Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B_hP18_190_gh_bf-001

This structure originally had the label A2B_hP18_190_gh_bf. Calls to that address will be redirected here.

If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)

Links to this page

https://aflow.org/p/DA98
or https://aflow.org/p/A2B_hP18_190_gh_bf-001
or PDF Version

Li$_{2}$Sb Structure: A2B_hP18_190_gh_bf-001

Picture of Structure; Click for Big Picture
Prototype Li$_{2}$Sb
AFLOW prototype label A2B_hP18_190_gh_bf-001
ICSD 100020
Pearson symbol hP18
Space group number 190
Space group symbol $P\overline{6}2c$
AFLOW prototype command aflow --proto=A2B_hP18_190_gh_bf-001
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak y_{4}$

\[ \begin{array}{ccc} \mathbf{a_{1}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{2}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{3}}&=&c \,\mathbf{\hat{z}} \end{array}\]

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}c \,\mathbf{\hat{z}}$ (2b) Sb I
$\mathbf{B_{2}}$ = $\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}c \,\mathbf{\hat{z}}$ (2b) Sb I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4f) Sb II
$\mathbf{B_{4}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Sb II
$\mathbf{B_{5}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (4f) Sb II
$\mathbf{B_{6}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Sb II
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ (6g) Li I
$\mathbf{B_{8}}$ = $x_{3} \, \mathbf{a}_{2}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ (6g) Li I
$\mathbf{B_{9}}$ = $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}$ = $- a x_{3} \,\mathbf{\hat{x}}$ (6g) Li I
$\mathbf{B_{10}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) Li I
$\mathbf{B_{11}}$ = $x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) Li I
$\mathbf{B_{12}}$ = $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) Li I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Li II
$\mathbf{B_{14}}$ = $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Li II
$\mathbf{B_{15}}$ = $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Li II
$\mathbf{B_{16}}$ = $y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Li II
$\mathbf{B_{17}}$ = $\left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Li II
$\mathbf{B_{18}}$ = $- x_{4} \, \mathbf{a}_{1}- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Li II

References

  • W. Müller, Darstellung und Struktur der Phase Li$_{2}$Sb, Z. Naturforsch. B 32, 357–359 (1977).

Prototype Generator

aflow --proto=A2B_hP18_190_gh_bf --params=$a,c/a,z_{2},x_{3},x_{4},y_{4}$

Species:

Running:

Output: