Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B3_hP30_170_2a_3a-001

This structure originally had the label A2B3_hP30_170_2a_3a. 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/G6GN
or https://aflow.org/p/A2B3_hP30_170_2a_3a-001
or PDF Version

Al$_{2}$S$_{3}$ Structure: A2B3_hP30_170_2a_3a-001

Picture of Structure; Click for Big Picture
Prototype Al$_{2}$S$_{3}$
AFLOW prototype label A2B3_hP30_170_2a_3a-001
ICSD 300213
Pearson symbol hP30
Space group number 170
Space group symbol $P6_5$
AFLOW prototype command aflow --proto=A2B3_hP30_170_2a_3a-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak y_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak y_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$

Other compounds with this structure

In$_{2}$S$_{3}$


  • This structure is the enantiomorph of the Al$_{2}$S$_{3}$ (A2B3_hP30_169_2a_3a) structure. The ICSD information is for that space group.
  • Space group $P6_{5}$ #170 allows an arbitary choice for the origin of the $z$-axis. We set this by taking $z_{4} = 0$ for the S-I site.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+y_{1} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{1} + y_{1}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{1} - y_{1}\right) \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (6a) Al I
$\mathbf{B_{2}}$ = $- y_{1} \, \mathbf{a}_{1}+\left(x_{1} - y_{1}\right) \, \mathbf{a}_{2}+\left(z_{1} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{1} - 2 y_{1}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{1} + 2\right) \,\mathbf{\hat{z}}$ (6a) Al I
$\mathbf{B_{3}}$ = $- \left(x_{1} - y_{1}\right) \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{1} - y_{1}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{1} \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (6a) Al I
$\mathbf{B_{4}}$ = $- x_{1} \, \mathbf{a}_{1}- y_{1} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{1} + y_{1}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{1} - y_{1}\right) \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6a) Al I
$\mathbf{B_{5}}$ = $y_{1} \, \mathbf{a}_{1}- \left(x_{1} - y_{1}\right) \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{1} + 2 y_{1}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{6}\right) \,\mathbf{\hat{z}}$ (6a) Al I
$\mathbf{B_{6}}$ = $\left(x_{1} - y_{1}\right) \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}+\left(z_{1} + \frac{5}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{1} - y_{1}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{1} \,\mathbf{\hat{y}}+\frac{1}{6}c \left(6 z_{1} + 5\right) \,\mathbf{\hat{z}}$ (6a) Al I
$\mathbf{B_{7}}$ = $x_{2} \, \mathbf{a}_{1}+y_{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{2} + y_{2}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{2} - y_{2}\right) \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (6a) Al II
$\mathbf{B_{8}}$ = $- y_{2} \, \mathbf{a}_{1}+\left(x_{2} - y_{2}\right) \, \mathbf{a}_{2}+\left(z_{2} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{2} - 2 y_{2}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{2} + 2\right) \,\mathbf{\hat{z}}$ (6a) Al II
$\mathbf{B_{9}}$ = $- \left(x_{2} - y_{2}\right) \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{2} - y_{2}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{2} \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (6a) Al II
$\mathbf{B_{10}}$ = $- x_{2} \, \mathbf{a}_{1}- y_{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{2} + y_{2}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{2} - y_{2}\right) \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6a) Al II
$\mathbf{B_{11}}$ = $y_{2} \, \mathbf{a}_{1}- \left(x_{2} - y_{2}\right) \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{2} + 2 y_{2}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{6}\right) \,\mathbf{\hat{z}}$ (6a) Al II
$\mathbf{B_{12}}$ = $\left(x_{2} - y_{2}\right) \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{5}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{2} - y_{2}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{2} \,\mathbf{\hat{y}}+\frac{1}{6}c \left(6 z_{2} + 5\right) \,\mathbf{\hat{z}}$ (6a) Al II
$\mathbf{B_{13}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (6a) S I
$\mathbf{B_{14}}$ = $- y_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} - 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{3} + 2\right) \,\mathbf{\hat{z}}$ (6a) S I
$\mathbf{B_{15}}$ = $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (6a) S I
$\mathbf{B_{16}}$ = $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6a) S I
$\mathbf{B_{17}}$ = $y_{3} \, \mathbf{a}_{1}- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{3} + 2 y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{6}\right) \,\mathbf{\hat{z}}$ (6a) S I
$\mathbf{B_{18}}$ = $\left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{5}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{1}{6}c \left(6 z_{3} + 5\right) \,\mathbf{\hat{z}}$ (6a) S I
$\mathbf{B_{19}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (6a) S II
$\mathbf{B_{20}}$ = $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{4} + 2\right) \,\mathbf{\hat{z}}$ (6a) S II
$\mathbf{B_{21}}$ = $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (6a) S II
$\mathbf{B_{22}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6a) S II
$\mathbf{B_{23}}$ = $y_{4} \, \mathbf{a}_{1}- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{4} + 2 y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{6}\right) \,\mathbf{\hat{z}}$ (6a) S II
$\mathbf{B_{24}}$ = $\left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{5}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+\frac{1}{6}c \left(6 z_{4} + 5\right) \,\mathbf{\hat{z}}$ (6a) S II
$\mathbf{B_{25}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6a) S III
$\mathbf{B_{26}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{5} + 2\right) \,\mathbf{\hat{z}}$ (6a) S III
$\mathbf{B_{27}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (6a) S III
$\mathbf{B_{28}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6a) S III
$\mathbf{B_{29}}$ = $y_{5} \, \mathbf{a}_{1}- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{6}\right) \,\mathbf{\hat{z}}$ (6a) S III
$\mathbf{B_{30}}$ = $\left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{5}{6}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+\frac{1}{6}c \left(6 z_{5} + 5\right) \,\mathbf{\hat{z}}$ (6a) S III

References

Found in

  • P. Villars and K. Cenzual, Pearson's Crystal Data – Crystal Structure Database for Inorganic Compounds (2013). ASM International.

Prototype Generator

aflow --proto=A2B3_hP30_170_2a_3a --params=$a,c/a,x_{1},y_{1},z_{1},x_{2},y_{2},z_{2},x_{3},y_{3},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5}$

Species:

Running:

Output: