AFLOW Prototype: AB2_tP12_115_j_egi-001
This structure originally had the label AB2_tP12_115_j_egi. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)
Links to this page
https://aflow.org/p/RJ86
or
https://aflow.org/p/AB2_tP12_115_j_egi-001
or
PDF Version
Prototype | HgI$_{2}$ |
AFLOW prototype label | AB2_tP12_115_j_egi-001 |
ICSD | none |
Pearson symbol | tP12 |
Space group number | 115 |
Space group symbol | $P\overline{4}m2$ |
AFLOW prototype command |
aflow --proto=AB2_tP12_115_j_egi-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak z_{4}$ |
averageof the Orange I and Orange II structures. The averaging procedure places the iodine I atoms only 0.36Å apart, so this is not a physical structure. We retain it as an example of a structure in space group $P\overline{4}m2$ #115.
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $z_{1} \, \mathbf{a}_{3}$ | = | $c z_{1} \,\mathbf{\hat{z}}$ | (2e) | I I |
$\mathbf{B_{2}}$ | = | $- z_{1} \, \mathbf{a}_{3}$ | = | $- c z_{1} \,\mathbf{\hat{z}}$ | (2e) | I I |
$\mathbf{B_{3}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (2g) | I II |
$\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- c z_{2} \,\mathbf{\hat{z}}$ | (2g) | I II |
$\mathbf{B_{5}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4i) | I III |
$\mathbf{B_{6}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4i) | I III |
$\mathbf{B_{7}}$ | = | $x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4i) | I III |
$\mathbf{B_{8}}$ | = | $- x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4i) | I III |
$\mathbf{B_{9}}$ | = | $x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+c z_{4} \,\mathbf{\hat{z}}$ | (4j) | Hg I |
$\mathbf{B_{10}}$ | = | $- x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+c z_{4} \,\mathbf{\hat{z}}$ | (4j) | Hg I |
$\mathbf{B_{11}}$ | = | $- x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (4j) | Hg I |
$\mathbf{B_{12}}$ | = | $x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (4j) | Hg I |