The Crystal Structure of KCuF₃

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The crystal structure of potassium trifluorocuprate (II) KCuF₃ has been determined by an X-ray analysis. The structure was refined by Fourier method. The crystals are tetragonal, a=\sqrt2a₀=5.855 and c=2c₀=7.852Å; space group D_4h¹⁸–I4⁄mcm, with four formula units (KCuF₃) in the unit cell, where a₀ and c₀ designate the lattice constants of the fundamental pseudo-perovskite structure. This superstructure is due to a displacement of fluorine ions along the copper-fluorine-copper bonds only in the c plane. The atoms are in the following positions: 4 K⁺ in (a): (0, 0, 0; \frac12, \frac12, \frac12)+(0, 0, \frac14; 0, 0, \frac34): 4 Cu²⁺ in (d): (0, 0, 0; \frac12, \frac12, \frac12)+(0, \frac12, 0; \frac12, 0, 0): 4 F⁻ in (b): (0, 0, 0; \frac12, \frac12, \frac12)+(0, \frac12, \frac14; \frac12, 0, \frac14): 8 F⁻ in (h): (0, 0, 0; \frac12, \frac12, \frac12)±(x, \frac12+x, 0; \frac12−x, x, 0) with x=0.228. In this structure a Cu²⁺ ion is surrounded by a distorted octahedron of F⁻ ions with copper-fluorine distances of 2.25, 1.96 and 1.89Å.