AFLOW Prototype: AB4C_tP12_137_a_g_b-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/Q06Z
or
https://aflow.org/p/AB4C_tP12_137_a_g_b-001
or
PDF Version
Prototype | BD$_{4}$K |
AFLOW prototype label | AB4C_tP12_137_a_g_b-001 |
ICSD | 99264 |
Pearson symbol | tP12 |
Space group number | 137 |
Space group symbol | $P4_2/nmc$ |
AFLOW prototype command |
aflow --proto=AB4C_tP12_137_a_g_b-001
--params=$a, \allowbreak c/a, \allowbreak y_{3}, \allowbreak z_{3}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2a) | B I |
$\mathbf{B_{2}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2a) | B I |
$\mathbf{B_{3}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2b) | K I |
$\mathbf{B_{4}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2b) | K I |
$\mathbf{B_{5}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+a y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (8g) | D I |
$\mathbf{B_{6}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (8g) | D I |
$\mathbf{B_{7}}$ | = | $- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8g) | D I |
$\mathbf{B_{8}}$ | = | $y_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a y_{3} \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8g) | D I |
$\mathbf{B_{9}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (8g) | D I |
$\mathbf{B_{10}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}- a y_{3} \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (8g) | D I |
$\mathbf{B_{11}}$ | = | $\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8g) | D I |
$\mathbf{B_{12}}$ | = | $- y_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a y_{3} \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8g) | D I |