AFLOW Prototype: AB3C4_hP16_194_c_af_ef-001
This structure originally had the label AB3C4_hP16_194_c_af_ef. Calls to that address will be redirected here.
If you are using this page, please cite:
M. J. Mehl, D. Hicks, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 1, Comp. Mat. Sci. 136, S1-S828 (2017). (doi=10.1016/j.commatsci.2017.01.017)
Links to this page
https://aflow.org/p/53FU
or
https://aflow.org/p/AB3C4_hP16_194_c_af_ef-001
or
PDF Version
Prototype | AlN$_{3}$Ti$_{4}$ |
AFLOW prototype label | AB3C4_hP16_194_c_af_ef-001 |
ICSD | 190338 |
Pearson symbol | hP16 |
Space group number | 194 |
Space group symbol | $P6_3/mmc$ |
AFLOW prototype command |
aflow --proto=AB3C4_hP16_194_c_af_ef-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}$ |
CAl$_{3}$Ti$_{4}$, CsYb$_{3}$Se$_{4}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | N I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2a) | N I |
$\mathbf{B_{3}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2c) | Al I |
$\mathbf{B_{4}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2c) | Al I |
$\mathbf{B_{5}}$ | = | $z_{3} \, \mathbf{a}_{3}$ | = | $c z_{3} \,\mathbf{\hat{z}}$ | (4e) | Ti I |
$\mathbf{B_{6}}$ | = | $\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4e) | Ti I |
$\mathbf{B_{7}}$ | = | $- z_{3} \, \mathbf{a}_{3}$ | = | $- c z_{3} \,\mathbf{\hat{z}}$ | (4e) | Ti I |
$\mathbf{B_{8}}$ | = | $- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4e) | Ti I |
$\mathbf{B_{9}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4f) | N II |
$\mathbf{B_{10}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | N II |
$\mathbf{B_{11}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (4f) | N II |
$\mathbf{B_{12}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | N II |
$\mathbf{B_{13}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (4f) | Ti II |
$\mathbf{B_{14}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Ti II |
$\mathbf{B_{15}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (4f) | Ti II |
$\mathbf{B_{16}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Ti II |