AFLOW Prototype: A12B_hP13_191_cdei_a-001
This structure originally had the label A12B_hP13_191_cdei_a. 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/FEUM
or
https://aflow.org/p/A12B_hP13_191_cdei_a-001
or
PDF Version
Prototype | Be$_{12}$Ti |
AFLOW prototype label | A12B_hP13_191_cdei_a-001 |
Strukturbericht designation | $D2_{a}$ |
ICSD | 58745 |
Pearson symbol | hP13 |
Space group number | 191 |
Space group symbol | $P6/mmm$ |
AFLOW prototype command |
aflow --proto=A12B_hP13_191_cdei_a-001
--params=$a, \allowbreak c/a, \allowbreak z_{4}, \allowbreak z_{5}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Ti I |
$\mathbf{B_{2}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}$ | (2c) | Be I |
$\mathbf{B_{3}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}$ | (2c) | Be I |
$\mathbf{B_{4}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2d) | Be II |
$\mathbf{B_{5}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2d) | Be II |
$\mathbf{B_{6}}$ | = | $z_{4} \, \mathbf{a}_{3}$ | = | $c z_{4} \,\mathbf{\hat{z}}$ | (2e) | Be III |
$\mathbf{B_{7}}$ | = | $- z_{4} \, \mathbf{a}_{3}$ | = | $- c z_{4} \,\mathbf{\hat{z}}$ | (2e) | Be III |
$\mathbf{B_{8}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6i) | Be IV |
$\mathbf{B_{9}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6i) | Be IV |
$\mathbf{B_{10}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+c z_{5} \,\mathbf{\hat{z}}$ | (6i) | Be IV |
$\mathbf{B_{11}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6i) | Be IV |
$\mathbf{B_{12}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6i) | Be IV |
$\mathbf{B_{13}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- c z_{5} \,\mathbf{\hat{z}}$ | (6i) | Be IV |