AFLOW Prototype: A2BC_oP16_53_eh_ab_g-001
This structure originally had the label A2BC_oP16_53_eh_ab_g. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M.J. Mehl, M. Esters, C. Oses, O. Levy, G.L.W. Hart, C. Toher, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 3, Comp. Mat. Sci. 199, 110450 (2021). (doi=10.1016/j.commatsci.2021.110450)
Links to this page
https://aflow.org/p/DCYY
or
https://aflow.org/p/A2BC_oP16_53_eh_ab_g-001
or
PDF Version
Prototype | F$_{2}$H$_{5}$N |
AFLOW prototype label | A2BC_oP16_53_eh_ab_g-001 |
Strukturbericht designation | $F5_{8}$ |
ICSD | 28893 |
Pearson symbol | oP16 |
Space group number | 53 |
Space group symbol | $Pmna$ |
AFLOW prototype command |
aflow --proto=A2BC_oP16_53_eh_ab_g-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak y_{4}, \allowbreak y_{5}, \allowbreak z_{5}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | H I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2a) | H I |
$\mathbf{B_{3}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}$ | (2b) | H II |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2b) | H II |
$\mathbf{B_{5}}$ | = | $x_{3} \, \mathbf{a}_{1}$ | = | $a x_{3} \,\mathbf{\hat{x}}$ | (4e) | F I |
$\mathbf{B_{6}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4e) | F I |
$\mathbf{B_{7}}$ | = | $- x_{3} \, \mathbf{a}_{1}$ | = | $- a x_{3} \,\mathbf{\hat{x}}$ | (4e) | F I |
$\mathbf{B_{8}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4e) | F I |
$\mathbf{B_{9}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (4g) | NH I |
$\mathbf{B_{10}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (4g) | NH I |
$\mathbf{B_{11}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (4g) | NH I |
$\mathbf{B_{12}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (4g) | NH I |
$\mathbf{B_{13}}$ | = | $y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $b y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (4h) | F II |
$\mathbf{B_{14}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4h) | F II |
$\mathbf{B_{15}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4h) | F II |
$\mathbf{B_{16}}$ | = | $- y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- b y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (4h) | F II |