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
or
PDF Version
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=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}$ |
CaIr$_{2}$P$_{2}$, SrIr$_{2}$P$_{2}$
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 |