Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3B2_oP20_52_de_cd-001

This structure originally had the label A3B2_oP20_52_de_cd. 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/12L3
or https://aflow.org/p/A3B2_oP20_52_de_cd-001
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Sr$_{2}$Bi$_{3}$ Structure: A3B2_oP20_52_de_cd-001

Picture of Structure; Click for Big Picture
Prototype Bi$_{3}$Sr$_{2}$
AFLOW prototype label A3B2_oP20_52_de_cd-001
ICSD 164987
Pearson symbol oP20
Space group number 52
Space group symbol $Pnna$
AFLOW prototype command aflow --proto=A3B2_oP20_52_de_cd-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+z_{1} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+c z_{1} \,\mathbf{\hat{z}}$ (4c) Sr I
$\mathbf{B_{2}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{2}b \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Sr I
$\mathbf{B_{3}}$ = $\frac{3}{4} \, \mathbf{a}_{1}- z_{1} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}- c z_{1} \,\mathbf{\hat{z}}$ (4c) Sr I
$\mathbf{B_{4}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{1}{2}b \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Sr I
$\mathbf{B_{5}}$ = $x_{2} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $a x_{2} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (4d) Bi I
$\mathbf{B_{6}}$ = $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (4d) Bi I
$\mathbf{B_{7}}$ = $- x_{2} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- a x_{2} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (4d) Bi I
$\mathbf{B_{8}}$ = $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (4d) Bi I
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (4d) Sr II
$\mathbf{B_{10}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (4d) Sr II
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (4d) Sr II
$\mathbf{B_{12}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (4d) Sr II
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8e) Bi II
$\mathbf{B_{14}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8e) Bi II
$\mathbf{B_{15}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8e) Bi II
$\mathbf{B_{16}}$ = $x_{4} \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}- b \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8e) Bi II
$\mathbf{B_{17}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (8e) Bi II
$\mathbf{B_{18}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (8e) Bi II
$\mathbf{B_{19}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8e) Bi II
$\mathbf{B_{20}}$ = $- x_{4} \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+b \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8e) Bi II

References

  • F. Merlo and M. L. Fornasini, Crystal structure of some phases and alloying behaviour in alkaline earths, europium and ytterbium pnictides, Mater. Res. Bull. 29, 149–154 (1994), doi:10.1016/0025-5408(94)90135-X.

Found in

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

Prototype Generator

aflow --proto=A3B2_oP20_52_de_cd --params=$a,b/a,c/a,z_{1},x_{2},x_{3},x_{4},y_{4},z_{4}$

Species:

Running:

Output: