Formation of M4Se4 cuboids (M = As, Sb, Bi) via secondary pnictogen–chalcogen interactions in the co-crystals MX3·SeP(p-FC6H4)3 (M = As, X = Br; M = Sb, X = Cl; M = Bi, X = Cl, Br)
作者:Fatma B. Alhanash、Nicholas A. Barnes、Alan K. Brisdon、Stephen M. Godfrey、Robin G. Pritchard
DOI:10.1039/c2dt31010d
日期:——
The reactions of the group 15 trihalides, MX3 (M = As, Sb, Bi; X = Cl, Br), with the phosphine selenide SeP(p-FC6H4)3 result in the formation of co-crystals of formula MX3·SeP(p-FC6H4)3. No reaction was observed with MI3 (M = As, Sb, Bi). The structures of MX3·SeP(p-FC6H4)3 (M = As, X = Br 2; M = Sb, X = Cl 3; M = Bi, X = Cl 5; M = Bi, X = Br 6) have been established, and are isomorphous, crystallising in the cubic I23 space group. All the structures feature a primary MX3 unit, which has three weak secondary Mâ¯Se interactions to SeP(p-FC6H4)3 molecules. However, each of these SeP(p-FC6H4)3 molecules bridges three MX3 molecules, resulting in the generation of an M4Se4 (M = As, Sb, Bi) distorted cuboid linked by the pnictogenâchalcogen interactions. Four opposing corners of the cuboid are occupied by the M atom (M = As, Sb, Bi) of an MX3 pyramid, and the other four by the selenium atom of the phosphine selenide.
第15族三卤化物MX3(M = As, Sb, Bi;X = Cl, Br)与膦硒化物SeP(p-FC6H4)3的反应导致了化学式为MX3·SeP(p-FC6H4)3的共晶体的形成。没有观察到与MI3(M = As, Sb, Bi)的反应。已确定了MX3·SeP(p-FC6H4)3(M = As, X = Br 2;M = Sb, X = Cl 3;M = Bi, X = Cl 5;M = Bi, X = Br 6)的结构,并且它们是同构的,结晶在立方I23空间群中。所有结构都具有一个主要MX3单元,该单元通过三个次级M…Se相互作用与SeP(p-FC6H4)3分子相连。然而,每个SeP(p-FC6H4)3分子桥接三个MX3分子,从而产生了一个由氮族-硫族相互作用连接的M4Se4(M = As, Sb, Bi)扭曲立方体。立方体的四个相对角被MX3金字塔的M原子(M = As, Sb, Bi)占据,另外四个角则被膦硒化物的硒原子占据。