Abstract

Black phosphorus can be prepared under low-pressure conditions at 873 K from red phosphorus via the addition of small quantities of gold, tin, and tin(IV) iodide. Au3SnP7, AuSn, and Sn4P3 were observed as additional phases. Tin(IV) iodide remains unreacted during the preparation process. The crystal structure of black phosphorus was redetermined from single crystals. P (295 K): a = 3.316(1) A, b = 10.484(2) A, c = 4.379(1) A, V = 152.24(6) A3, space group Cmce (No. 64). Solid-state 31P MAS NMR spectroscopy and X-ray powder diffraction were performed to substantiate the high crystal quality of black phosphorus. A possible mechanism for the formation is discussed in terms of the comparable structural features of black phosphorus and Au3SnP7. Thermodynamic calculations showed that the only relevant gas-phase species, P4, and the transport reactions are not suitable for the preparation of orthorhombic black phosphorus at temperatures above 773 K. A kinetically controlled mechanism must be favored instead of a thermodynamically controlled formation. The new preparation method of black phosphorus represents an easy and effective way to avoid complicated preparative setups, toxic catalysts, or "dirty" flux methods and is of general interest in elemental chemistry.

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