AFLOW Prototype: A3B9CD2_hP15_164_ad_ei_b_d-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/4LGS
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https://aflow.org/p/A3B9CD2_hP15_164_ad_ei_b_d-001
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PDF Version
Prototype | Ba$_{3}$O$_{9}$SrTa$_{2}$ |
AFLOW prototype label | A3B9CD2_hP15_164_ad_ei_b_d-001 |
ICSD | 27496 |
Pearson symbol | hP15 |
Space group number | 164 |
Space group symbol | $P\overline{3}m1$ |
AFLOW prototype command |
aflow --proto=A3B9CD2_hP15_164_ad_ei_b_d-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak x_{6}, \allowbreak z_{6}$ |
Ba$_{3}$CoNb$_{2}$O$_{9}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Ba I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (1b) | Sr I |
$\mathbf{B_{3}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (2d) | Ba II |
$\mathbf{B_{4}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (2d) | Ba II |
$\mathbf{B_{5}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (2d) | Ta I |
$\mathbf{B_{6}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (2d) | Ta I |
$\mathbf{B_{7}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ | (3e) | O I |
$\mathbf{B_{8}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ | (3e) | O I |
$\mathbf{B_{9}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}$ | (3e) | O I |
$\mathbf{B_{10}}$ | = | $x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6i) | O II |
$\mathbf{B_{11}}$ | = | $x_{6} \, \mathbf{a}_{1}+2 x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6i) | O II |
$\mathbf{B_{12}}$ | = | $- 2 x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6i) | O II |
$\mathbf{B_{13}}$ | = | $- x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6i) | O II |
$\mathbf{B_{14}}$ | = | $2 x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6i) | O II |
$\mathbf{B_{15}}$ | = | $- x_{6} \, \mathbf{a}_{1}- 2 x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6i) | O II |