Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A3B_hP24_194_hk_bf-002

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/E109
or https://aflow.org/p/A3B_hP24_194_hk_bf-002
or PDF Version

Hexagonal PuAl$_{3}$ Structure: A3B_hP24_194_hk_bf-002

Picture of Structure; Click for Big Picture
Prototype Al$_{3}$Pu
AFLOW prototype label A3B_hP24_194_hk_bf-002
ICSD 58141
Pearson symbol hP24
Space group number 194
Space group symbol $P6_3/mmc$
AFLOW prototype command aflow --proto=A3B_hP24_194_hk_bf-002
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak z_{4}$

  • PuAl$_{3}$ can also be found in a rhombohedral form which is isostructural with BaPb$_{3}$. (Runnals, 1965)
  • Hexagonal PuAl$_{3}$ and VCo$_{3}$ have the same AFLOW prototype label, A3B_hP24_194_hk_bf. They are generated by the same symmetry operations with different sets of parameters (--params) specified in their corresponding CIF files.

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

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) Pu I
$\mathbf{B_{2}}$ = $\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}c \,\mathbf{\hat{z}}$ (2b) Pu 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) Pu II
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Pu II
$\mathbf{B_{5}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (4f) Pu II
$\mathbf{B_{6}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4f) Pu II
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}+2 x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Al I
$\mathbf{B_{8}}$ = $- 2 x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Al I
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6h) Al I
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}- 2 x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Al I
$\mathbf{B_{11}}$ = $2 x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Al I
$\mathbf{B_{12}}$ = $- x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6h) Al I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+2 x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{14}}$ = $- 2 x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{15}}$ = $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{16}}$ = $- x_{4} \, \mathbf{a}_{1}- 2 x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{17}}$ = $2 x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{18}}$ = $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{19}}$ = $2 x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{20}}$ = $- x_{4} \, \mathbf{a}_{1}- 2 x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{21}}$ = $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\sqrt{3}a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{22}}$ = $- 2 x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{23}}$ = $x_{4} \, \mathbf{a}_{1}+2 x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12k) Al II
$\mathbf{B_{24}}$ = $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \sqrt{3}a x_{4} \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12k) Al II

References

  • A. C. Larson, D. T. Cromer, and C. K. Stambaugh, The Crystal Structure of PuAl$_{3}$, Acta Cryst. 10, 443–446 (1957), doi:10.1107/S0365110X57001504.
  • O. J. C. Runnalls and R. R. Boucher, The crystal structure of rhombohedral PuAl$_{3}$, Acta Cryst. 19, 184–186 (1965), doi:10.1107/S0365110X65003055.

Found in

  • A. Jain, S. Ping, G. Hautier, W. Chen, W. D. Richards, S. Dacek, S. Cholia, D. Gunter, D. Skinner, G. Ceder, and K. A. Persson, Commentary: The Materials Project: A materials genome approach to accelerating materials innovation, APL Materials 1, 011002 (2013), doi:10.1063/1.4812323.

Prototype Generator

aflow --proto=A3B_hP24_194_hk_bf --params=$a,c/a,z_{2},x_{3},x_{4},z_{4}$

Species:

Running:

Output: