AFLOW Prototype: A2B3_tP20_116_adi_ej-001
This structure originally had the label A2B3_tP20_116_bci_fj. 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/RJFD
or
https://aflow.org/p/A2B3_tP20_116_adi_ej-001
or
PDF Version
Prototype | Ru$_{2}$Sn$_{3}$ |
AFLOW prototype label | A2B3_tP20_116_adi_ej-001 |
ICSD | 105994 |
Pearson symbol | tP20 |
Space group number | 116 |
Space group symbol | $P\overline{4}c2$ |
AFLOW prototype command |
aflow --proto=A2B3_tP20_116_adi_ej-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2a) | Ru I |
$\mathbf{B_{2}}$ | = | $\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2a) | Ru I |
$\mathbf{B_{3}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}$ | (2d) | Ru II |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2d) | Ru II |
$\mathbf{B_{5}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (4e) | Sn I |
$\mathbf{B_{6}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (4e) | Sn I |
$\mathbf{B_{7}}$ | = | $x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (4e) | Sn I |
$\mathbf{B_{8}}$ | = | $- x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (4e) | Sn I |
$\mathbf{B_{9}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4i) | Ru III |
$\mathbf{B_{10}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- c z_{4} \,\mathbf{\hat{z}}$ | (4i) | Ru III |
$\mathbf{B_{11}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4i) | Ru III |
$\mathbf{B_{12}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4i) | Ru III |
$\mathbf{B_{13}}$ | = | $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $a x_{5} \,\mathbf{\hat{x}}+a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (8j) | Sn II |
$\mathbf{B_{14}}$ | = | $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $- a x_{5} \,\mathbf{\hat{x}}- a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (8j) | Sn II |
$\mathbf{B_{15}}$ | = | $y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $a y_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (8j) | Sn II |
$\mathbf{B_{16}}$ | = | $- y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- a y_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (8j) | Sn II |
$\mathbf{B_{17}}$ | = | $x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a x_{5} \,\mathbf{\hat{x}}- a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Sn II |
$\mathbf{B_{18}}$ | = | $- x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{5} \,\mathbf{\hat{x}}+a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Sn II |
$\mathbf{B_{19}}$ | = | $y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a y_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Sn II |
$\mathbf{B_{20}}$ | = | $- y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a y_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (8j) | Sn II |