AFLOW Prototype: A2B3_hR10_148_abc_f-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/CVKE
or
https://aflow.org/p/A2B3_hR10_148_abc_f-001
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PDF Version
Prototype | Cr$_{2}$S$_{3}$ |
AFLOW prototype label | A2B3_hR10_148_abc_f-001 |
ICSD | 16721 |
Pearson symbol | hR10 |
Space group number | 148 |
Space group symbol | $R\overline{3}$ |
AFLOW prototype command |
aflow --proto=A2B3_hR10_148_abc_f-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$ |
--hex
. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Cr I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (1b) | Cr II |
$\mathbf{B_{3}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ | = | $c x_{3} \,\mathbf{\hat{z}}$ | (2c) | Cr III |
$\mathbf{B_{4}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ | = | $- c x_{3} \,\mathbf{\hat{z}}$ | (2c) | Cr III |
$\mathbf{B_{5}}$ | = | $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} - 2 y_{4} + z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S I |
$\mathbf{B_{6}}$ | = | $z_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+y_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(y_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(2 x_{4} - y_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S I |
$\mathbf{B_{7}}$ | = | $y_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} + y_{4} - 2 z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S I |
$\mathbf{B_{8}}$ | = | $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - 2 y_{4} + z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S I |
$\mathbf{B_{9}}$ | = | $- z_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- y_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(y_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(2 x_{4} - y_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S I |
$\mathbf{B_{10}}$ | = | $- y_{4} \, \mathbf{a}_{1}- z_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} + y_{4} - 2 z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S I |