AFLOW Prototype: A5B6_hP22_194_ach_gh-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/09JW
or
https://aflow.org/p/A5B6_hP22_194_ach_gh-001
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PDF Version
Prototype | Sn$_{5}$Ti$_{6}$ |
AFLOW prototype label | A5B6_hP22_194_ach_gh-001 |
ICSD | 660295 |
Pearson symbol | hP22 |
Space group number | 194 |
Space group symbol | $P6_3/mmc$ |
AFLOW prototype command |
aflow --proto=A5B6_hP22_194_ach_gh-001
--params=$a, \allowbreak c/a, \allowbreak x_{4}, \allowbreak x_{5}$ |
V$_{6}$Ga$_{5}$
1/4should be at zero and vice versa.
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | Sn I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2a) | Sn I |
$\mathbf{B_{3}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2c) | Sn II |
$\mathbf{B_{4}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2c) | Sn II |
$\mathbf{B_{5}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ | (6g) | Ti I |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ | (6g) | Ti I |
$\mathbf{B_{7}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}$ | (6g) | Ti I |
$\mathbf{B_{8}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (6g) | Ti I |
$\mathbf{B_{9}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (6g) | Ti I |
$\mathbf{B_{10}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (6g) | Ti I |
$\mathbf{B_{11}}$ | = | $x_{4} \, \mathbf{a}_{1}+2 x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Sn III |
$\mathbf{B_{12}}$ | = | $- 2 x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Sn III |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Sn III |
$\mathbf{B_{14}}$ | = | $- x_{4} \, \mathbf{a}_{1}- 2 x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Sn III |
$\mathbf{B_{15}}$ | = | $2 x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Sn III |
$\mathbf{B_{16}}$ | = | $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Sn III |
$\mathbf{B_{17}}$ | = | $x_{5} \, \mathbf{a}_{1}+2 x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{18}}$ | = | $- 2 x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{19}}$ | = | $x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{20}}$ | = | $- x_{5} \, \mathbf{a}_{1}- 2 x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{21}}$ | = | $2 x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{22}}$ | = | $- x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ti II |