Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3BC2_oP24_62_3c_c_2c-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.

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https://aflow.org/p/WK7H
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K$_{2}$CuCl$_{3}$ Structure: A3BC2_oP24_62_3c_c_2c-001

Picture of Structure; Click for Big Picture
Prototype Cl$_{3}$CuK$_{2}$
AFLOW prototype label A3BC2_oP24_62_3c_c_2c-001
ICSD 30656
Pearson symbol oP24
Space group number 62
Space group symbol $Pnma$
AFLOW prototype command aflow --proto=A3BC2_oP24_62_3c_c_2c-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak z_{6}$

Other compounds with this structure

Ba$_{2}$ZnS$_{3}$,  Cs$_{2}$AgCl$_{3}$,  Cs$_{2}$AgI$_{3}$


  • (Brink, 1949) present the structure in the $Pnam$ setting of space group #62. We used FINDSYM to convert this to the standard $Pnma$ setting.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $a x_{1} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{2}}$ = $- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{3}}$ = $- x_{1} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{1} \, \mathbf{a}_{3}$ = $- a x_{1} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{4}}$ = $\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{5}}$ = $x_{2} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $a x_{2} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{6}}$ = $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{7}}$ = $- x_{2} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $- a x_{2} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{8}}$ = $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{10}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{12}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4c) Cu I
$\mathbf{B_{14}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cu I
$\mathbf{B_{15}}$ = $- x_{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (4c) Cu I
$\mathbf{B_{16}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cu I
$\mathbf{B_{17}}$ = $x_{5} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4c) K I
$\mathbf{B_{18}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) K I
$\mathbf{B_{19}}$ = $- x_{5} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (4c) K I
$\mathbf{B_{20}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) K I
$\mathbf{B_{21}}$ = $x_{6} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (4c) K II
$\mathbf{B_{22}}$ = $- \left(x_{6} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{6} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) K II
$\mathbf{B_{23}}$ = $- x_{6} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (4c) K II
$\mathbf{B_{24}}$ = $\left(x_{6} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{6} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) K II

References

  • C. Brink and C. H. MacGillavry, The crystal structure of K$_{2}$CuCl$_{3}$ and isomorphous substances, Acta Cryst. 2, 158–163 (1949), doi:10.1107/S0365110X49000436.

Prototype Generator

aflow --proto=A3BC2_oP24_62_3c_c_2c --params=$a,b/a,c/a,x_{1},z_{1},x_{2},z_{2},x_{3},z_{3},x_{4},z_{4},x_{5},z_{5},x_{6},z_{6}$

Species:

Running:

Output: