AFLOW Prototype: A3B_hP24_151_3c_2a-001
This structure originally had the label A3B_hP24_151_3c_2a. Calls to that address will be redirected here.
If you are using this page, please cite:
M. J. Mehl, D. Hicks, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 1, Comp. Mat. Sci. 136, S1-S828 (2017). (doi=10.1016/j.commatsci.2017.01.017)
Links to this page
https://aflow.org/p/5EPG
or
https://aflow.org/p/A3B_hP24_151_3c_2a-001
or
PDF Version
Prototype | Cl$_{3}$Cr |
AFLOW prototype label | A3B_hP24_151_3c_2a-001 |
Strukturbericht designation | $D0_{4}$ |
ICSD | 33578 |
Pearson symbol | hP24 |
Space group number | 151 |
Space group symbol | $P3_112$ |
AFLOW prototype command |
aflow --proto=A3B_hP24_151_3c_2a-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$ |
CrBr$_{3}$, CrI$_{3}$, $\alpha$-AlCl$_{3}$, $\alpha$-RuCl$_{3}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{1} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3a) | Cr I |
$\mathbf{B_{2}}$ | = | $x_{1} \, \mathbf{a}_{1}+2 x_{1} \, \mathbf{a}_{2}+\frac{2}{3} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{1} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ | (3a) | Cr I |
$\mathbf{B_{3}}$ | = | $- 2 x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}$ | = | $- \frac{3}{2}a x_{1} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}$ | (3a) | Cr I |
$\mathbf{B_{4}}$ | = | $x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{2} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3a) | Cr II |
$\mathbf{B_{5}}$ | = | $x_{2} \, \mathbf{a}_{1}+2 x_{2} \, \mathbf{a}_{2}+\frac{2}{3} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{2} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ | (3a) | Cr II |
$\mathbf{B_{6}}$ | = | $- 2 x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}$ | = | $- \frac{3}{2}a x_{2} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}$ | (3a) | Cr II |
$\mathbf{B_{7}}$ | = | $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (6c) | Cl I |
$\mathbf{B_{8}}$ | = | $- y_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{3} - 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (6c) | Cl I |
$\mathbf{B_{9}}$ | = | $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{3} + 2\right) \,\mathbf{\hat{z}}$ | (6c) | Cl I |
$\mathbf{B_{10}}$ | = | $- y_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{3} - 2\right) \,\mathbf{\hat{z}}$ | (6c) | Cl I |
$\mathbf{B_{11}}$ | = | $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{3} + 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (6c) | Cl I |
$\mathbf{B_{12}}$ | = | $x_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (6c) | Cl I |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (6c) | Cl II |
$\mathbf{B_{14}}$ | = | $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (6c) | Cl II |
$\mathbf{B_{15}}$ | = | $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{4} + 2\right) \,\mathbf{\hat{z}}$ | (6c) | Cl II |
$\mathbf{B_{16}}$ | = | $- y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{4} - 2\right) \,\mathbf{\hat{z}}$ | (6c) | Cl II |
$\mathbf{B_{17}}$ | = | $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{4} + 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (6c) | Cl II |
$\mathbf{B_{18}}$ | = | $x_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (6c) | Cl II |
$\mathbf{B_{19}}$ | = | $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6c) | Cl III |
$\mathbf{B_{20}}$ | = | $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (6c) | Cl III |
$\mathbf{B_{21}}$ | = | $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{5} + 2\right) \,\mathbf{\hat{z}}$ | (6c) | Cl III |
$\mathbf{B_{22}}$ | = | $- y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{5} - 2\right) \,\mathbf{\hat{z}}$ | (6c) | Cl III |
$\mathbf{B_{23}}$ | = | $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ | (6c) | Cl III |
$\mathbf{B_{24}}$ | = | $x_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6c) | Cl III |