Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A4B2C_tP28_135_gh_h_d-001

This structure originally had the label A4B2C_tP28_135_gh_h_d. 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/20MM
or https://aflow.org/p/A4B2C_tP28_135_gh_h_d-001
or PDF Version

ZnSb$_{2}$O$_{4}$ Structure: A4B2C_tP28_135_gh_h_d-001

Picture of Structure; Click for Big Picture
Prototype O$_{4}$Sb$_{2}$Zn
AFLOW prototype label A4B2C_tP28_135_gh_h_d-001
ICSD 647391
Pearson symbol tP28
Space group number 135
Space group symbol $P4_2/mbc$
AFLOW prototype command aflow --proto=A4B2C_tP28_135_gh_h_d-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak x_{4}, \allowbreak y_{4}$

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

Basis vectors

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

References

  • S. Ståhl, The crystal structure of ZnSb$_{2}$O$_{4}$ and isomorphous compounds, Ark. Kem. Mineral. Geol. 17B, 1–7 (1943).

Found in

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

Prototype Generator

aflow --proto=A4B2C_tP28_135_gh_h_d --params=$a,c/a,x_{2},x_{3},y_{3},x_{4},y_{4}$

Species:

Running:

Output: