Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: ABC4D_hP14_173_a_b_bc_b-001

This structure originally had the label ABC4D_hP14_173_a_b_bc_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/F86C
or https://aflow.org/p/ABC4D_hP14_173_a_b_bc_b-001
or PDF Version

LiKSO$_{4}$ ($H1_{4}$) Structure: ABC4D_hP14_173_a_b_bc_b-001

Picture of Structure; Click for Big Picture
Prototype KLiO$_{4}$S
AFLOW prototype label ABC4D_hP14_173_a_b_bc_b-001
Strukturbericht designation $H1_{4}$
ICSD 48136
Pearson symbol hP14
Space group number 173
Space group symbol $P6_3$
AFLOW prototype command aflow --proto=ABC4D_hP14_173_a_b_bc_b-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$

  • As there is no center of symmetry in the structure, the point $z = 0$ can be set arbitrarily. Following (Bhakay-Tamhane, 1984) we set $z_{1} = 0$.
  • While Struckturbericht symbols $H1_{1}$, $H1_{2}$ and $H1_{5}$ are identical with $S1_{1}$, $S1_{2}$ and $S1_{5}$, respectively, there is no connection between this structure and $S1_{4}$, garnet (A2B3C12D3_cI160_230_a_c_h_d).

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $z_{1} \, \mathbf{a}_{3}$ = $c z_{1} \,\mathbf{\hat{z}}$ (2a) K I
$\mathbf{B_{2}}$ = $\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2a) K I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (2b) Li I
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2b) Li I
$\mathbf{B_{5}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (2b) O I
$\mathbf{B_{6}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2b) O I
$\mathbf{B_{7}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (2b) S I
$\mathbf{B_{8}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2b) S I
$\mathbf{B_{9}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) O II
$\mathbf{B_{10}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) O II
$\mathbf{B_{11}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) O II
$\mathbf{B_{12}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) O II
$\mathbf{B_{13}}$ = $y_{5} \, \mathbf{a}_{1}- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) O II
$\mathbf{B_{14}}$ = $\left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) O II

References

  • S. Bhakay-Tamhane, A. Sequiera, and R. Chidambaram, Structure of lithium potassium sulphate, LiKSO$_{4}$: a neutron diffraction study, Acta Crystallogr. Sect. C 40, 1648–1651 (1984), doi:10.1107/S0108270184009045.

Prototype Generator

aflow --proto=ABC4D_hP14_173_a_b_bc_b --params=$a,c/a,z_{1},z_{2},z_{3},z_{4},x_{5},y_{5},z_{5}$

Species:

Running:

Output: