Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A13B5_hP36_194_a2e4f_b2f-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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Na$_{13}$Pb$_{5}$ (γ-NaPb) Structure: A13B5_hP36_194_a2e4f_b2f-001

Picture of Structure; Click for Big Picture
Prototype Na$_{13}$Pb$_{5}$
AFLOW prototype label A13B5_hP36_194_a2e4f_b2f-001
ICSD none
Pearson symbol hP36
Space group number 194
Space group symbol $P6_3/mmc$
AFLOW prototype command aflow --proto=A13B5_hP36_194_a2e4f_b2f-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak z_{6}, \allowbreak z_{7}, \allowbreak z_{8}, \allowbreak z_{9}, \allowbreak z_{10}$

  • (Weston, 1957) determined the space group, lattice constants, and positions of the lead atoms in Na$_{13}$Pb$_{5}$, but were unable to place the sodium atoms. Using the First-Principles Assisted Structure Solution (FPASS) method, (Ward, 2015) determined a set of positions for the sodium atoms consistent with the available experimental data. To our knowledge no further experimental study of Na$_{13}$Pb$_{5}$ has taken place.
  • There is no ICSD entry for this structure, but (Ward, 2015) provide the CIF for this structure in their supplementary material.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (2a) Na I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}c \,\mathbf{\hat{z}}$ (2a) Na I
$\mathbf{B_{3}}$ = $\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}c \,\mathbf{\hat{z}}$ (2b) Pb I
$\mathbf{B_{4}}$ = $\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}c \,\mathbf{\hat{z}}$ (2b) Pb I
$\mathbf{B_{5}}$ = $z_{3} \, \mathbf{a}_{3}$ = $c z_{3} \,\mathbf{\hat{z}}$ (4e) Na II
$\mathbf{B_{6}}$ = $\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Na II
$\mathbf{B_{7}}$ = $- z_{3} \, \mathbf{a}_{3}$ = $- c z_{3} \,\mathbf{\hat{z}}$ (4e) Na II
$\mathbf{B_{8}}$ = $- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Na II
$\mathbf{B_{9}}$ = $z_{4} \, \mathbf{a}_{3}$ = $c z_{4} \,\mathbf{\hat{z}}$ (4e) Na III
$\mathbf{B_{10}}$ = $\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Na III
$\mathbf{B_{11}}$ = $- z_{4} \, \mathbf{a}_{3}$ = $- c z_{4} \,\mathbf{\hat{z}}$ (4e) Na III
$\mathbf{B_{12}}$ = $- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Na III
$\mathbf{B_{13}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4f) Na IV
$\mathbf{B_{14}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na IV
$\mathbf{B_{15}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (4f) Na IV
$\mathbf{B_{16}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na IV
$\mathbf{B_{17}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (4f) Na V
$\mathbf{B_{18}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na V
$\mathbf{B_{19}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (4f) Na V
$\mathbf{B_{20}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na V
$\mathbf{B_{21}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (4f) Na VI
$\mathbf{B_{22}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na VI
$\mathbf{B_{23}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ (4f) Na VI
$\mathbf{B_{24}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{7} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{7} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na VI
$\mathbf{B_{25}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ (4f) Na VII
$\mathbf{B_{26}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na VII
$\mathbf{B_{27}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{8} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{8} \,\mathbf{\hat{z}}$ (4f) Na VII
$\mathbf{B_{28}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{8} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{8} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Na VII
$\mathbf{B_{29}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ (4f) Pb II
$\mathbf{B_{30}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Pb II
$\mathbf{B_{31}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{9} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{9} \,\mathbf{\hat{z}}$ (4f) Pb II
$\mathbf{B_{32}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{9} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{9} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Pb II
$\mathbf{B_{33}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ (4f) Pb III
$\mathbf{B_{34}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{10} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{10} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Pb III
$\mathbf{B_{35}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{10} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{10} \,\mathbf{\hat{z}}$ (4f) Pb III
$\mathbf{B_{36}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{10} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{10} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Pb III

References

  • L. Ward, K. Michel, and C. Wolverton, Three new crystal structures in the Na-Pb system: solving structures without additional experimental input, Acta Crystallogr. Sect. A 71, 542–548 (2015), doi:10.1107/S2053273315012516.
  • N. E. Weston and D. P. Shoemaker, The crystal structures of three phases in the Na-Pb system, Acta Cryst. 10, 775 (1957), doi:10.1107/S0365110X57002649. International Union of Crystallography Abstracts, pp. 735-863.

Prototype Generator

aflow --proto=A13B5_hP36_194_a2e4f_b2f --params=$a,c/a,z_{3},z_{4},z_{5},z_{6},z_{7},z_{8},z_{9},z_{10}$

Species:

Running:

Output: