The valence problem of Pd4Br4Te3

Chemistry. 2007;13(35):9882-91. doi: 10.1002/chem.200700658.

Abstract

Pd(4)Br(4)Te(3) was prepared from Pd, Te, and PdBr(2) at 700 K. Its structure was determined by single-crystal X-ray diffraction to be triclinic, P$\bar 1$, Pearson symbol aP22; a=842.5(2), b=845.0(3), c=864.8(3) pm; alpha=82.55(3), beta=73.36(2), gamma=88.80(2) degrees ; Z=2. The Br and Te atoms are arranged according to the motif of cubic closest-packed spheres in which every 15th position is vacant; the Pd atoms occupy 8/15 of the octahedral voids. The symmetry relations with the packing of spheres are derived. Prominent structural units are hollow cuboctahedral [(PdBrTe)(6)] units, the Pd atoms are positioned near the centers of the square faces of the Br(6)Te(6) cuboctahedra; the cuboctahedra and double-octahedral Pd(2)Br(4)Te(6) units are connected to strands by sharing triangular Te(3) faces. The strands are condensed by common Br atoms into layered assemblies. Conspicuously close Te--Te contacts in the Te(3) triangles indicate attractive Te--Te interactions. The valence puzzle is resolved by the formula Pd(+II)(4)Br(-I)(4)Te(-4/3)(3). Positive Te--Te Mulliken orbital populations and the Pd--K, Br--K, and Te--L(III) XANES spectra of Pd(4)Br(4)Te(3) referenced to the spectra of PdBr(2), K(2)PdBr(6), PdTe, and PdTe(2) are in accord with attractive Te--Te interactions. The measured semiconducting and diamagnetic properties are compatible with the derived picture of chemical bonding in Pd(4)Br(4)Te(3).