Synthesis and Properties of Ph 3 Sn substituted Tellurates – The Crystal Structures of trans ‐[(Ph 3 SnO) 4 Te(OH) 2 ] and trans ‐[(Ph 3 SnO) 2 Te(OMe) 4 ]

The reaction of Te(OH) 6 with Ph 3 SnOH in ethanol leads to the formation of trans ‐[(Ph 3 SnO) 4 Te(OH) 2 ] ( 1 ). Compound 1 crystallizes triclinic in the space group P \bar{1} with a = 996.6(2) pm, b = 1365.4(3) pm, c = 1368.2(3) pm and α = 71.15(2)°, β = 71.48(2)°, γ = 74.81(3)° (at 220 K). The molecular structure of 1 consists of a tellurium atom, which is coordinated nearly octahedrally by four Ph 3 SnO units and two hydroxyl groups that are trans to each other. The Te–O bond lengths are in the range of 190.5(2) and 193.7(2) pm. Treatment of 1 with methanol under reflux yields trans ‐[(Ph 3 SnO) 2 Te(OMe) 4 ] ( 2 ). Compound 2 crystallizes triclinic in the space group P \bar{1} with a = 1012.8(1) pm, b = 1422.4(2) pm, c = 1618.1(2) pm, and α = 100.44(1)°, β = 107.92(1)°, γ = 110.66(1)° (at 220 K). 2 forms centrosymmetric molecules in which the tellurium atom is surrounded nearly octahedrally by four methoxy groups and two trans arranged Ph 3 SnO units. The Te–O bond lengths of 187.9(3)–194.5(3) pm are similar to those observed in 1 .