Answers are grounded in this paper and cite their sources.
BaAg 2 SnSe 4, BaCu 2 SnSe 4, and Ba 3 Cu 2 Sn 3 Se 10 were prepared by directly reacting the elements in stoichiometric ratios at 800 °C, followed by long heating between 600 and 650 °C. BaAg 2 SnSe 4 crystallizes in the BaAg 2 SnS 4 type, space group I 222, BaCu 2 SnSe 4 in the SrCu 2 GeSe 4 type, space group Ama 2, and Ba 3 Cu 2 Sn 3 Se 10 in a new structure type, space group P 2 1 / n . All three structures comprise almost undistorted SnSe 4 tetrahedra and pairs of the group 11 elements (M = Cu and Ag). While both the M atoms are bonded to four Se atoms in each case, the deviations from an MSe 4 tetrahedron are severe, and distinctly different within this series. BaAg 2 SnSe 4, BaCu 2 SnSe 4, and Ba 3 Cu 2 Sn 3 Se 10 are black semiconductors, with computed band gaps of 0.2, 0.7, and 1.2 eV, respectively. The experimental gaps for the former two are 0.24 and 0.48 eV, respectively. The electrical conductivities consistently decrease with increasing band gap.
No takes yet. Share an insight, caveat, or question.
Assoud et al. (2005) studied this question.
Synapse has enriched 10 related papers. Consider them for comparative context: