AFLOW Prototype: A2B3_oP20_62_2c_3c-001
This structure originally had the label A2B3_oP20_62_2c_3c. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)
Links to this page
https://aflow.org/p/KS86
or
https://aflow.org/p/A2B3_oP20_62_2c_3c-001
or
PDF Version
Prototype | C$_{2}$Cr$_{3}$ |
AFLOW prototype label | A2B3_oP20_62_2c_3c-001 |
Strukturbericht designation | $D5_{10}$ |
Mineral name | tongbaite |
ICSD | 15086 |
Pearson symbol | oP20 |
Space group number | 62 |
Space group symbol | $Pnma$ |
AFLOW prototype command |
aflow --proto=A2B3_oP20_62_2c_3c-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}$ |
Hf$_{3}$P$_{2}$, La$_{2}$S$_{3}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ | = | $a x_{1} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ | (4c) | C I |
$\mathbf{B_{2}}$ | = | $- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | C I |
$\mathbf{B_{3}}$ | = | $- x_{1} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{1} \, \mathbf{a}_{3}$ | = | $- a x_{1} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ | (4c) | C I |
$\mathbf{B_{4}}$ | = | $\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | C I |
$\mathbf{B_{5}}$ | = | $x_{2} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $a x_{2} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (4c) | C II |
$\mathbf{B_{6}}$ | = | $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | C II |
$\mathbf{B_{7}}$ | = | $- x_{2} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ | = | $- a x_{2} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ | (4c) | C II |
$\mathbf{B_{8}}$ | = | $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | C II |
$\mathbf{B_{9}}$ | = | $x_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (4c) | Cr I |
$\mathbf{B_{10}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Cr I |
$\mathbf{B_{11}}$ | = | $- x_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (4c) | Cr I |
$\mathbf{B_{12}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Cr I |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4c) | Cr II |
$\mathbf{B_{14}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Cr II |
$\mathbf{B_{15}}$ | = | $- x_{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (4c) | Cr II |
$\mathbf{B_{16}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Cr II |
$\mathbf{B_{17}}$ | = | $x_{5} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $a x_{5} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (4c) | Cr III |
$\mathbf{B_{18}}$ | = | $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Cr III |
$\mathbf{B_{19}}$ | = | $- x_{5} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- a x_{5} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (4c) | Cr III |
$\mathbf{B_{20}}$ | = | $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Cr III |