Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: ABC2D6_mP20_13_e_e_g_3g-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/NDG5
or https://aflow.org/p/ABC2D6_mP20_13_e_e_g_3g-001
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AgCrP$_{2}$S$_{6}$ Structure: ABC2D6_mP20_13_e_e_g_3g-001

Picture of Structure; Click for Big Picture
Prototype AgCrP$_{2}$S$_{6}$
AFLOW prototype label ABC2D6_mP20_13_e_e_g_3g-001
ICSD 36456
Pearson symbol mP20
Space group number 13
Space group symbol $P2/c$
AFLOW prototype command aflow --proto=ABC2D6_mP20_13_e_e_g_3g-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak \beta, \allowbreak y_{1}, \allowbreak y_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$

  • (Selter, 2021) give the coordinates of this structure in the $P2/a$ setting of space group #13. We used FINDSYM to transform this to the standard $P2/c$ setting.
  • (Selter, 2021) does not yet have an ICSD entry. We list the entry from the earlier work of (Colombet, 1983).

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

Basis vectors

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

References

  • S. Selter, Y. Shemerliuk, B. Büchner, and S. Aswartham, Crystal Growth of the Quasi-2D Quarternary Compound AgCrP$_{2}$S$_{6}$ by Chemical Vapor Transport, Crystals 11, 500 (2021), doi:10.3390/cryst11050500.
  • P. Colombet, A. Leblanc, M. Danot, and J. Rouxel, Coordinance inhabituelle de l'argent dans un sufur lamellaire a sous-reseau magnetique 1D: le compose Ag$_{0.5}$Cr$_{0.5}$PS$_{3}$, Nouveau J. Chim. 7, 333–338 (1983).

Prototype Generator

aflow --proto=ABC2D6_mP20_13_e_e_g_3g --params=$a,b/a,c/a,\beta,y_{1},y_{2},x_{3},y_{3},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6}$

Species:

Running:

Output: