AFLOW Prototype: A2B3_mC20_12_2i_3i-003
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/8W6Z
or
https://aflow.org/p/A2B3_mC20_12_2i_3i-003
or
PDF Version
Prototype | As$_{2}$Te$_{3}$ |
AFLOW prototype label | A2B3_mC20_12_2i_3i-003 |
ICSD | 196146 |
Pearson symbol | mC20 |
Space group number | 12 |
Space group symbol | $C2/m$ |
AFLOW prototype command |
aflow --proto=A2B3_mC20_12_2i_3i-003
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak \beta, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}$ |
--params
) specified in their corresponding CIF files. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ | = | $\left(a x_{1} + c z_{1} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{1} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | As I |
$\mathbf{B_{2}}$ | = | $- x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}- z_{1} \, \mathbf{a}_{3}$ | = | $- \left(a x_{1} + c z_{1} \cos{\beta}\right) \,\mathbf{\hat{x}}- c z_{1} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | As I |
$\mathbf{B_{3}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\left(a x_{2} + c z_{2} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{2} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | As II |
$\mathbf{B_{4}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ | = | $- \left(a x_{2} + c z_{2} \cos{\beta}\right) \,\mathbf{\hat{x}}- c z_{2} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | As II |
$\mathbf{B_{5}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\left(a x_{3} + c z_{3} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{3} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | Te I |
$\mathbf{B_{6}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $- \left(a x_{3} + c z_{3} \cos{\beta}\right) \,\mathbf{\hat{x}}- c z_{3} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | Te I |
$\mathbf{B_{7}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\left(a x_{4} + c z_{4} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{4} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | Te II |
$\mathbf{B_{8}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- \left(a x_{4} + c z_{4} \cos{\beta}\right) \,\mathbf{\hat{x}}- c z_{4} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | Te II |
$\mathbf{B_{9}}$ | = | $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\left(a x_{5} + c z_{5} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{5} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | Te III |
$\mathbf{B_{10}}$ | = | $- x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- \left(a x_{5} + c z_{5} \cos{\beta}\right) \,\mathbf{\hat{x}}- c z_{5} \sin{\beta} \,\mathbf{\hat{z}}$ | (4i) | Te III |