Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3B4C3_cI40_220_a_c_b-001

If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.

Links to this page

https://aflow.org/p/GET1
or https://aflow.org/p/A3B4C3_cI40_220_a_c_b-001
or PDF Version

Y$_{3}$Au$_{3}$Sb$_{4}$ Structure: A3B4C3_cI40_220_a_c_b-001

Picture of Structure; Click for Big Picture
Prototype Au$_{3}$Sb$_{4}$Y$_{3}$
AFLOW prototype label A3B4C3_cI40_220_a_c_b-001
ICSD 957
Pearson symbol cI40
Space group number 220
Space group symbol $I\overline{4}3d$
AFLOW prototype command aflow --proto=A3B4C3_cI40_220_a_c_b-001
--params=$a, \allowbreak x_{3}$

Other compounds with this structure

Ce$_{3}$Pd$_{3}$Bi$_{4}$,  Ce$_{3}$Pt$_{3}$Bi$_{4}$,  Dy$_{3}$Au$_{3}$Sb$_{4}$,  Dy$_{3}$Cu$_{3}$Sb$_{4}$,  Er$_{3}$Au$_{3}$Sb$_{4}$,  Gd$_{3}$Au$_{3}$Sb$_{4}$,  Hf$_{3}$Ni$_{3}$Sb$_{4}$,  Ho$_{3}$Au$_{3}$Sb$_{4}$,  La$_{3}$Cu$_{3}$Bi$_{4}$,  Lu$_{3}$Au$_{3}$Sb$_{4}$,  Nd$_{3}$Au$_{3}$Sb$_{4}$,  Sm$_{3}$Au$_{3}$Sb$_{4}$,  Sm$_{3}$Cu$_{3}$Sb$_{4}$,  Tb$_{3}$Au$_{3}$Sb$_{4}$,  Tb$_{3}$Cu$_{3}$Sb$_{4}$,  Tm$_{3}$Au$_{3}$Sb$_{4}$,  U$_{3}$Ni$_{3}$Sb$_{4}$,  U$_{3}$NiAs$_{4}$


  • (Lu, 2008) describe this as a filled Th$_{3}$P$_{4}$ ($D7_{3}$) structure, and, like its parent, there can be considerable disorder on the transition metal site. This is seen in the U$_{3}$NiAs$_{4}$ compound, where 2/3 of the nickel (2b) sites are vacant.

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

Basis vectors

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

References

Found in

  • Y.-M. Lu, F. Fan, C.-B. Cai, S.-X. Cao, and J.-C. Zhang, Y$_{3}$Au$_{3}$Sb$_{4}$ type structure La$_{3}$Cu$_{3}$Bi$_{4}$: synthesis, structure and property, J. Shanghai Univ. (Engl. Ed.) 12, 486–488 (2008), doi:10.1007/s 11741-008-0604-2.

Prototype Generator

aflow --proto=A3B4C3_cI40_220_a_c_b --params=$a,x_{3}$

Species:

Running:

Output: