AFLOW Prototype: ABC2_tI16_141_a_b_e-003
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/XJE5
or
https://aflow.org/p/ABC2_tI16_141_a_b_e-003
or
PDF Version
Prototype | FeLiO$_{2}$ |
AFLOW prototype label | ABC2_tI16_141_a_b_e-003 |
ICSD | 174085 |
Pearson symbol | tI16 |
Space group number | 141 |
Space group symbol | $I4_1/amd$ |
AFLOW prototype command |
aflow --proto=ABC2_tI16_141_a_b_e-003
--params=$a, \allowbreak c/a, \allowbreak z_{3}$ |
ErLiO$_{2}$, $\delta$-LiAlO$_{2}$, NaGdO$_{2}$, NdNaO$_{2}$, InLiO$_{2}$
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $\frac{7}{8} \, \mathbf{a}_{1}+\frac{1}{8} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ | (4a) | Fe I |
$\mathbf{B_{2}}$ | = | $\frac{1}{8} \, \mathbf{a}_{1}+\frac{7}{8} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{1}{4}a \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ | (4a) | Fe I |
$\mathbf{B_{3}}$ | = | $\frac{5}{8} \, \mathbf{a}_{1}+\frac{3}{8} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{3}{8}c \,\mathbf{\hat{z}}$ | (4b) | Li I |
$\mathbf{B_{4}}$ | = | $\frac{3}{8} \, \mathbf{a}_{1}+\frac{5}{8} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{8}c \,\mathbf{\hat{z}}$ | (4b) | Li I |
$\mathbf{B_{5}}$ | = | $\left(z_{3} + \frac{1}{4}\right) \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (8e) | O I |
$\mathbf{B_{6}}$ | = | $z_{3} \, \mathbf{a}_{1}+\left(z_{3} + \frac{1}{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{3} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8e) | O I |
$\mathbf{B_{7}}$ | = | $- \left(z_{3} - \frac{3}{4}\right) \, \mathbf{a}_{1}- z_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (8e) | O I |
$\mathbf{B_{8}}$ | = | $- z_{3} \, \mathbf{a}_{1}- \left(z_{3} - \frac{3}{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{1}{4}a \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (8e) | O I |