Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A16B2C_hP19_164_2d2i_d_a-001

This structure originally had the label A16B2C_hP19_164_2d2i_d_b. 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/LLDQ
or https://aflow.org/p/A16B2C_hP19_164_2d2i_d_a-001
or PDF Version

Predicted Li$_{2}$MgH$_{16}$ 300 GPa Structure: A16B2C_hP19_164_2d2i_d_a-001

Picture of Structure; Click for Big Picture
Prototype H$_{16}$Li$_{2}$Mg
AFLOW prototype label A16B2C_hP19_164_2d2i_d_a-001
ICSD none
Pearson symbol hP19
Space group number 164
Space group symbol $P\overline{3}m1$
AFLOW prototype command aflow --proto=A16B2C_hP19_164_2d2i_d_a-001
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak z_{6}$


\[ \begin{array}{ccc} \mathbf{a_{1}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{2}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{3}}&=&c \,\mathbf{\hat{z}} \end{array}\]

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (1a) Mg 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) H I
$\mathbf{B_{3}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (2d) H I
$\mathbf{B_{4}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (2d) H II
$\mathbf{B_{5}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (2d) H II
$\mathbf{B_{6}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (2d) Li I
$\mathbf{B_{7}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (2d) Li I
$\mathbf{B_{8}}$ = $x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6i) H III
$\mathbf{B_{9}}$ = $x_{5} \, \mathbf{a}_{1}+2 x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6i) H III
$\mathbf{B_{10}}$ = $- 2 x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6i) H III
$\mathbf{B_{11}}$ = $- x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (6i) H III
$\mathbf{B_{12}}$ = $2 x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (6i) H III
$\mathbf{B_{13}}$ = $- x_{5} \, \mathbf{a}_{1}- 2 x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (6i) H III
$\mathbf{B_{14}}$ = $x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6i) H IV
$\mathbf{B_{15}}$ = $x_{6} \, \mathbf{a}_{1}+2 x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6i) H IV
$\mathbf{B_{16}}$ = $- 2 x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6i) H IV
$\mathbf{B_{17}}$ = $- x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (6i) H IV
$\mathbf{B_{18}}$ = $2 x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (6i) H IV
$\mathbf{B_{19}}$ = $- x_{6} \, \mathbf{a}_{1}- 2 x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{6} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (6i) H IV

References

  • Y. Sun, J. Lv, Y. Xie, H. Liu, and Y. Ma, Route to a Superconducting Phase above Room Temperature in Electron-Doped Hydride Compounds under High Pressure, Phys. Rev. Lett. 123, 097001 (2019), doi:10.1103/PhysRevLett.123.097001.

Prototype Generator

aflow --proto=A16B2C_hP19_164_2d2i_d_a --params=$a,c/a,z_{2},z_{3},z_{4},x_{5},z_{5},x_{6},z_{6}$

Species:

Running:

Output: