AFLOW Prototype: A2B3_tP20_136_j_cfg-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/T3N4
or
https://aflow.org/p/A2B3_tP20_136_j_cfg-001
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PDF Version
Prototype | Hg$_{2}$Na$_{3}$ |
AFLOW prototype label | A2B3_tP20_136_j_cfg-001 |
ICSD | 104328 |
Pearson symbol | tP20 |
Space group number | 136 |
Space group symbol | $P4_2/mnm$ |
AFLOW prototype command |
aflow --proto=A2B3_tP20_136_j_cfg-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak z_{4}$ |
Al$_{2}$Zr$_{3}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}$ | (4c) | Na I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4c) | Na I |
$\mathbf{B_{3}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4c) | Na I |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}$ | (4c) | Na I |
$\mathbf{B_{5}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}$ | = | $a x_{2} \,\mathbf{\hat{x}}+a x_{2} \,\mathbf{\hat{y}}$ | (4f) | Na II |
$\mathbf{B_{6}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}$ | = | $- a x_{2} \,\mathbf{\hat{x}}- a x_{2} \,\mathbf{\hat{y}}$ | (4f) | Na II |
$\mathbf{B_{7}}$ | = | $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4f) | Na II |
$\mathbf{B_{8}}$ | = | $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4f) | Na II |
$\mathbf{B_{9}}$ | = | $x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}$ | = | $a x_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}$ | (4g) | Na III |
$\mathbf{B_{10}}$ | = | $- x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}$ | = | $- a x_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}$ | (4g) | Na III |
$\mathbf{B_{11}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4g) | Na III |
$\mathbf{B_{12}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4g) | Na III |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (8j) | Hg I |
$\mathbf{B_{14}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (8j) | Hg I |
$\mathbf{B_{15}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Hg I |
$\mathbf{B_{16}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Hg I |
$\mathbf{B_{17}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Hg I |
$\mathbf{B_{18}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Hg I |
$\mathbf{B_{19}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (8j) | Hg I |
$\mathbf{B_{20}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (8j) | Hg I |