AFLOW Prototype: ABC3_tP10_100_b_a_bc-001
This structure originally had the label ABC3_tP10_100_b_a_bc. 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/80PK
or
https://aflow.org/p/ABC3_tP10_100_b_a_bc-001
or
PDF Version
Prototype | N(NH$_{4}$)O$_{3}$ |
AFLOW prototype label | ABC3_tP10_100_b_a_bc-001 |
Strukturbericht designation | $G0_{9}$ |
ICSD | none |
Pearson symbol | tP10 |
Space group number | 100 |
Space group symbol | $P4bm$ |
AFLOW prototype command |
aflow --proto=ABC3_tP10_100_b_a_bc-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}$ |
Phase | Temperature °C | Strukturbericht | Page | |
I | 125 - 170 | $G0_{8}$ | AB_cP2_221_a_b-001 | |
II | 84 - 125 | $G0_{9}$ | ABC3_tP10_100_b_a_bc | (this structure) |
III | 32 - 84 | $G0_{10}$ | ABC3_oP20_62_c_c_cd-002 | |
IV | -18 - 32 | $G0_{11}$ | A4B2C3_oP18_59_ef_ab_af-001 | |
V | $< -18$ | A4B2C3_tP72_77_8d_ab2c2d_6d2-001 |
is almost bound in two orientations (in opposite directions).He then gives two possible orientations for the nitrate. We present the first orientation here. The second orientation is obtained by taking $z_{3} \rightarrow - z_{3}$ and $z_{4} \rightarrow - z_{4}$.
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) | NH I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ | (2a) | NH I |
$\mathbf{B_{3}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+c z_{2} \,\mathbf{\hat{z}}$ | (2b) | N I |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (2b) | N I |
$\mathbf{B_{5}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+c z_{3} \,\mathbf{\hat{z}}$ | (2b) | O I |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (2b) | O I |
$\mathbf{B_{7}}$ | = | $x_{4} \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4c) | O II |
$\mathbf{B_{8}}$ | = | $- x_{4} \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4c) | O II |
$\mathbf{B_{9}}$ | = | $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4c) | O II |
$\mathbf{B_{10}}$ | = | $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (4c) | O II |