Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2BC6_hP18_182_f_b_gh-001

This structure originally had the label A2BC6_hP18_182_f_b_gh. 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/6VW9
or https://aflow.org/p/A2BC6_hP18_182_f_b_gh-001
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BaAl$_{2}$O$_{4}$ ($H2_{8}$) Structure: A2BC6_hP18_182_f_b_gh-001

Picture of Structure; Click for Big Picture
Prototype Al$_{2}$BaO$_{4}$
AFLOW prototype label A2BC6_hP18_182_f_b_gh-001
Strukturbericht designation $H2_{8}$
ICSD 15728
Pearson symbol hP18
Space group number 182
Space group symbol $P6_322$
AFLOW prototype command aflow --proto=A2BC6_hP18_182_f_b_gh-001
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}$

  • The structure of BaAl$_{2}$O$_{4}$ was originally determined by (Wallmark, 1937) and designated $H2_{8}$ by (Gottfried, 1940). This had two oxygen atoms at the (2c) Wyckoff position, $\pm (1/3, 2/3, z)$. (Perrotta, 1968) redetermined the structure and found that the best fit to external data required the (2c) oxygen atoms to be distributed among the (6h) sites. Thus each of the (6h) sites in this structure is 1/3 occupied.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}c \,\mathbf{\hat{z}}$ (2b) Ba I
$\mathbf{B_{2}}$ = $\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}c \,\mathbf{\hat{z}}$ (2b) Ba I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4f) Al I
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Al I
$\mathbf{B_{5}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (4f) Al I
$\mathbf{B_{6}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Al I
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ (6g) O I
$\mathbf{B_{8}}$ = $x_{3} \, \mathbf{a}_{2}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ (6g) O I
$\mathbf{B_{9}}$ = $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}$ = $- a x_{3} \,\mathbf{\hat{x}}$ (6g) O I
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) O I
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) O I
$\mathbf{B_{12}}$ = $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (6g) O I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+2 x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) O II
$\mathbf{B_{14}}$ = $- 2 x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) O II
$\mathbf{B_{15}}$ = $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) O II
$\mathbf{B_{16}}$ = $- x_{4} \, \mathbf{a}_{1}- 2 x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) O II
$\mathbf{B_{17}}$ = $2 x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) O II
$\mathbf{B_{18}}$ = $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) O II

References

  • A. J. Perrotta and J. V. Smith, The Crystal Structure of BaAl$_{2}$O$_{4}$, Bull. Soc. franç. de Minëral. Cristal. 91, 85–87 (1968), doi:10.3406/bulmi.1968.6190.
  • S. Wallmark and A. Westgren, X-Ray Analysis of Barium Aluminates, Ark. Kem. Mineral. Geol. B 12 (1937).
  • C. Gottfried, ed., Strukturbericht Band V 1937 (Akademische Verlagsgesellschaft M. B. H., Leipzig, 1940).

Prototype Generator

aflow --proto=A2BC6_hP18_182_f_b_gh --params=$a,c/a,z_{2},x_{3},x_{4}$

Species:

Running:

Output: