AFLOW Prototype: A2B_hP9_150_ef_ad-001
This structure originally had the label A2B_hP9_150_ef_bd. 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/6CGG
or
https://aflow.org/p/A2B_hP9_150_ef_ad-001
or
PDF Version
Prototype | Fe$_{2}$P |
AFLOW prototype label | A2B_hP9_150_ef_ad-001 |
Strukturbericht designation | $C22$ |
ICSD | none |
Pearson symbol | hP9 |
Space group number | 150 |
Space group symbol | $P321$ |
AFLOW prototype command |
aflow --proto=A2B_hP9_150_ef_ad-001
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}$ |
generally accepted for years, has recently been shown to be incorrect.(Vol I., 360, also in Hendricks, 1930). The corrected structure, as given in Pearson's Handbook, is given in the revised Fe$_{2}$P page. When $z_{2}$ is set to zero this structure reverts to the revised Fe$_{2}$P structure.
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | P I |
$\mathbf{B_{2}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (2d) | P II |
$\mathbf{B_{3}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ | (2d) | P II |
$\mathbf{B_{4}}$ | = | $x_{3} \, \mathbf{a}_{1}$ | = | $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ | (3e) | Fe I |
$\mathbf{B_{5}}$ | = | $x_{3} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ | (3e) | Fe I |
$\mathbf{B_{6}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}$ | = | $- a x_{3} \,\mathbf{\hat{x}}$ | (3e) | Fe I |
$\mathbf{B_{7}}$ | = | $x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (3f) | Fe II |
$\mathbf{B_{8}}$ | = | $x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (3f) | Fe II |
$\mathbf{B_{9}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (3f) | Fe II |