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Abstract:
The reaction of iridium powder with an excess of selenium and SeBr4 yielded lustrous, vermillion crystals of the mononuclear iridium complex [IrBr3(SeBr2)(3)]. The transition metal is coordinated octahedrally by three SeBr2 and three bromide ligands with facial or meridional configuration. Three different modifications were obtained under similar conditions: a-fac-IrBr3(SeBr2)(3), space group P (1) over bar, with a = 789.4(1) pm, b = 830.4(1) pm, c = 1334.4(1) pm, alpha = 81.634(5)degrees, beta = 84.948(5)degrees, gamma = 67.616(4)degrees; m-fac-IrBr3(SeBr2)(3), space group P2(1)/n, with a = 1205.3(1) pm, b = 962.4(1) pm, c = 1383.9(1) pm, beta = 91.114(3)degrees; mer-IrBr3(SeBr2)(3), space group P2(1)/n with a = 859.7(1) pm, b = 1284.3(1) pm, c = 1437.5(1) pm, beta = 94.427(3)degrees. A lower bromine content in the starting composition resulted in shiny, deep-red crystals of [Se-9(IrBr3)(2)]. X-ray diffraction on a single-crystal revealed a tetragonal lattice (space group I4(1)/a) with a = 1245.4(1) pm and c = 2486.8(1) pm at 296(1) K. In the [Se-9(IrBr3)(2)] complex, a crown-shaped uncharged Se-9 ring coordinates two iridium(III) cations as a bridging bis-tridentate ligand. Three terminal bromide ions complete the distorted octahedral coordination of each transition metal atom.
