Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3B2C2_tI28_122_ad_c_d-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/1YSS
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Mn$_{1.4}$PtSn Structure: A3B2C2_tI28_122_ad_c_d-001

Picture of Structure; Click for Big Picture
Prototype Mn$_{1.4}$PtSn
AFLOW prototype label A3B2C2_tI28_122_ad_c_d-001
ICSD 11061
Pearson symbol tI28
Space group number 122
Space group symbol $I\overline{4}2d$
AFLOW prototype command aflow --proto=A3B2C2_tI28_122_ad_c_d-001
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}$

  • (Vir, 2019) found that the Mn (4a) site (in the layer with the platinum (8c) atoms) has 20% vacancies, giving the noted departure from the nominal Mn$_{3}$Pt$_{2}$Sn$_{2}$ stoichiometry.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (4a) Mn I
$\mathbf{B_{2}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (4a) Mn I
$\mathbf{B_{3}}$ = $z_{2} \, \mathbf{a}_{1}+z_{2} \, \mathbf{a}_{2}$ = $c z_{2} \,\mathbf{\hat{z}}$ (8c) Pt I
$\mathbf{B_{4}}$ = $- z_{2} \, \mathbf{a}_{1}- z_{2} \, \mathbf{a}_{2}$ = $- c z_{2} \,\mathbf{\hat{z}}$ (8c) Pt I
$\mathbf{B_{5}}$ = $- \left(z_{2} - \frac{3}{4}\right) \, \mathbf{a}_{1}- \left(z_{2} - \frac{1}{4}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (8c) Pt I
$\mathbf{B_{6}}$ = $\left(z_{2} + \frac{3}{4}\right) \, \mathbf{a}_{1}+\left(z_{2} + \frac{1}{4}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (8c) Pt I
$\mathbf{B_{7}}$ = $\frac{3}{8} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{8}\right) \, \mathbf{a}_{2}+\left(x_{3} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ (8d) Mn II
$\mathbf{B_{8}}$ = $\frac{7}{8} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{8}\right) \, \mathbf{a}_{2}- \left(x_{3} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ (8d) Mn II
$\mathbf{B_{9}}$ = $- \left(x_{3} - \frac{7}{8}\right) \, \mathbf{a}_{1}+\frac{1}{8} \, \mathbf{a}_{2}- \left(x_{3} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{4}a \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ (8d) Mn II
$\mathbf{B_{10}}$ = $\left(x_{3} + \frac{7}{8}\right) \, \mathbf{a}_{1}+\frac{5}{8} \, \mathbf{a}_{2}+\left(x_{3} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ (8d) Mn II
$\mathbf{B_{11}}$ = $\frac{3}{8} \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{8}\right) \, \mathbf{a}_{2}+\left(x_{4} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ (8d) Sn I
$\mathbf{B_{12}}$ = $\frac{7}{8} \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{8}\right) \, \mathbf{a}_{2}- \left(x_{4} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ (8d) Sn I
$\mathbf{B_{13}}$ = $- \left(x_{4} - \frac{7}{8}\right) \, \mathbf{a}_{1}+\frac{1}{8} \, \mathbf{a}_{2}- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{4}a \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ (8d) Sn I
$\mathbf{B_{14}}$ = $\left(x_{4} + \frac{7}{8}\right) \, \mathbf{a}_{1}+\frac{5}{8} \, \mathbf{a}_{2}+\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ (8d) Sn I

References

  • P. Vir, N. Kumar, H. Borrmann, B. Jamijansuren, G. Kreiner, C. Shekhar, and C. Felser, Tetragonal Superstructure of the Antiskyrmion Hosting Heusler Compound Mn$_{1.4}$PtSn, Chem. Mater. 31, 5876–5880 (2019), doi:10.1021/acs.chemmater.9b02013.

Prototype Generator

aflow --proto=A3B2C2_tI28_122_ad_c_d --params=$a,c/a,z_{2},x_{3},x_{4}$

Species:

Running:

Output: