zr(μ-η2-N2H)Zr[p2n2](μ-H) but not the thermodynamically more favorable [p2n2]Zr(μ-NH)2Zr[p2n2]

Harold Basch, Djamaladdin G. Musaev, Keiji Morokuma

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Abstract

The mechanism of reaction of binuclear zirconium dinitrogen complex [p2n2]Zr(μ-η2-N2)Zr[p2n2] (1, where p2n2 = (PH3)2(NH2)2 as a model of the experimentally used P2N2 = PhP(CH2SiMe2NSiMe2CH2)2-PPh ligand), with a hydrogen molecule has been studied by using the density functional method. It is shown that this reaction proceeds via (i) activation of the H-H σ-bond via a 'metathesis-like' transition state where simultaneously Zr-H and N-H bonds are formed and the H-H and one of the N-N π-bonds are broken, to produce the diazenidohydride complex 3, [p2n2]Zr(μ-η2-NNH)Zr(H)[p2n2], and (ii) migration of the Zr-bonded hydride ligand to a position bridging the two Zr atoms to form the diazenido- μ-hydride complex 7, [p2n2]Zr(μ-η2-NNH)Zr[p2n2](μ-H). The entire reaction is calculated to be exothermic by 13-15 kcal/mol. The rate- determining step of this reaction is found to be the activation of the H-H bond, which occurs with a 21-kcal/mol barrier. The experimentally observed diazenido-μ-hydride complex 7, [p2n2]Zr(μ-η2-NNH)Zr[p2n2](μ-H), is not the lowest energy structure in the potential energy surface. The hydrazono complex [p2n2]Zr(μ-NNH2)Zr[p2n2] (with a bridging NH2) and the hydrado complex [p2n2]Zr(μ-NHNH)Zr[p2n2] (with two bridging NH units) are calculated to be more stable than the diazenido-μ-hydride complex 7 by about 50 kcal/mol. However, these complexes cannot be generated by the reaction of 1 + H2 at ambient laboratory conditions because of very high (nearly 60 kcal/mol) barriers separating them from 7.

Original languageEnglish
Pages (from-to)5754-5761
Number of pages8
JournalJournal of the American Chemical Society
Volume121
Issue number24
DOIs
StatePublished - 23 Jun 1999

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