Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3B4_hP14_176_h_ch-002

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/FUM3
or https://aflow.org/p/A3B4_hP14_176_h_ch-002
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Nb$_{3}$Te$_{4}$ Structure: A3B4_hP14_176_h_ch-002

Picture of Structure; Click for Big Picture
Prototype Nb$_{3}$Te$_{4}$
AFLOW prototype label A3B4_hP14_176_h_ch-002
ICSD 16606
Pearson symbol hP14
Space group number 176
Space group symbol $P6_3/m$
AFLOW prototype command aflow --proto=A3B4_hP14_176_h_ch-002
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak y_{2}, \allowbreak x_{3}, \allowbreak y_{3}$

Other compounds with this structure

Nb$_{3}$S$_{4}$,  Nb$_{3}$Se$_{4}$


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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (2c) Te I
$\mathbf{B_{2}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (2c) Te I
$\mathbf{B_{3}}$ = $x_{2} \, \mathbf{a}_{1}+y_{2} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{2} + y_{2}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{2} - y_{2}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Nb I
$\mathbf{B_{4}}$ = $- y_{2} \, \mathbf{a}_{1}+\left(x_{2} - y_{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{2} - 2 y_{2}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Nb I
$\mathbf{B_{5}}$ = $- \left(x_{2} - y_{2}\right) \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{2} - y_{2}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{2} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Nb I
$\mathbf{B_{6}}$ = $- x_{2} \, \mathbf{a}_{1}- y_{2} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{2} + y_{2}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{2} - y_{2}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Nb I
$\mathbf{B_{7}}$ = $y_{2} \, \mathbf{a}_{1}- \left(x_{2} - y_{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{2} + 2 y_{2}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Nb I
$\mathbf{B_{8}}$ = $\left(x_{2} - y_{2}\right) \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{2} - y_{2}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{2} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Nb I
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Te II
$\mathbf{B_{10}}$ = $- y_{3} \, \mathbf{a}_{1}+\left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{3} - 2 y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Te II
$\mathbf{B_{11}}$ = $- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Te II
$\mathbf{B_{12}}$ = $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{3} + y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{3} - y_{3}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Te II
$\mathbf{B_{13}}$ = $y_{3} \, \mathbf{a}_{1}- \left(x_{3} - y_{3}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{3} + 2 y_{3}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Te II
$\mathbf{B_{14}}$ = $\left(x_{3} - y_{3}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{3} - y_{3}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Te II

References

  • K. Selte and A. Kjekshus, The crystal structures of Nb$_{3}$Se$_{4}$ and Nb$_{3}$Te$_{4}$, Acta Cryst. 17, 1568–1572 (1964), doi:10.1107/S0365110X64003875.

Found in

  • K. Suzuki, M. Ichihara, I. Nakada, and Y. Ishihara, Superlattice structure of Nb$_{3}$Te$_{4}$ at low temperatures, Solid State Commun. 52, 743–746 (1984), doi:10.1016/0038-1098(84)90402-2.

Prototype Generator

aflow --proto=A3B4_hP14_176_h_ch --params=$a,c/a,x_{2},y_{2},x_{3},y_{3}$

Species:

Running:

Output: