Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: ABC4_hP18_189_f_g_fgj-001

If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.

Links to this page

https://aflow.org/p/6BNK
or https://aflow.org/p/ABC4_hP18_189_f_g_fgj-001
or PDF Version

CsCrF$_{4}$ Structure: ABC4_hP18_189_f_g_fgj-001

Picture of Structure; Click for Big Picture
Prototype CrCsF$_{4}$
AFLOW prototype label ABC4_hP18_189_f_g_fgj-001
ICSD 2278
Pearson symbol hP18
Space group number 189
Space group symbol $P\overline{6}2m$
AFLOW prototype command aflow --proto=ABC4_hP18_189_f_g_fgj-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak x_{5}, \allowbreak y_{5}$

\[ \begin{array}{ccc} \mathbf{a_{1}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{2}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{3}}&=&c \,\mathbf{\hat{z}} \end{array}\]

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}$ = $\frac{1}{2}a x_{1} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}$ (3f) Cr I
$\mathbf{B_{2}}$ = $x_{1} \, \mathbf{a}_{2}$ = $\frac{1}{2}a x_{1} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}$ (3f) Cr I
$\mathbf{B_{3}}$ = $- x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}$ = $- a x_{1} \,\mathbf{\hat{x}}$ (3f) Cr I
$\mathbf{B_{4}}$ = $x_{2} \, \mathbf{a}_{1}$ = $\frac{1}{2}a x_{2} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}$ (3f) F I
$\mathbf{B_{5}}$ = $x_{2} \, \mathbf{a}_{2}$ = $\frac{1}{2}a x_{2} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}$ (3f) F I
$\mathbf{B_{6}}$ = $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}$ = $- a x_{2} \,\mathbf{\hat{x}}$ (3f) F I
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3g) Cs I
$\mathbf{B_{8}}$ = $x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3g) Cs I
$\mathbf{B_{9}}$ = $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3g) Cs I
$\mathbf{B_{10}}$ = $x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3g) F II
$\mathbf{B_{11}}$ = $x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3g) F II
$\mathbf{B_{12}}$ = $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3g) F II
$\mathbf{B_{13}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}$ (6j) F III
$\mathbf{B_{14}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}$ (6j) F III
$\mathbf{B_{15}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}$ (6j) F III
$\mathbf{B_{16}}$ = $y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}$ (6j) F III
$\mathbf{B_{17}}$ = $\left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}$ (6j) F III
$\mathbf{B_{18}}$ = $- x_{5} \, \mathbf{a}_{1}- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}$ (6j) F III

References

  • D. Babel and G. Knoke, Strukturen caesiumhaltiger Fluoride. IV. Die KristallStruktur von CsCrF$_{4}$ - ein neuer Tetrafluorometallat‐Typ mit Kettenstruktur, Z. Anorganische und Allgemeine Chemie 442, 151–162 (1978), doi:10.1002/zaac.19784420119.

Found in

  • H. Manaka, Y. Hirai, Y. Hachigo, M. Mitsunaga, M. Ito, and N. Terada, Spin-Liquid State Study of Equilateral Triangle S=3/2 Spin Tubes Formed in CsCrF$_{4}$, J. Phys. Soc. Jpn. 78, 093701 (2009), doi:10.1143/JPSJ.78.093701.

Prototype Generator

aflow --proto=ABC4_hP18_189_f_g_fgj --params=$a,c/a,x_{1},x_{2},x_{3},x_{4},x_{5},y_{5}$

Species:

Running:

Output: