Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB2CD8E2_hP14_147_a_d_b_dg_d-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/B64B
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Na$_{2}$BaNi(PO$_{4}$)$_{2}$ Structure: AB2CD8E2_hP14_147_a_d_b_dg_d-001

Picture of Structure; Click for Big Picture
Prototype BaNa$_{2}$NiO$_{8}$P$_{2}$
AFLOW prototype label AB2CD8E2_hP14_147_a_d_b_dg_d-001
ICSD none
Pearson symbol hP14
Space group number 147
Space group symbol $P\overline{3}$
AFLOW prototype command aflow --proto=AB2CD8E2_hP14_147_a_d_b_dg_d-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (1a) Ba I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}c \,\mathbf{\hat{z}}$ (1b) Ni I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (2d) Na I
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (2d) Na I
$\mathbf{B_{5}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (2d) O I
$\mathbf{B_{6}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (2d) O I
$\mathbf{B_{7}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (2d) P I
$\mathbf{B_{8}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (2d) P I
$\mathbf{B_{9}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6g) O II
$\mathbf{B_{10}}$ = $- y_{6} \, \mathbf{a}_{1}+\left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} - 2 y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6g) O II
$\mathbf{B_{11}}$ = $- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6g) O II
$\mathbf{B_{12}}$ = $- x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (6g) O II
$\mathbf{B_{13}}$ = $y_{6} \, \mathbf{a}_{1}- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{6} + 2 y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (6g) O II
$\mathbf{B_{14}}$ = $\left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (6g) O II

References

  • N. Li, Q. Huang, A. Brassington, X. Y. Yue, W. J. Chu, S. K. Guang, X. H. Zhou, P. Gao, E. X. Feng, H. B. Cao, E. S. Choi, Y. Sun, Q. J. Li, X. Zhao, H. D. Zhou, and X. F. Sun, Quantum spin state transitions in spin-1 equilateral triangular lattice antiferromagnet Na$_{2}$BaNi(PO$_{4}$)$_{2}$, Phys. Rev. B 104, 104403 (2021), doi:10.1103/PhysRevB.104.104403.

Prototype Generator

aflow --proto=AB2CD8E2_hP14_147_a_d_b_dg_d --params=$a,c/a,z_{3},z_{4},z_{5},x_{6},y_{6},z_{6}$

Species:

Running:

Output: