Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB2C2_hP15_154_a_ab_c-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/3CT8
or https://aflow.org/p/AB2C2_hP15_154_a_ab_c-001
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EuIr$_{2}$P$_{2}$ Structure: AB2C2_hP15_154_a_ab_c-001

Picture of Structure; Click for Big Picture
Prototype EuIr$_{2}$P$_{2}$
AFLOW prototype label AB2C2_hP15_154_a_ab_c-001
ICSD 73530
Pearson symbol hP15
Space group number 154
Space group symbol $P3_221$
AFLOW prototype command aflow --proto=AB2C2_hP15_154_a_ab_c-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$

Other compounds with this structure

CaIr$_{2}$P$_{2}$,  SrIr$_{2}$P$_{2}$


  • We shifted the origin used by (Lux, 1993) so that the europium atoms are on a (3a) Wyckoff position.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{1} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ (3a) Eu I
$\mathbf{B_{2}}$ = $x_{1} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{1} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{1} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ (3a) Eu I
$\mathbf{B_{3}}$ = $- x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}$ = $- a x_{1} \,\mathbf{\hat{x}}$ (3a) Eu I
$\mathbf{B_{4}}$ = $x_{2} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{2} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{2}{3}c \,\mathbf{\hat{z}}$ (3a) Ir I
$\mathbf{B_{5}}$ = $x_{2} \, \mathbf{a}_{2}+\frac{1}{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{2} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{2} \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ (3a) Ir I
$\mathbf{B_{6}}$ = $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}$ = $- a x_{2} \,\mathbf{\hat{x}}$ (3a) Ir I
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (3b) Ir II
$\mathbf{B_{8}}$ = $x_{3} \, \mathbf{a}_{2}+\frac{5}{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{5}{6}c \,\mathbf{\hat{z}}$ (3b) Ir II
$\mathbf{B_{9}}$ = $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (3b) Ir II
$\mathbf{B_{10}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (6c) P I
$\mathbf{B_{11}}$ = $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{3}c \left(3 z_{4} + 2\right) \,\mathbf{\hat{z}}$ (6c) P I
$\mathbf{B_{12}}$ = $- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (6c) P I
$\mathbf{B_{13}}$ = $y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} + y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{4} - y_{4}\right) \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (6c) P I
$\mathbf{B_{14}}$ = $\left(x_{4} - y_{4}\right) \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{3}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{4} - 2 y_{4}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{3}\right) \,\mathbf{\hat{z}}$ (6c) P I
$\mathbf{B_{15}}$ = $- x_{4} \, \mathbf{a}_{1}- \left(x_{4} - y_{4}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{2}{3}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{4} - y_{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{4} \,\mathbf{\hat{y}}- \frac{1}{3}c \left(3 z_{4} - 2\right) \,\mathbf{\hat{z}}$ (6c) P I

References

  • C. Lux, A. Mewis, S. Junk, A. Gruetz, and G. Michels, Kristallstrukturen und Eigenschaften neuer ternärer iridiumphosphide, J. Alloys Compd. 200, 135–139 (1993), doi:10.1016/0925-8388(93)90483-4.

Found in


Prototype Generator

aflow --proto=AB2C2_hP15_154_a_ab_c --params=$a,c/a,x_{1},x_{2},x_{3},x_{4},y_{4},z_{4}$

Species:

Running:

Output: