Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB3C_hP30_193_g_gk_bd-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.

Links to this page

https://aflow.org/p/AE3Q
or https://aflow.org/p/AB3C_hP30_193_g_gk_bd-001
or PDF Version

Ordered TmBO$_{3}$ Structure: AB3C_hP30_193_g_gk_bd-001

Picture of Structure; Click for Big Picture
Prototype BO$_{3}$Tm
AFLOW prototype label AB3C_hP30_193_g_gk_bd-001
ICSD 27942
Pearson symbol hP30
Space group number 193
Space group symbol $P6_3/mcm$
AFLOW prototype command aflow --proto=AB3C_hP30_193_g_gk_bd-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak x_{5}, \allowbreak z_{5}$

Other compounds with this structure

DyBO$_{3}$,  ErBO$_{3}$,  EuBO$_{3}$,  GdBO$_{3}$,  HoBO$_{3}$,  LuBO$_{3}$,  SmBO$_{3}$,  YBO$_{3}$,  YbBO$_{3}$


  • (Newnham, 1963) found two possible structures for TmBO$_{3}$ and YBO$_{3}$:
  • There are several problems with this structure:
    • The ICSD entry 27942 places this structure in space group $P\overline{6}c2$ #188 even though the positions are such that the structure can be resolved in the higher symmetry $P6_{3}/mcm$ space group, and
    • the positions of the O I atoms are such that the O-O distance is less than 1Å.
    • The ICSD entry gives what seem to be reasonable O-O distances, so we use those coordinates, using AFLOW to transform the structure into space group $P6_{3}/mcm$.
    .

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

Basis vectors

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

References

  • R. E. Newnham, M. J. Redman, and R. P. Santoro, Crystal Structure of Yttrium and Other Rareā€Earth Borates, J. Am. Ceram. Soc. 46, 253–256 (1963), doi:10.1111/j.1151-2916.1963.tb11721.x.

Prototype Generator

aflow --proto=AB3C_hP30_193_g_gk_bd --params=$a,c/a,x_{3},x_{4},x_{5},z_{5}$

Species:

Running:

Output: