Y$_{3}$Au$_{3}$Sb$_{4}$ Structure: A3B4C3_cI40_220_a_c_b-001
| 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
- A. E. Dwight, Yttrium-gold-antimony Y$_{3}$Au$_{3}$Sb$_{4}$, Acta Crystallogr. Sect. B 33, 1579–1581 (1977), doi:10.1107/S0567740877006530.
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=$a,x_{3}$