Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A17B2_hR19_166_cegh_c-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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https://aflow.org/p/3JAE
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Nb$_{2}$Be$_{17}$ Structure: A17B2_hR19_166_cegh_c-001

Picture of Structure; Click for Big Picture
Prototype Be$_{17}$Nb$_{2}$
AFLOW prototype label A17B2_hR19_166_cegh_c-001
ICSD 58724
Pearson symbol hR19
Space group number 166
Space group symbol $R\overline{3}m$
AFLOW prototype command aflow --proto=A17B2_hR19_166_cegh_c-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{4}, \allowbreak x_{5}, \allowbreak z_{5}$

Other compounds with this structure

$\alpha$-Be$_{17}$Hf$_{2}$,  Be$_{17}$Ta$_{2}$,  $\alpha$-Be$_{17}$Ti$_{2}$,  Be$_{17}$Zr$_{2}$


  • Hexagonal settings of this structure can be obtained with the option --hex.

\[ \begin{array}{ccc} \mathbf{a_{1}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}\\\mathbf{a_{2}}&=&\frac{1}{\sqrt{3}}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}\\\mathbf{a_{3}}&=&- \frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}} \end{array}\]

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}+x_{1} \, \mathbf{a}_{3}$ = $c x_{1} \,\mathbf{\hat{z}}$ (2c) Be I
$\mathbf{B_{2}}$ = $- x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}- x_{1} \, \mathbf{a}_{3}$ = $- c x_{1} \,\mathbf{\hat{z}}$ (2c) Be I
$\mathbf{B_{3}}$ = $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ = $c x_{2} \,\mathbf{\hat{z}}$ (2c) Nb I
$\mathbf{B_{4}}$ = $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ = $- c x_{2} \,\mathbf{\hat{z}}$ (2c) Nb I
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- \frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ (3e) Be II
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ (3e) Be II
$\mathbf{B_{7}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ (3e) Be II
$\mathbf{B_{8}}$ = $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \left(2 x_{4} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \left(6 x_{4} + 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (6g) Be III
$\mathbf{B_{9}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \left(2 x_{4} + 1\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \left(6 x_{4} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (6g) Be III
$\mathbf{B_{10}}$ = $- x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (6g) Be III
$\mathbf{B_{11}}$ = $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- \frac{1}{4}a \left(2 x_{4} + 1\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \left(6 x_{4} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (6g) Be III
$\mathbf{B_{12}}$ = $\frac{1}{2} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ = $- \frac{1}{4}a \left(2 x_{4} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \left(6 x_{4} + 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (6g) Be III
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ (6g) Be III
$\mathbf{B_{14}}$ = $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (6h) Be IV
$\mathbf{B_{15}}$ = $z_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (6h) Be IV
$\mathbf{B_{16}}$ = $x_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{\sqrt{3}}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (6h) Be IV
$\mathbf{B_{17}}$ = $- z_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (6h) Be IV
$\mathbf{B_{18}}$ = $- x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (6h) Be IV
$\mathbf{B_{19}}$ = $- x_{5} \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ = $\frac{1}{\sqrt{3}}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ (6h) Be IV

References

  • A. Zalkin, D. E. Sands, and O. H. Krikorian, The crystal structure of Nb$_{2}$Be$_{17}$, Acta Cryst. 12, 713–715 (1967), doi:10.1107/S0365110X59002110.

Prototype Generator

aflow --proto=A17B2_hR19_166_cegh_c --params=$a,c/a,x_{1},x_{2},x_{4},x_{5},z_{5}$

Species:

Running:

Output: