AFLOW Prototype: A3B2C5_oP40_62_3c_2c_5c-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/72A0
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https://aflow.org/p/A3B2C5_oP40_62_3c_2c_5c-001
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PDF Version
Prototype | Ni$_{3}$Ta$_{2}$Te$_{5}$ |
AFLOW prototype label | A3B2C5_oP40_62_3c_2c_5c-001 |
ICSD | none |
Pearson symbol | oP40 |
Space group number | 62 |
Space group symbol | $Pnma$ |
AFLOW prototype command |
aflow --proto=A3B2C5_oP40_62_3c_2c_5c-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak z_{7}, \allowbreak x_{8}, \allowbreak z_{8}, \allowbreak x_{9}, \allowbreak z_{9}, \allowbreak x_{10}, \allowbreak z_{10}$ |
Ta$_{2}$Pd$_{3}$Te$_{5}$
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) | Ni 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) | Ni 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) | Ni 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) | Ni 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) | Ni 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) | Ni 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) | Ni 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) | Ni 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) | Ni III |
$\mathbf{B_{10}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Ni III |
$\mathbf{B_{11}}$ | = | $- x_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (4c) | Ni III |
$\mathbf{B_{12}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Ni III |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4c) | Ta I |
$\mathbf{B_{14}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Ta I |
$\mathbf{B_{15}}$ | = | $- x_{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (4c) | Ta I |
$\mathbf{B_{16}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Ta I |
$\mathbf{B_{17}}$ | = | $x_{5} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $a x_{5} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (4c) | Ta II |
$\mathbf{B_{18}}$ | = | $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Ta II |
$\mathbf{B_{19}}$ | = | $- x_{5} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- a x_{5} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (4c) | Ta II |
$\mathbf{B_{20}}$ | = | $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Ta II |
$\mathbf{B_{21}}$ | = | $x_{6} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $a x_{6} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (4c) | Te I |
$\mathbf{B_{22}}$ | = | $- \left(x_{6} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{6} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te I |
$\mathbf{B_{23}}$ | = | $- x_{6} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $- a x_{6} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (4c) | Te I |
$\mathbf{B_{24}}$ | = | $\left(x_{6} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{6} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te I |
$\mathbf{B_{25}}$ | = | $x_{7} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $a x_{7} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (4c) | Te II |
$\mathbf{B_{26}}$ | = | $- \left(x_{7} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{7} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te II |
$\mathbf{B_{27}}$ | = | $- x_{7} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ | = | $- a x_{7} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ | (4c) | Te II |
$\mathbf{B_{28}}$ | = | $\left(x_{7} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{7} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{7} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{7} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te II |
$\mathbf{B_{29}}$ | = | $x_{8} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $a x_{8} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ | (4c) | Te III |
$\mathbf{B_{30}}$ | = | $- \left(x_{8} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{8} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te III |
$\mathbf{B_{31}}$ | = | $- x_{8} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{8} \, \mathbf{a}_{3}$ | = | $- a x_{8} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{8} \,\mathbf{\hat{z}}$ | (4c) | Te III |
$\mathbf{B_{32}}$ | = | $\left(x_{8} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{8} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{8} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{8} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te III |
$\mathbf{B_{33}}$ | = | $x_{9} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $a x_{9} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ | (4c) | Te IV |
$\mathbf{B_{34}}$ | = | $- \left(x_{9} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{9} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te IV |
$\mathbf{B_{35}}$ | = | $- x_{9} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{9} \, \mathbf{a}_{3}$ | = | $- a x_{9} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{9} \,\mathbf{\hat{z}}$ | (4c) | Te IV |
$\mathbf{B_{36}}$ | = | $\left(x_{9} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{9} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{9} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{9} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te IV |
$\mathbf{B_{37}}$ | = | $x_{10} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $a x_{10} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ | (4c) | Te V |
$\mathbf{B_{38}}$ | = | $- \left(x_{10} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{10} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{10} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{10} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te V |
$\mathbf{B_{39}}$ | = | $- x_{10} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{10} \, \mathbf{a}_{3}$ | = | $- a x_{10} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{10} \,\mathbf{\hat{z}}$ | (4c) | Te V |
$\mathbf{B_{40}}$ | = | $\left(x_{10} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{10} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{10} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{10} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Te V |