Hexagonal Ti$_{6}$Sn$_{5}$ Structure: A5B6_hP22_194_ach_gh-001

Picture of Structure; Click for Big Picture
Prototype Sn$_{5}$Ti$_{6}$
AFLOW prototype label A5B6_hP22_194_ach_gh-001
ICSD 660295
CCDC 1768105
Pearson symbol hP22
Space group number 194
Space group symbol $P6_3/mmc$
AFLOW prototype command aflow --proto=A5B6_hP22_194_ach_gh-001
--params=$a, \allowbreak c/a, \allowbreak x_{4}, \allowbreak x_{5}$

Other compounds with this structure

V$_{6}$Ga$_{5}$


  • Hexagonal Ti$_{6}$Sn$_{5}$ coexists with orthorhombic Ti$_{6}$Sn$_{5}$, which takes on the Nb$_{6}$Sn$_{5}$ structure. (Oni, 2014)
  • (van Vucht, 1964) state that the Sn-I atom is on the (4b) Wyckoff position, but the coordinates given are for the (2a) position. This fits with their figure of the structure, although the atoms they label at height 1/4 should be at zero and vice versa.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (2a) Sn I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}c \,\mathbf{\hat{z}}$ (2a) Sn I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (2c) Sn II
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (2c) Sn II
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{1}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ (6g) Ti I
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ (6g) Ti I
$\mathbf{B_{7}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}$ (6g) Ti I
$\mathbf{B_{8}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) Ti I
$\mathbf{B_{9}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) Ti I
$\mathbf{B_{10}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) Ti I
$\mathbf{B_{11}}$ = $x_{4} \, \mathbf{a}_{1}+2 x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Sn III
$\mathbf{B_{12}}$ = $- 2 x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Sn III
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Sn III
$\mathbf{B_{14}}$ = $- x_{4} \, \mathbf{a}_{1}- 2 x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Sn III
$\mathbf{B_{15}}$ = $2 x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Sn III
$\mathbf{B_{16}}$ = $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Sn III
$\mathbf{B_{17}}$ = $x_{5} \, \mathbf{a}_{1}+2 x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Ti II
$\mathbf{B_{18}}$ = $- 2 x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Ti II
$\mathbf{B_{19}}$ = $x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{5} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Ti II
$\mathbf{B_{20}}$ = $- x_{5} \, \mathbf{a}_{1}- 2 x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Ti II
$\mathbf{B_{21}}$ = $2 x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Ti II
$\mathbf{B_{22}}$ = $- x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{5} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Ti II

References

  • J. H. N. van Vucht, H. A. C. M. Bruning, H. C. Donkersloot, and A. H. G. de Mesquita, The System Vanadium-Gallium, Phillips Res. Repts. 19, 407–421 (1964).
  • A. A. Oni, D. Hook, J. P. Maria, and J. M. LeBeau, Phase coexistence in Ti$_{6}$Sn$_{5}$ intermetallics, Intermetallics 51, 48–52 (2014), doi:10.1016/j.intermet.2014.03.002.

First cited in

  • 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, Comp. Mat. Sci. 240, 112988 (2024). (doi=10.1016/j.commatsci.2021.110450)

Geometry files


Prototype Generator

aflow --proto=A5B6_hP22_194_ach_gh --params=$a,c/a,x_{4},x_{5}$

Species:

Running:

Output: