Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A6B2C_tP18_128_eh_d_a-001

This structure originally had the label A6B2C_tP18_128_eh_d_a. 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/AD9M
or https://aflow.org/p/A6B2C_tP18_128_eh_d_a-001
or PDF Version

Phase II K$_{2}$SnCl$_{6}$ Structure: A6B2C_tP18_128_eh_d_a-001

Picture of Structure; Click for Big Picture
Prototype Cl$_{6}$K$_{2}$Sn
AFLOW prototype label A6B2C_tP18_128_eh_d_a-001
ICSD 1669
Pearson symbol tP18
Space group number 128
Space group symbol $P4/mnc$
AFLOW prototype command aflow --proto=A6B2C_tP18_128_eh_d_a-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}$


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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (2a) Sn I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (2a) Sn I
$\mathbf{B_{3}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (4d) K I
$\mathbf{B_{4}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (4d) K I
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (4d) K I
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (4d) K I
$\mathbf{B_{7}}$ = $z_{3} \, \mathbf{a}_{3}$ = $c z_{3} \,\mathbf{\hat{z}}$ (4e) Cl I
$\mathbf{B_{8}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Cl I
$\mathbf{B_{9}}$ = $- z_{3} \, \mathbf{a}_{3}$ = $- c z_{3} \,\mathbf{\hat{z}}$ (4e) Cl I
$\mathbf{B_{10}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Cl I
$\mathbf{B_{11}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}$ = $a x_{4} \,\mathbf{\hat{x}}+a y_{4} \,\mathbf{\hat{y}}$ (8h) Cl II
$\mathbf{B_{12}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}$ = $- a x_{4} \,\mathbf{\hat{x}}- a y_{4} \,\mathbf{\hat{y}}$ (8h) Cl II
$\mathbf{B_{13}}$ = $- y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}$ = $- a y_{4} \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}$ (8h) Cl II
$\mathbf{B_{14}}$ = $y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}$ = $a y_{4} \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}$ (8h) Cl II
$\mathbf{B_{15}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cl II
$\mathbf{B_{16}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cl II
$\mathbf{B_{17}}$ = $\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cl II
$\mathbf{B_{18}}$ = $- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cl II

References

  • H. Boysen and A. W. Hewat, A neutron powder investigation of the structural changes in K$_{2}$SnCl$_{6}$, Acta Crystallogr. Sect. B 34, 1412–1418 (1978), doi:10.1107/S0567740878005816.

Found in

  • P. Villars and K. Cenzual, Pearson's Crystal Data – Crystal Structure Database for Inorganic Compounds (2013). ASM International.

Prototype Generator

aflow --proto=A6B2C_tP18_128_eh_d_a --params=$a,c/a,z_{3},x_{4},y_{4}$

Species:

Running:

Output: