AFLOW Prototype: AB2C6D_hP20_163_c_f_i_a-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/UZBW
or
https://aflow.org/p/AB2C6D_hP20_163_c_f_i_a-001
or
PDF Version
Prototype | GeNa$_{2}$O$_{6}$Te |
AFLOW prototype label | AB2C6D_hP20_163_c_f_i_a-001 |
ICSD | 88789 |
Pearson symbol | hP20 |
Space group number | 163 |
Space group symbol | $P\overline{3}1c$ |
AFLOW prototype command |
aflow --proto=AB2C6D_hP20_163_c_f_i_a-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$ |
Li$_{2}$MnTeO$_{6}$, Na$_{2}$SnTeO$_{6}$
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) | Te I |
$\mathbf{B_{2}}$ | = | $\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2a) | Te I |
$\mathbf{B_{3}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2c) | Ge I |
$\mathbf{B_{4}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2c) | Ge I |
$\mathbf{B_{5}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (4f) | Na I |
$\mathbf{B_{6}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na I |
$\mathbf{B_{7}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (4f) | Na I |
$\mathbf{B_{8}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na I |
$\mathbf{B_{9}}$ | = | $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{10}}$ | = | $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{11}}$ | = | $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{12}}$ | = | $- y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{13}}$ | = | $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{4} + 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{14}}$ | = | $x_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{15}}$ | = | $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{16}}$ | = | $y_{4} \, \mathbf{a}_{1}- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(- x_{4} + 2 y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{17}}$ | = | $\left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{18}}$ | = | $y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{19}}$ | = | $\left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12i) | O I |
$\mathbf{B_{20}}$ | = | $- x_{4} \, \mathbf{a}_{1}- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12i) | O I |