Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB3_hP24_185_c_ab2c-001

This structure originally had the label AB3_hP24_185_c_ab2c. Calls to that address will be redirected here.

If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)

Links to this page

https://aflow.org/p/3ERL
or https://aflow.org/p/AB3_hP24_185_c_ab2c-001
or PDF Version

Na$_{3}$As Structure: AB3_hP24_185_c_ab2c-001

Picture of Structure; Click for Big Picture
Prototype AsNa$_{3}$
AFLOW prototype label AB3_hP24_185_c_ab2c-001
ICSD 79586
Pearson symbol hP24
Space group number 185
Space group symbol $P6_3cm$
AFLOW prototype command aflow --proto=AB3_hP24_185_c_ab2c-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}$

Other compounds with this structure

Cu$_{3}$P


  • (Hafner, 1994) state that this is a correction of the $D0_{18}$ Na$_{3}$As structure, but see the discussion about Cu$_{3}$P and other similar compounds.
  • Space group $P6_{3}cm$ #185 allows an arbitary choice of the origin of the $z$-axis. Here we set $z_{3} = 1/4$ for the arsenic (6c) atoms.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $z_{1} \, \mathbf{a}_{3}$ = $c z_{1} \,\mathbf{\hat{z}}$ (2a) Na I
$\mathbf{B_{2}}$ = $\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2a) Na I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4b) Na II
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4b) Na II
$\mathbf{B_{5}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4b) Na II
$\mathbf{B_{6}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4b) Na II
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (6c) As I
$\mathbf{B_{8}}$ = $x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (6c) As I
$\mathbf{B_{9}}$ = $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+c z_{3} \,\mathbf{\hat{z}}$ (6c) As I
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) As I
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) As I
$\mathbf{B_{12}}$ = $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) As I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (6c) Na III
$\mathbf{B_{14}}$ = $x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (6c) Na III
$\mathbf{B_{15}}$ = $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+c z_{4} \,\mathbf{\hat{z}}$ (6c) Na III
$\mathbf{B_{16}}$ = $- x_{4} \, \mathbf{a}_{1}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) Na III
$\mathbf{B_{17}}$ = $- x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) Na III
$\mathbf{B_{18}}$ = $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) Na III
$\mathbf{B_{19}}$ = $x_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) Na IV
$\mathbf{B_{20}}$ = $x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) Na IV
$\mathbf{B_{21}}$ = $- x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) Na IV
$\mathbf{B_{22}}$ = $- x_{5} \, \mathbf{a}_{1}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) Na IV
$\mathbf{B_{23}}$ = $- x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) Na IV
$\mathbf{B_{24}}$ = $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) Na IV

References


Prototype Generator

aflow --proto=AB3_hP24_185_c_ab2c --params=$a,c/a,z_{1},z_{2},x_{3},z_{3},x_{4},z_{4},x_{5},z_{5}$

Species:

Running:

Output: