Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB6C_tP32_86_c_3g_d-001

This structure originally had the label AB6C_tP32_86_d_3g_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/NR43
or https://aflow.org/p/AB6C_tP32_86_c_3g_d-001
or PDF Version

NaSb(OH)$_{6}$ ($J1_{11}$) Structure: AB6C_tP32_86_c_3g_d-001

Picture of Structure; Click for Big Picture
Prototype Na(OH)$_{6}$Sb
AFLOW prototype label AB6C_tP32_86_c_3g_d-001
Strukturbericht designation $J1_{11}$
ICSD 4211
Pearson symbol tP32
Space group number 86
Space group symbol $P4_2/n$
AFLOW prototype command aflow --proto=AB6C_tP32_86_c_3g_d-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$

Other compounds with this structure

AgSb(OH)$_{6}$


  • The atomic positions were originally determined using setting 1 of space group $P4_{2}/n$ #86. We used FINDSYM to change the origin to setting 2.
  • The sites labeled O are actually OH radicals.
  • Although the replacement of fluorine by OH does not affect the shape of the Sb-(F,OH)$_{6}$ ions, it has a profound effect on the structure, as can be seen by looking at NaSbF$_{6}$ and NaSbF$_{4}$(OH)$_{2}$ ($J1_{12}$).

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (4c) Na I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}$ (4c) Na I
$\mathbf{B_{3}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (4c) Na I
$\mathbf{B_{4}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (4c) Na I
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}c \,\mathbf{\hat{z}}$ (4d) Sc I
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (4d) Sc I
$\mathbf{B_{7}}$ = $\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \,\mathbf{\hat{y}}$ (4d) Sc I
$\mathbf{B_{8}}$ = $\frac{1}{2} \, \mathbf{a}_{1}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}$ (4d) Sc I
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+a y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{10}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{11}}$ = $- y_{3} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a y_{3} \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{12}}$ = $\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{13}}$ = $- x_{3} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}- a y_{3} \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{14}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{15}}$ = $y_{3} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a y_{3} \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{16}}$ = $- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{17}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+a y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{18}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{19}}$ = $- y_{4} \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a y_{4} \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{20}}$ = $\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{21}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}- a y_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{22}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{23}}$ = $y_{4} \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a y_{4} \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{24}}$ = $- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O II
$\mathbf{B_{25}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8g) O III
$\mathbf{B_{26}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (8g) O III
$\mathbf{B_{27}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a y_{5} \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O III
$\mathbf{B_{28}}$ = $\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O III
$\mathbf{B_{29}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (8g) O III
$\mathbf{B_{30}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (8g) O III
$\mathbf{B_{31}}$ = $y_{5} \, \mathbf{a}_{1}- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a y_{5} \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O III
$\mathbf{B_{32}}$ = $- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O III

References

  • T. Asai, Refinement of the Crystal Structure of Sodium Hexahydroxoantimonate(V), NaSb(OH)$_{6}$, Bull. Chem. Soc. Japan 48, 2677–2679 (1975), doi:10.1246/bcsj.48.2677.

Found in

  • F. Hoffmann, M. Sartor, and M. Fröba, The Fascination of Crystals and Symmetry (2014). NASB(OH)6.

Prototype Generator

aflow --proto=AB6C_tP32_86_c_3g_d --params=$a,c/a,x_{3},y_{3},z_{3},x_{4},y_{4},z_{4},x_{5},y_{5},z_{5}$

Species:

Running:

Output: