AFLOW Prototype: A4B5_tP36_92_2b_a2b-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/C3GS
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https://aflow.org/p/A4B5_tP36_92_2b_a2b-001
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PDF Version
Prototype | Si$_{4}$Zr$_{5}$ |
AFLOW prototype label | A4B5_tP36_92_2b_a2b-001 |
ICSD | 43214 |
Pearson symbol | tP36 |
Space group number | 92 |
Space group symbol | $P4_12_12$ |
AFLOW prototype command |
aflow --proto=A4B5_tP36_92_2b_a2b-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak y_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$ |
Ce$_{5}$Si$_{4}$, Hf$_{5}$Si$_{4}$, La$_{5}$Si$_{4}$, Nb$_{5}$Si$_{4}$, Pr$_{5}$Si$_{4}$, Ti$_{5}$Si$_{4}$, Zr$_{5}$Ge$_{4}$, Pr$_{5}$Si$_{2}$Ge$_{2}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}$ | = | $a x_{1} \,\mathbf{\hat{x}}+a x_{1} \,\mathbf{\hat{y}}$ | (4a) | Zr I |
$\mathbf{B_{2}}$ | = | $- x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a x_{1} \,\mathbf{\hat{x}}- a x_{1} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4a) | Zr I |
$\mathbf{B_{3}}$ | = | $- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (4a) | Zr I |
$\mathbf{B_{4}}$ | = | $\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (4a) | Zr I |
$\mathbf{B_{5}}$ | = | $x_{2} \, \mathbf{a}_{1}+y_{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $a x_{2} \,\mathbf{\hat{x}}+a y_{2} \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{6}}$ | = | $- x_{2} \, \mathbf{a}_{1}- y_{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{2} \,\mathbf{\hat{x}}- a y_{2} \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{7}}$ | = | $- \left(y_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(y_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{8}}$ | = | $\left(y_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{2} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(y_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{9}}$ | = | $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{2} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{2} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{10}}$ | = | $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{2} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{2} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{11}}$ | = | $y_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ | = | $a y_{2} \,\mathbf{\hat{x}}+a x_{2} \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{12}}$ | = | $- y_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a y_{2} \,\mathbf{\hat{x}}- a x_{2} \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Si I |
$\mathbf{B_{13}}$ | = | $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}+a y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{14}}$ | = | $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}- a y_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{15}}$ | = | $- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{16}}$ | = | $\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{17}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{18}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{3} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{19}}$ | = | $y_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $a y_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{20}}$ | = | $- y_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a y_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Si II |
$\mathbf{B_{21}}$ | = | $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+a y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{22}}$ | = | $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- a y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{23}}$ | = | $- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{24}}$ | = | $\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{25}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{26}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{27}}$ | = | $y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $a y_{4} \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{28}}$ | = | $- y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a y_{4} \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr II |
$\mathbf{B_{29}}$ | = | $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $a x_{5} \,\mathbf{\hat{x}}+a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (8b) | Zr III |
$\mathbf{B_{30}}$ | = | $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{5} \,\mathbf{\hat{x}}- a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr III |
$\mathbf{B_{31}}$ | = | $- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr III |
$\mathbf{B_{32}}$ | = | $\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr III |
$\mathbf{B_{33}}$ | = | $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr III |
$\mathbf{B_{34}}$ | = | $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{3}{4}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr III |
$\mathbf{B_{35}}$ | = | $y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $a y_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (8b) | Zr III |
$\mathbf{B_{36}}$ | = | $- y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a y_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8b) | Zr III |