Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB3C_hP30_194_bf_hk_af-001

If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.

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Hexagonal BaTiO$_{3}$ Structure: AB3C_hP30_194_bf_hk_af-001

Picture of Structure; Click for Big Picture
Prototype BaO$_{3}$Ti
AFLOW prototype label AB3C_hP30_194_bf_hk_af-001
ICSD 75240
Pearson symbol hP30
Space group number 194
Space group symbol $P6_3/mmc$
AFLOW prototype command aflow --proto=AB3C_hP30_194_bf_hk_af-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak x_{6}, \allowbreak z_{6}$

Other compounds with this structure

Ba(Co$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Cr$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Fe$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Ir$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Mn$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Os$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Pd$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Pt$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Rh$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(Ru$_{x}$Ti$_{1-x}$)O$_{3}$,  Ba(V$_{x}$Ti$_{1-x}$)O$_{3}$


  • The perovskite BaTiO$_{3}$ undergoes a variety of temperature driven phase transitions. (Shirane, 1957) The first three structures are ferroelectric:
  • Hexagonal BaTiO$_{3}$ (this structure) can be stabilized by alloying the titanium sites with other transition metals. (Dickson, 1961) The pure structure has been grown at 1853K and cooled to room temperature. (Akimo, 1994) In some cases there is a deficiency of oxygen atoms. (Dickson, 1961)
  • This pure sample was grown at 1853K and cooled to room temperature.
  • If the transition metal atoms have the composition M1M2$_{2}$ and are ordered on the (2a) and (4f) sites, this becomes the hexagonal Ba$_{3}$CoIr$_{2}$O$_{9}$ structure.

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

Basis vectors

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

References

  • G. Shirane, H. Danner, and R. Pepinsky, Neutron Diffraction Study of Orthorhombic BaTiO$_{3}$, Phys. Rev. 105, 856–860 (1957), doi:10.1103/PhysRev.105.856.
  • J. G. Dickson, L. Katz, and R. Ward, Compounds with the Hexagonal Barium Titanate Structure, J. Am. Chem. Soc. 83, 3026–3029 (1961), doi:10.1021/ja01475a012.
  • A. W. Hewat, Structure of rhombohedral ferroelectric barium titanate, Ferroelectrics 6, 215–218 (1974), doi:10.1080/00150197408243970.
  • J. Akimoto, Y. Gotoh, and Y. Oosawa, Refinement of Hexagonal BaTiO$_{3}$, Acta Crystallogr. Sect. C 50, 160–161 (1994), doi:10.1107/S0108270193008637.

Prototype Generator

aflow --proto=AB3C_hP30_194_bf_hk_af --params=$a,c/a,z_{3},z_{4},x_{5},x_{6},z_{6}$

Species:

Running:

Output: