AFLOW Prototype: AB7_hR16_166_c_c2h-001
This structure originally had the label AB7_hR16_166_c_c2h. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M.J. Mehl, M. Esters, C. Oses, O. Levy, G.L.W. Hart, C. Toher, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 3, Comp. Mat. Sci. 199, 110450 (2021). (doi=10.1016/j.commatsci.2021.110450)
Links to this page
https://aflow.org/p/56NK
or
https://aflow.org/p/AB7_hR16_166_c_c2h-001
or
PDF Version
Prototype | TaTi$_{7}$ |
AFLOW prototype label | AB7_hR16_166_c_c2h-001 |
ICSD | none |
Pearson symbol | hR16 |
Space group number | 166 |
Space group symbol | $R\overline{3}m$ |
AFLOW prototype command |
aflow --proto=AB7_hR16_166_c_c2h-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}+x_{1} \, \mathbf{a}_{3}$ | = | $c x_{1} \,\mathbf{\hat{z}}$ | (2c) | Ta I |
$\mathbf{B_{2}}$ | = | $- x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}- x_{1} \, \mathbf{a}_{3}$ | = | $- c x_{1} \,\mathbf{\hat{z}}$ | (2c) | Ta I |
$\mathbf{B_{3}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ | = | $c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Ti I |
$\mathbf{B_{4}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ | = | $- c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Ti I |
$\mathbf{B_{5}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{6}}$ | = | $z_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{7}}$ | = | $x_{3} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ | = | $- \frac{1}{\sqrt{3}}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{8}}$ | = | $- z_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{9}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{10}}$ | = | $- x_{3} \, \mathbf{a}_{1}- z_{3} \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{\sqrt{3}}a \left(x_{3} - z_{3}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{3} + z_{3}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti II |
$\mathbf{B_{11}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti III |
$\mathbf{B_{12}}$ | = | $z_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti III |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{\sqrt{3}}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti III |
$\mathbf{B_{14}}$ | = | $- z_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti III |
$\mathbf{B_{15}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti III |
$\mathbf{B_{16}}$ | = | $- x_{4} \, \mathbf{a}_{1}- z_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{\sqrt{3}}a \left(x_{4} - z_{4}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{4} + z_{4}\right) \,\mathbf{\hat{z}}$ | (6h) | Ti III |