AFLOW Prototype: ABC4D_hP14_156_ac_bc_ab2d_ab-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/Z1Y1
or
https://aflow.org/p/ABC4D_hP14_156_ac_bc_ab2d_ab-001
or
PDF Version
Prototype | KNaO$_{4}$S |
AFLOW prototype label | ABC4D_hP14_156_ac_bc_ab2d_ab-001 |
ICSD | 133733 |
Pearson symbol | hP14 |
Space group number | 156 |
Space group symbol | $P3m1$ |
AFLOW prototype command |
aflow --proto=ABC4D_hP14_156_ac_bc_ab2d_ab-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak z_{6}, \allowbreak z_{7}, \allowbreak z_{8}, \allowbreak x_{9}, \allowbreak z_{9}, \allowbreak x_{10}, \allowbreak z_{10}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $z_{1} \, \mathbf{a}_{3}$ | = | $c z_{1} \,\mathbf{\hat{z}}$ | (1a) | K I |
$\mathbf{B_{2}}$ | = | $z_{2} \, \mathbf{a}_{3}$ | = | $c z_{2} \,\mathbf{\hat{z}}$ | (1a) | O I |
$\mathbf{B_{3}}$ | = | $z_{3} \, \mathbf{a}_{3}$ | = | $c z_{3} \,\mathbf{\hat{z}}$ | (1a) | S I |
$\mathbf{B_{4}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (1b) | Na I |
$\mathbf{B_{5}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (1b) | O II |
$\mathbf{B_{6}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (1b) | S II |
$\mathbf{B_{7}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (1c) | K II |
$\mathbf{B_{8}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ | (1c) | Na II |
$\mathbf{B_{9}}$ | = | $x_{9} \, \mathbf{a}_{1}- x_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{9} \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ | (3d) | O III |
$\mathbf{B_{10}}$ | = | $x_{9} \, \mathbf{a}_{1}+2 x_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{9} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{9} \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ | (3d) | O III |
$\mathbf{B_{11}}$ | = | $- 2 x_{9} \, \mathbf{a}_{1}- x_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{9} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{9} \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ | (3d) | O III |
$\mathbf{B_{12}}$ | = | $x_{10} \, \mathbf{a}_{1}- x_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{10} \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ | (3d) | O IV |
$\mathbf{B_{13}}$ | = | $x_{10} \, \mathbf{a}_{1}+2 x_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{10} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{10} \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ | (3d) | O IV |
$\mathbf{B_{14}}$ | = | $- 2 x_{10} \, \mathbf{a}_{1}- x_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{10} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{10} \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ | (3d) | O IV |