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

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 Sn10[Si(SiMe3)3]6(1) was synthesized from a reaction of a SnBr emulsion with LiSi(SiMe3)3. Using the phosphane PnBu3as a donor component during the synthesis of the monohalide solution, an isolable SnIhalide solution is obtained. The reaction of such an isolable SnCl solution with LiSi(SiMe3)3gives the anionic metalloid cluster compound {Sn10[Si(SiMe3)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 5might be used for further applications in the field of subsequent build‐up reactions.