Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B3_oC40_39_2d_2c2d-001

This structure originally had the label A2B3_oC40_39_2d_2c2d. 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/4YG0
or https://aflow.org/p/A2B3_oC40_39_2d_2c2d-001
or PDF Version

Ta$_{3}$S$_{2}$ Structure: A2B3_oC40_39_2d_2c2d-001

Picture of Structure; Click for Big Picture
Prototype S$_{2}$Ta$_{3}$
AFLOW prototype label A2B3_oC40_39_2d_2c2d-001
ICSD 71143
Pearson symbol oC40
Space group number 39
Space group symbol $Aem2$
AFLOW prototype command aflow --proto=A2B3_oC40_39_2d_2c2d-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$

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

Basis vectors

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

References

  • S. J. Kim, K. S. Nanjundaswamy, and T. Hughbanks, Single-crystal structure of tantalum sulfide (Ta$_{3}$S$_{2}$). Structure and bonding in the Ta$_{6}$S$_{n}$ (n = 1,3,4,5?) pentagonal-antiprismatic chain compounds, Inorg. Chem. 30, 159–164 (1991), doi:10.1021/ic00002a004.

Found in

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

Prototype Generator

aflow --proto=A2B3_oC40_39_2d_2c2d --params=$a,b/a,c/a,x_{1},z_{1},x_{2},z_{2},x_{3},y_{3},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6}$

Species:

Running:

Output: