{Sn₁₀[Si(SiMe₃)₃]₅}^– : An Anionic Metalloid Tin Cluster from an Isolable Sn^I Halide Solution

Abstract

Abstract The disproportionation reaction of the subvalent metastable halide SnBr proved to be a powerful synthetic method for the synthesis of metalloid cluster compounds of tin. Hence the neutral metalloid cluster compound Sn 10 [Si(SiMe 3 ) 3 ] 6 ( 1 ) was synthesized from a reaction of a SnBr emulsion with LiSi(SiMe 3 ) 3 . Using the phosphane P n Bu 3 as a donor component during the synthesis of the monohalide solution, an isolable Sn I halide solution is obtained. The reaction of such an isolable SnCl solution with LiSi(SiMe 3 ) 3 gives the anionic metalloid cluster compound {Sn 10 [Si(SiMe 3 ) 3 ] 5 } – ( 5 ) in 63 % yield, where only five tin atoms are bound to one ligand and where a centaur polyhedral arrangement of the ten tin atoms is present. Due to the open structure 5 might be used for further applications in the field of subsequent build‐up reactions.