AFLOW Prototype: A2B2C_oP20_55_2g_2g_h-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/93YX
or
https://aflow.org/p/A2B2C_oP20_55_2g_2g_h-001
or
PDF Version
Prototype | B$_{2}$C$_{2}$Sc |
AFLOW prototype label | A2B2C_oP20_55_2g_2g_h-001 |
ICSD | 23834 |
Pearson symbol | oP20 |
Space group number | 55 |
Space group symbol | $Pbam$ |
AFLOW prototype command |
aflow --proto=A2B2C_oP20_55_2g_2g_h-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak y_{1}, \allowbreak x_{2}, \allowbreak y_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak x_{5}, \allowbreak y_{5}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+y_{1} \, \mathbf{a}_{2}$ | = | $a x_{1} \,\mathbf{\hat{x}}+b y_{1} \,\mathbf{\hat{y}}$ | (4g) | B I |
$\mathbf{B_{2}}$ | = | $- x_{1} \, \mathbf{a}_{1}- y_{1} \, \mathbf{a}_{2}$ | = | $- a x_{1} \,\mathbf{\hat{x}}- b y_{1} \,\mathbf{\hat{y}}$ | (4g) | B I |
$\mathbf{B_{3}}$ | = | $- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{1} + \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{1} + \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | B I |
$\mathbf{B_{4}}$ | = | $\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{1} - \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{1} - \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | B I |
$\mathbf{B_{5}}$ | = | $x_{2} \, \mathbf{a}_{1}+y_{2} \, \mathbf{a}_{2}$ | = | $a x_{2} \,\mathbf{\hat{x}}+b y_{2} \,\mathbf{\hat{y}}$ | (4g) | B II |
$\mathbf{B_{6}}$ | = | $- x_{2} \, \mathbf{a}_{1}- y_{2} \, \mathbf{a}_{2}$ | = | $- a x_{2} \,\mathbf{\hat{x}}- b y_{2} \,\mathbf{\hat{y}}$ | (4g) | B II |
$\mathbf{B_{7}}$ | = | $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{2} + \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{2} + \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | B II |
$\mathbf{B_{8}}$ | = | $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{2} - \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | B II |
$\mathbf{B_{9}}$ | = | $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}$ | = | $a x_{3} \,\mathbf{\hat{x}}+b y_{3} \,\mathbf{\hat{y}}$ | (4g) | C I |
$\mathbf{B_{10}}$ | = | $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}$ | = | $- a x_{3} \,\mathbf{\hat{x}}- b y_{3} \,\mathbf{\hat{y}}$ | (4g) | C I |
$\mathbf{B_{11}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | C I |
$\mathbf{B_{12}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | C I |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}$ | = | $a x_{4} \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}$ | (4g) | C II |
$\mathbf{B_{14}}$ | = | $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}$ | (4g) | C II |
$\mathbf{B_{15}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | C II |
$\mathbf{B_{16}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (4g) | C II |
$\mathbf{B_{17}}$ | = | $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a x_{5} \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4h) | Sc I |
$\mathbf{B_{18}}$ | = | $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a x_{5} \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4h) | Sc I |
$\mathbf{B_{19}}$ | = | $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4h) | Sc I |
$\mathbf{B_{20}}$ | = | $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4h) | Sc I |