AFLOW Prototype: A2BC_hR12_166_g_d_ac-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/AFH9
or
https://aflow.org/p/A2BC_hR12_166_g_d_ac-001
or
PDF Version
Prototype | BCTh$_{2}$ |
AFLOW prototype label | A2BC_hR12_166_g_d_ac-001 |
ICSD | 68414 |
Pearson symbol | hR12 |
Space group number | 166 |
Space group symbol | $R\overline{3}m$ |
AFLOW prototype command |
aflow --proto=A2BC_hR12_166_g_d_ac-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{4}$ |
$\beta$-UB$_{2}$C (HT)
--hex
. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Th I |
$\mathbf{B_{2}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ | = | $c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Th II |
$\mathbf{B_{3}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ | = | $- c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Th II |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (3d) | C I |
$\mathbf{B_{5}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (3d) | C I |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (3d) | C I |
$\mathbf{B_{7}}$ | = | $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \left(2 x_{4} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \left(6 x_{4} + 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | B I |
$\mathbf{B_{8}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \left(2 x_{4} + 1\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \left(6 x_{4} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | B I |
$\mathbf{B_{9}}$ | = | $- x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | B I |
$\mathbf{B_{10}}$ | = | $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \left(2 x_{4} + 1\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \left(6 x_{4} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | B I |
$\mathbf{B_{11}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \left(2 x_{4} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \left(6 x_{4} + 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | B I |
$\mathbf{B_{12}}$ | = | $x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | B I |