Abstract
The reduction of Mo2[O2CMe]4 by sodium amalgam in tetrahydrofuran (thf) in the presence of excess of PMe3 under hydrogen (3 atm) yields the dimeric complex Mo2H2(µ-H)2(PMe3)6(1), whose structure has been determined by X-ray crystallography. The compound is orthorhombic, space group Pbca, with a= 12.133(6), b= 14.892(4), c= 17.692(5)A, and Z= 4. The structure has been refined to an R of 0.0362 for 2 388 observed [l > 1.5σ(l)] diffractometer data. The centrosymmetric molecule has an edge-shared (H, H) bioctahedral structure with a Mo–Mo distance of 2.194(3)A. Of the three phosphine ligands on each metal, one is approximately perpendicular to the Mo2H2 plane and trans to the terminal hydride with Mo–P 2.430(3)A, and two lie in the Mo2H2 plane; differences in the Mo–P distances to these phosphines [2.430, 2.417(3)A] suggest possible asymmetry in the Mo–H–Mo bridges, although the low precision with which the H-atom positions are located does not allow an unequivocal decision. Reduction of the complex W2Cl4(PMe3)4 in thf by sodium amalgam under hydrogen leads to a species best formulated as W2H4(µ-H)(µ-PMe2)(PMe3)5(2). This complex is monoclinic, space group P21/c, with a= 9.931 (5), b= 12.023(3), c= 26.605(4)A, β= 91.66(3)°, and Z= 4. The structure was refined to an R value of 0.0525 for 4 218 observed reflections. The molecule appears to be a pentahydride of the binuclear unit (Me3P)2[graphic omitted](PMe3)3 although it has not been possible to locate the metal-bonded hydrogen atoms. The two metal atoms give two sets of W–P bond lengths, with those for the metal carrying three terminal phosphines slightly shorter [2.355–2.464(5)A] than those to the metal carrying two phosphines [2.433–2.502(6)A]. Interaction of MMe5(M = Nb or Ta) and PMe3 gives MMe5(PMe3)2; the niobium compound reacts with hydrogen to give NbH5(PMe3)4.
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