The crystal structure of potassium trifluorocuprate (II) KCuF
3 has been determined by an X-ray analysis. The structure was refined by Fourier method. The crystals are tetragonal,
a=\sqrt2
a0=5.855 and
c=2
c0=7.852Å; space group
D4h18–
I4⁄
mcm, with four formula units (KCuF
3) 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 Cu
2+ 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 Cu
2+ ion is surrounded by a distorted octahedron of F
− ions with copper-fluorine distances of 2.25, 1.96 and 1.89Å.
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