Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A4BC2D_oP32_62_2cd_c_d_c-001

This structure originally had the label A4BC2D_oP32_62_2cd_c_d_c. 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/ZADN
or https://aflow.org/p/A4BC2D_oP32_62_2cd_c_d_c-001
or PDF Version

K$_{2}$SnCl$_{4}\cdot$H$_{2}$O Structure: A4BC2D_oP32_62_2cd_c_d_c-001

Picture of Structure; Click for Big Picture
Prototype Cl$_{4}$(H$_{2}$O)K$_{2}$Sn
AFLOW prototype label A4BC2D_oP32_62_2cd_c_d_c-001
ICSD 14219
Pearson symbol oP32
Space group number 62
Space group symbol $Pnma$
AFLOW prototype command aflow --proto=A4BC2D_oP32_62_2cd_c_d_c-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$

  • This is not the version of the structure given in (Hermann, 1939) as Strukturbericht $E3_{3}$. That structure was was determined by (Brasseur, 1939).
  • A later refinement of the structure by (Ye, 2013) located the hydrogen atoms and is otherwise in agreement with (Kamenar, 1962).
  • (Kamenar, 1962) did not give the positions of atoms in the water molecules and so we only give the positions of the oxygen atoms.
  • (Kamenar, 1962) gives the crystal structure and atomic positions in the $Pbnm$ setting of space group #62. We have used FINDSYM to show the structure in the standard $Pnma$ setting.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $a x_{1} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{2}}$ = $- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{3}}$ = $- x_{1} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{1} \, \mathbf{a}_{3}$ = $- a x_{1} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{4}}$ = $\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{5}}$ = $x_{2} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $a x_{2} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{6}}$ = $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{7}}$ = $- x_{2} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $- a x_{2} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{8}}$ = $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (4c) H I
$\mathbf{B_{10}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) H I
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (4c) H I
$\mathbf{B_{12}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) H I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4c) Sn I
$\mathbf{B_{14}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Sn I
$\mathbf{B_{15}}$ = $- x_{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (4c) Sn I
$\mathbf{B_{16}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Sn I
$\mathbf{B_{17}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{18}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{19}}$ = $- x_{5} \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+b \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{20}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{21}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- b y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{22}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{23}}$ = $x_{5} \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}- b \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{24}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) Cl III
$\mathbf{B_{25}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (8d) K I
$\mathbf{B_{26}}$ = $- \left(x_{6} - \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{6} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) K I
$\mathbf{B_{27}}$ = $- x_{6} \, \mathbf{a}_{1}+\left(y_{6} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}+b \left(y_{6} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (8d) K I
$\mathbf{B_{28}}$ = $\left(x_{6} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{6} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{6} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{6} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) K I
$\mathbf{B_{29}}$ = $- x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (8d) K I
$\mathbf{B_{30}}$ = $\left(x_{6} + \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{6} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) K I
$\mathbf{B_{31}}$ = $x_{6} \, \mathbf{a}_{1}- \left(y_{6} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}- b \left(y_{6} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (8d) K I
$\mathbf{B_{32}}$ = $- \left(x_{6} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{6} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{6} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{6} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8d) K I

References

  • B. Kamenar and D. Grdenić, The crystal structure of potassium chloride trichlorostannite hydrate, KCl, KSnCl$_{3}$, H$_{2}$O, J. Inorg. Nucl. Chem. 24, 1039–1045 (1962), doi:10.1016/0022-1902(62)80247-4.
  • K. Herrmann, ed., Strukturbericht Band VII 1939 (Akademische Verlagsgesellschaft M. B. H., Leipzig, 1943).
  • H. Brasseur and A. de Rassenfosse, The Crystal Structure of Hydrated Potassium Chlorostannite, Z. Kristallogr. 101, 389–395 (1939), doi:10.1524/zkri.1939.101.1.396.
  • F. Ye and H. Reuter, Redetermination of dipotassium trichloridostannate(II) chloride monohydrate, Acta Crystallogr. Sect. E 69, i10 (2013), doi:10.1107/S1600536813001372.

Prototype Generator

aflow --proto=A4BC2D_oP32_62_2cd_c_d_c --params=$a,b/a,c/a,x_{1},z_{1},x_{2},z_{2},x_{3},z_{3},x_{4},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6}$

Species:

Running:

Output: