AFLOW Prototype: A3B_hP8_194_af_c-001
This structure originally had the label A3B_hP8_194_af_c. 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/72PY
or
https://aflow.org/p/A3B_hP8_194_af_c-001
or
PDF Version
Prototype | B$_{3}$Re |
AFLOW prototype label | A3B_hP8_194_af_c-001 |
ICSD | 24361 |
Pearson symbol | hP8 |
Space group number | 194 |
Space group symbol | $P6_3/mmc$ |
AFLOW prototype command |
aflow --proto=A3B_hP8_194_af_c-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}$ |
TcB${3}$, CaNi$_{2}$Si, GdPt$_{2}$Sn
falsely assigned,and that the true structure is ReB$_{2}$, which is identical to this structure with the (2a) boron atoms removed.
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | B I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2a) | B I |
$\mathbf{B_{3}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2c) | Re I |
$\mathbf{B_{4}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2c) | Re 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) | B 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) | B 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) | B 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) | B II |