AFLOW Prototype: AB3C_hP30_194_bf_hk_af-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/VHHA
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https://aflow.org/p/AB3C_hP30_194_bf_hk_af-001
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PDF Version
Prototype | BaO$_{3}$Ti |
AFLOW prototype label | AB3C_hP30_194_bf_hk_af-001 |
ICSD | 75240 |
Pearson symbol | hP30 |
Space group number | 194 |
Space group symbol | $P6_3/mmc$ |
AFLOW prototype command |
aflow --proto=AB3C_hP30_194_bf_hk_af-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak x_{6}, \allowbreak z_{6}$ |
Ba(Co$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Cr$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Fe$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Ir$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Mn$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Os$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Pd$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Pt$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Rh$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(Ru$_{x}$Ti$_{1-x}$)O$_{3}$, Ba(V$_{x}$Ti$_{1-x}$)O$_{3}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | Ti I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2a) | Ti I |
$\mathbf{B_{3}}$ | = | $\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2b) | Ba I |
$\mathbf{B_{4}}$ | = | $\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2b) | Ba 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) | Ba II |
$\mathbf{B_{6}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Ba II |
$\mathbf{B_{7}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (4f) | Ba II |
$\mathbf{B_{8}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Ba II |
$\mathbf{B_{9}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4f) | Ti II |
$\mathbf{B_{10}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Ti II |
$\mathbf{B_{11}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (4f) | Ti II |
$\mathbf{B_{12}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Ti II |
$\mathbf{B_{13}}$ | = | $x_{5} \, \mathbf{a}_{1}+2 x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | O I |
$\mathbf{B_{14}}$ | = | $- 2 x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | O I |
$\mathbf{B_{15}}$ | = | $x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | O I |
$\mathbf{B_{16}}$ | = | $- x_{5} \, \mathbf{a}_{1}- 2 x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | O I |
$\mathbf{B_{17}}$ | = | $2 x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | O I |
$\mathbf{B_{18}}$ | = | $- x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | O I |
$\mathbf{B_{19}}$ | = | $x_{6} \, \mathbf{a}_{1}+2 x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{20}}$ | = | $- 2 x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{21}}$ | = | $x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{22}}$ | = | $- x_{6} \, \mathbf{a}_{1}- 2 x_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{23}}$ | = | $2 x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{24}}$ | = | $- x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{25}}$ | = | $2 x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{26}}$ | = | $- x_{6} \, \mathbf{a}_{1}- 2 x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{27}}$ | = | $- x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{28}}$ | = | $- 2 x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{29}}$ | = | $x_{6} \, \mathbf{a}_{1}+2 x_{6} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12k) | O II |
$\mathbf{B_{30}}$ | = | $x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{6} \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (12k) | O II |