Answers are grounded in this paper and cite their sources.
The title compound, (Bn 2 SnTe) 3, Bn = CH 2 C 6 H 5, was prepared in high yield by treating Bn 2 SnCl 2 with (NH 4 ) 2 Te. Details of the synthesis and characterization ( 1 H, 13 C, 119 Sn, and 125 Te NMR; IR, single-crystal XRD, and elemental analyses) are presented. (Bn 2 SnTe) 3 serves as an efficient single-source precursor to phase-pure cubic SnTe under mild conditions
No takes yet. Share an insight, caveat, or question.
Boudjouk et al. (1999) studied this question.
Synapse has enriched 10 related papers. Consider them for comparative context: