AFLOW Prototype: A2B3C2_cI28_199_a_b_a-002
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/VYHG
or
https://aflow.org/p/A2B3C2_cI28_199_a_b_a-002
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PDF Version
Prototype | Cl$_{2}$Hg$_{3}$S$_{2}$ |
AFLOW prototype label | A2B3C2_cI28_199_a_b_a-002 |
Mineral name | corderoite |
ICSD | 27399 |
Pearson symbol | cI28 |
Space group number | 199 |
Space group symbol | $I2_13$ |
AFLOW prototype command |
aflow --proto=A2B3C2_cI28_199_a_b_a-002
--params=$a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}$ |
Hg$_{3}$S$_{2}$F$_{2}$, Hg$_{3}$S$_{2}$I$_{2}$, Hg$_{3}$Se$_{2}$F$_{2}$, Hg$_{3}$Se$_{2}$Cl$_{2}$, Hg$_{3}$Te$_{2}$Br$_{2}$, Hg$_{3}$Te$_{2}$Cl$_{2}$, K$_{2}$Pb$_{2}$O$_{3}$, K$_{2}$Sn$_{2}$O$_{3}$, Pd$_{3}$S$_{2}$Bi$_{2}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $2 x_{1} \, \mathbf{a}_{1}+2 x_{1} \, \mathbf{a}_{2}+2 x_{1} \, \mathbf{a}_{3}$ | = | $a x_{1} \,\mathbf{\hat{x}}+a x_{1} \,\mathbf{\hat{y}}+a x_{1} \,\mathbf{\hat{z}}$ | (8a) | Cl I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- \left(2 x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{1} \,\mathbf{\hat{x}}- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+a x_{1} \,\mathbf{\hat{z}}$ | (8a) | Cl I |
$\mathbf{B_{3}}$ | = | $- \left(2 x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{1} \,\mathbf{\hat{y}}- a x_{1} \,\mathbf{\hat{z}}$ | (8a) | Cl I |
$\mathbf{B_{4}}$ | = | $- \left(2 x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $a x_{1} \,\mathbf{\hat{x}}- a x_{1} \,\mathbf{\hat{y}}- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8a) | Cl I |
$\mathbf{B_{5}}$ | = | $2 x_{2} \, \mathbf{a}_{1}+2 x_{2} \, \mathbf{a}_{2}+2 x_{2} \, \mathbf{a}_{3}$ | = | $a x_{2} \,\mathbf{\hat{x}}+a x_{2} \,\mathbf{\hat{y}}+a x_{2} \,\mathbf{\hat{z}}$ | (8a) | S I |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- \left(2 x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{2} \,\mathbf{\hat{x}}- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+a x_{2} \,\mathbf{\hat{z}}$ | (8a) | S I |
$\mathbf{B_{7}}$ | = | $- \left(2 x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{2} \,\mathbf{\hat{y}}- a x_{2} \,\mathbf{\hat{z}}$ | (8a) | S I |
$\mathbf{B_{8}}$ | = | $- \left(2 x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $a x_{2} \,\mathbf{\hat{x}}- a x_{2} \,\mathbf{\hat{y}}- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8a) | S I |
$\mathbf{B_{9}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{4}\right) \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ | (12b) | Hg I |
$\mathbf{B_{10}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{4}\right) \, \mathbf{a}_{2}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ | (12b) | Hg I |
$\mathbf{B_{11}}$ | = | $x_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\left(x_{3} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}$ | (12b) | Hg I |
$\mathbf{B_{12}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- \left(x_{3} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ | (12b) | Hg I |
$\mathbf{B_{13}}$ | = | $\left(x_{3} + \frac{1}{4}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{y}}+a x_{3} \,\mathbf{\hat{z}}$ | (12b) | Hg I |
$\mathbf{B_{14}}$ | = | $- \left(x_{3} - \frac{1}{4}\right) \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}- a x_{3} \,\mathbf{\hat{z}}$ | (12b) | Hg I |