AFLOW Prototype: A8B_hP27_180_2ik_d-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/T338
or
https://aflow.org/p/A8B_hP27_180_2ik_d-001
or
PDF Version
Prototype | C$_{8}$Cs |
AFLOW prototype label | A8B_hP27_180_2ik_d-001 |
ICSD | none |
Pearson symbol | hP27 |
Space group number | 180 |
Space group symbol | $P6_222$ |
AFLOW prototype command |
aflow --proto=A8B_hP27_180_2ik_d-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (3d) | Cs I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (3d) | Cs I |
$\mathbf{B_{3}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{5}{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{5}{6}c \,\mathbf{\hat{z}}$ | (3d) | Cs I |
$\mathbf{B_{4}}$ | = | $x_{2} \, \mathbf{a}_{1}+2 x_{2} \, \mathbf{a}_{2}$ | = | $\frac{3}{2}a x_{2} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}$ | (6i) | C I |
$\mathbf{B_{5}}$ | = | $- 2 x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}+\frac{2}{3} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{2} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ | (6i) | C I |
$\mathbf{B_{6}}$ | = | $x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{2} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (6i) | C I |
$\mathbf{B_{7}}$ | = | $- x_{2} \, \mathbf{a}_{1}- 2 x_{2} \, \mathbf{a}_{2}$ | = | $- \frac{3}{2}a x_{2} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}$ | (6i) | C I |
$\mathbf{B_{8}}$ | = | $2 x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+\frac{2}{3} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{2} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ | (6i) | C I |
$\mathbf{B_{9}}$ | = | $- x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{2} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (6i) | C I |
$\mathbf{B_{10}}$ | = | $x_{3} \, \mathbf{a}_{1}+2 x_{3} \, \mathbf{a}_{2}$ | = | $\frac{3}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ | (6i) | C II |
$\mathbf{B_{11}}$ | = | $- 2 x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{2}{3} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ | (6i) | C II |
$\mathbf{B_{12}}$ | = | $x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (6i) | C II |
$\mathbf{B_{13}}$ | = | $- x_{3} \, \mathbf{a}_{1}- 2 x_{3} \, \mathbf{a}_{2}$ | = | $- \frac{3}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ | (6i) | C II |
$\mathbf{B_{14}}$ | = | $2 x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{2}{3} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ | (6i) | C II |
$\mathbf{B_{15}}$ | = | $- x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (6i) | C II |
$\mathbf{B_{16}}$ | = | $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{17}}$ | = | $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{4} + 2\right) \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{18}}$ | = | $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{19}}$ | = | $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{20}}$ | = | $y_{4} \, \mathbf{a}_{1}- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{4} + 2 y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{4} + 2\right) \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{21}}$ | = | $\left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{22}}$ | = | $y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{4} - 2\right) \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{23}}$ | = | $\left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{24}}$ | = | $- x_{4} \, \mathbf{a}_{1}- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{25}}$ | = | $- y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{4} - 2\right) \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{26}}$ | = | $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{4} + 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (12k) | C III |
$\mathbf{B_{27}}$ | = | $x_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (12k) | C III |