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{Sn 10 [Si(SiMe 3 ) 3 ] 5 } – : An Anionic Metalloid Tin Cluster from an Isolable Sn I Halide Solution
Author(s) -
Schrenk Claudio,
Helmlinger Jens,
Schnepf Andreas
Publication year - 2012
Publication title -
zeitschrift für anorganische und allgemeine chemie
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.354
H-Index - 66
eISSN - 1521-3749
pISSN - 0044-2313
DOI - 10.1002/zaac.201100364
Subject(s) - tin , metalloid , halide , cluster (spacecraft) , disproportionation , chemistry , ligand (biochemistry) , yield (engineering) , group 2 organometallic chemistry , crystallography , antimony , inorganic chemistry , molecule , materials science , metal , organic chemistry , catalysis , biochemistry , receptor , computer science , programming language , metallurgy
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.