AFLOW Prototype: AB6C8_hR15_148_a_f_cf-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/CY1Y
or
https://aflow.org/p/AB6C8_hR15_148_a_f_cf-001
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PDF Version
Prototype | BaMo$_{6}$S$_{8}$ |
AFLOW prototype label | AB6C8_hR15_148_a_f_cf-001 |
ICSD | none |
Pearson symbol | hR15 |
Space group number | 148 |
Space group symbol | $R\overline{3}$ |
AFLOW prototype command |
aflow --proto=AB6C8_hR15_148_a_f_cf-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$ |
CaMo$_{6}$S$_{8}$, CeMo$_{6}$S$_{8}$, CeMo$_{6}$Se$_{8}$, EuMo$_{6}$S$_{8}$, LaMo$_{6}$S$_{8}$, LaMo$_{6}$Se$_{8}$, NdMo$_{6}$S$_{8}$, NdMo$_{6}$Se$_{8}$, PbMo$_{6}$S$_{8}$, PrMo$_{6}$S$_{8}$, PrMo$_{6}$Se$_{8}$, SmMo$_{6}$S$_{8}$, SmMo$_{6}$Se$_{8}$, SrMo$_{6}$S$_{8}$
--hex
. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Ba 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) | S I |
$\mathbf{B_{3}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ | = | $- c x_{2} \,\mathbf{\hat{z}}$ | (2c) | S I |
$\mathbf{B_{4}}$ | = | $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{3} - 2 y_{3} + z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{3} + y_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6f) | Mo I |
$\mathbf{B_{5}}$ | = | $z_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+y_{3} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(y_{3} - z_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(2 x_{3} - y_{3} - z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{3} + y_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6f) | Mo I |
$\mathbf{B_{6}}$ | = | $y_{3} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{3} + y_{3} - 2 z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{3} + y_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6f) | Mo I |
$\mathbf{B_{7}}$ | = | $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{3} - 2 y_{3} + z_{3}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{3} + y_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6f) | Mo I |
$\mathbf{B_{8}}$ | = | $- z_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- y_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(y_{3} - z_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(2 x_{3} - y_{3} - z_{3}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{3} + y_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6f) | Mo I |
$\mathbf{B_{9}}$ | = | $- y_{3} \, \mathbf{a}_{1}- z_{3} \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{3} + y_{3} - 2 z_{3}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{3} + y_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6f) | Mo I |
$\mathbf{B_{10}}$ | = | $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} - 2 y_{4} + z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S II |
$\mathbf{B_{11}}$ | = | $z_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+y_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(y_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(2 x_{4} - y_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S II |
$\mathbf{B_{12}}$ | = | $y_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} + y_{4} - 2 z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S II |
$\mathbf{B_{13}}$ | = | $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - 2 y_{4} + z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S II |
$\mathbf{B_{14}}$ | = | $- z_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- y_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(y_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(2 x_{4} - y_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S II |
$\mathbf{B_{15}}$ | = | $- y_{4} \, \mathbf{a}_{1}- z_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} + y_{4} - 2 z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(x_{4} + y_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6f) | S II |