Almost all the trihalides of titanium, zirconium and hafnium can be prepared by the high-temperature reduction of the corresponding tetrahalide with the metal. Incomplete reaction and contamination of the product with excess metal often occurs.[1]
Alternatively, zirconium(III) bromide crystallises from a solution of zirconium(III) in aluminium tribromide. The solution is prepared by reducing a eutectic solution of ZrBr4 in liquid AlBr3 at a temperature of 230–300 °C with metallic zirconium or aluminium.[2][3]
The crystal structure of zirconium(III) bromide is based on hexagonal close packing of bromide ions with one third of the octahedralinterstices occupied by Zr3+ ions.[1] The structure consists of parallel chains of face-sharing {ZrBr6} octahedra with equally spaced metal atoms. There is some elongation of the octahedra along the metal-metal axis, partly due to metal-metal repulsion.[3] ZrCl3, ZrBr3 and ZrI3 all adopt the β-TiCl3 structure,[1] but the elongation of octahedra is most pronounced in the chloride, moderate in the bromide and negligible in the iodide.[3]
^Larsen, E. M.; Moyer, James W.; Gil-Arnao, Francisco; Camp, Michael J. (1974). "Synthesis of crystalline zirconium trihalides by reduction of tetrahalides in molten aluminum halides. Nonreduction of hafnium". Inorg. Chem. 13 (3): 574–581. doi:10.1021/ic50133a015.
^ abcLarsen, Edwin M.; Wrazel, Julie S.; Hoard, Laurence G. (1982). "Single-crystal structures of ZrX3 (X = Cl−, Br−, I−) and ZrI3.40 synthesized in low-temperature aluminum halide melts". Inorg. Chem.21 (7): 2619–2624. doi:10.1021/ic00137a018.