AFLOW Prototype: A5B3C_hP18_186_2a3b_2ab_b-001
This structure originally had the label A5B3C_hP18_186_2a3b_2ab_b. 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/KHLK
or
https://aflow.org/p/A5B3C_hP18_186_2a3b_2ab_b-001
or
PDF Version
Prototype | Al$_{5}$C$_{3}$N |
AFLOW prototype label | A5B3C_hP18_186_2a3b_2ab_b-001 |
Strukturbericht designation | $E9_{4}$ |
ICSD | 14398 |
Pearson symbol | hP18 |
Space group number | 186 |
Space group symbol | $P6_3mc$ |
AFLOW prototype command |
aflow --proto=A5B3C_hP18_186_2a3b_2ab_b-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak z_{6}, \allowbreak z_{7}, \allowbreak z_{8}, \allowbreak z_{9}$ |
U$_{2}$Al$_{3}$C$_{4}$
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) | Al II |
$\mathbf{B_{4}}$ | = | $\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | Al II |
$\mathbf{B_{5}}$ | = | $z_{3} \, \mathbf{a}_{3}$ | = | $c z_{3} \,\mathbf{\hat{z}}$ | (2a) | C I |
$\mathbf{B_{6}}$ | = | $\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | C I |
$\mathbf{B_{7}}$ | = | $z_{4} \, \mathbf{a}_{3}$ | = | $c z_{4} \,\mathbf{\hat{z}}$ | (2a) | C II |
$\mathbf{B_{8}}$ | = | $\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | C II |
$\mathbf{B_{9}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (2b) | Al III |
$\mathbf{B_{10}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Al III |
$\mathbf{B_{11}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (2b) | Al IV |
$\mathbf{B_{12}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Al IV |
$\mathbf{B_{13}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (2b) | Al V |
$\mathbf{B_{14}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Al V |
$\mathbf{B_{15}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ | (2b) | C III |
$\mathbf{B_{16}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | C III |
$\mathbf{B_{17}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ | (2b) | N I |
$\mathbf{B_{18}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | N I |