AFLOW Prototype: AB4C_tP12_127_a_eg_c-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/QJN8
or
https://aflow.org/p/AB4C_tP12_127_a_eg_c-001
or
PDF Version
Prototype | AlF$_{4}$Rb |
AFLOW prototype label | AB4C_tP12_127_a_eg_c-001 |
ICSD | 54121 |
Pearson symbol | tP12 |
Space group number | 127 |
Space group symbol | $P4/mbm$ |
AFLOW prototype command |
aflow --proto=AB4C_tP12_127_a_eg_c-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak x_{4}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | Al I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}$ | (2a) | Al I |
$\mathbf{B_{3}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2c) | Rb I |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2c) | Rb I |
$\mathbf{B_{5}}$ | = | $z_{3} \, \mathbf{a}_{3}$ | = | $c z_{3} \,\mathbf{\hat{z}}$ | (4e) | F I |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (4e) | F I |
$\mathbf{B_{7}}$ | = | $- z_{3} \, \mathbf{a}_{3}$ | = | $- c z_{3} \,\mathbf{\hat{z}}$ | (4e) | F I |
$\mathbf{B_{8}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (4e) | F I |
$\mathbf{B_{9}}$ | = | $x_{4} \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $a x_{4} \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | F II |
$\mathbf{B_{10}}$ | = | $- x_{4} \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | F II |
$\mathbf{B_{11}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}$ | (4g) | F II |
$\mathbf{B_{12}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}$ | (4g) | F II |