AFLOW Prototype: A2B4C5_hP22_186_ab_4b_a4b-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/7JMK
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https://aflow.org/p/A2B4C5_hP22_186_ab_4b_a4b-001
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PDF Version
Prototype | Al$_{2}$N$_{2}$Ti$_{2}$ |
AFLOW prototype label | A2B4C5_hP22_186_ab_4b_a4b-001 |
ICSD | 52643 |
Pearson symbol | hP22 |
Space group number | 186 |
Space group symbol | $P6_3mc$ |
AFLOW prototype command |
aflow --proto=A2B4C5_hP22_186_ab_4b_a4b-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak z_{6}, \allowbreak z_{7}, \allowbreak z_{8}, \allowbreak z_{9}, \allowbreak z_{10}, \allowbreak z_{11}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $z_{1} \, \mathbf{a}_{3}$ | = | $c z_{1} \,\mathbf{\hat{z}}$ | (2a) | Al I |
$\mathbf{B_{2}}$ | = | $\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | Al I |
$\mathbf{B_{3}}$ | = | $z_{2} \, \mathbf{a}_{3}$ | = | $c z_{2} \,\mathbf{\hat{z}}$ | (2a) | Ti I |
$\mathbf{B_{4}}$ | = | $\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | Ti I |
$\mathbf{B_{5}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (2b) | Al II |
$\mathbf{B_{6}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Al II |
$\mathbf{B_{7}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (2b) | Ni I |
$\mathbf{B_{8}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ni I |
$\mathbf{B_{9}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (2b) | Ni II |
$\mathbf{B_{10}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ni II |
$\mathbf{B_{11}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (2b) | Ni III |
$\mathbf{B_{12}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ni III |
$\mathbf{B_{13}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (2b) | Ni IV |
$\mathbf{B_{14}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ni IV |
$\mathbf{B_{15}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ | (2b) | Ti II |
$\mathbf{B_{16}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ti II |
$\mathbf{B_{17}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ | (2b) | Ti III |
$\mathbf{B_{18}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ti III |
$\mathbf{B_{19}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ | (2b) | Ti IV |
$\mathbf{B_{20}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{10} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{10} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ti IV |
$\mathbf{B_{21}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{11} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{11} \,\mathbf{\hat{z}}$ | (2b) | Ti V |
$\mathbf{B_{22}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{11} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{11} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ti V |