Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B13C6D_hP44_182_bc_a2i_i_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.

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Na$_{2}$Co$_{2}$TeO$_{6}$ Structure: A2B13C6D_hP44_182_bc_a2i_i_d-001

Picture of Structure; Click for Big Picture
Prototype Co$_{2}$Na$_{2}$O$_{6}$Te
AFLOW prototype label A2B13C6D_hP44_182_bc_a2i_i_d-001
ICSD 245537
Pearson symbol hP44
Space group number 182
Space group symbol $P6_322$
AFLOW prototype command aflow --proto=A2B13C6D_hP44_182_bc_a2i_i_d-001
--params=$a, \allowbreak c/a, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak y_{7}, \allowbreak z_{7}$

Other compounds with this structure

Na$_{2}$Zn$_{2}$TeO$_{6}$


  • (Viciu, 2007) find that the four sodium atoms in the primitive cell are distributed over the 26 sites available: The (2a) site has 11% occupancy, the first (Na-II) (12i) site has 8.1% occupancy, and the second (Na-III) has 23.4% occupancy. (Xiao, 2019) present a slightly different structure, where all of the sodium atoms are on a (6g) site [ $(x,0,0)$, $x = 0.32$ ], which is 1/3 vacant. In both cases all of the sodium atoms are near the $z = 0$ and $z = 1/2$ planes.
  • We have shifted the origin by 1/2 {\bf a$_3$} from that of (Viciu, 2007).

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

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) Co I
$\mathbf{B_{4}}$ = $\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}c \,\mathbf{\hat{z}}$ (2b) Co I
$\mathbf{B_{5}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (2c) Co II
$\mathbf{B_{6}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (2c) Co II
$\mathbf{B_{7}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (2d) Te I
$\mathbf{B_{8}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (2d) Te I
$\mathbf{B_{9}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{10}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{11}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{12}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{13}}$ = $y_{5} \, \mathbf{a}_{1}- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{14}}$ = $\left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{15}}$ = $y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{16}}$ = $\left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{17}}$ = $- x_{5} \, \mathbf{a}_{1}- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{18}}$ = $- y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{19}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{20}}$ = $x_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na II
$\mathbf{B_{21}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{22}}$ = $- y_{6} \, \mathbf{a}_{1}+\left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} - 2 y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{23}}$ = $- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{24}}$ = $- x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{25}}$ = $y_{6} \, \mathbf{a}_{1}- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{6} + 2 y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{26}}$ = $\left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{27}}$ = $y_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{28}}$ = $\left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} - 2 y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{29}}$ = $- x_{6} \, \mathbf{a}_{1}- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{30}}$ = $- y_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{31}}$ = $- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{6} + 2 y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{32}}$ = $x_{6} \, \mathbf{a}_{1}+\left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) Na III
$\mathbf{B_{33}}$ = $x_{7} \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{7} + y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{7} - y_{7}\right) \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{34}}$ = $- y_{7} \, \mathbf{a}_{1}+\left(x_{7} - y_{7}\right) \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{7} - 2 y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{35}}$ = $- \left(x_{7} - y_{7}\right) \, \mathbf{a}_{1}- x_{7} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{7} - y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{7} \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{36}}$ = $- x_{7} \, \mathbf{a}_{1}- y_{7} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{7} + y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{7} - y_{7}\right) \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{37}}$ = $y_{7} \, \mathbf{a}_{1}- \left(x_{7} - y_{7}\right) \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{7} + 2 y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{7} \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{38}}$ = $\left(x_{7} - y_{7}\right) \, \mathbf{a}_{1}+x_{7} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{7} - y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{7} \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{39}}$ = $y_{7} \, \mathbf{a}_{1}+x_{7} \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{7} + y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{7} - y_{7}\right) \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{40}}$ = $\left(x_{7} - y_{7}\right) \, \mathbf{a}_{1}- y_{7} \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{7} - 2 y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{7} \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{41}}$ = $- x_{7} \, \mathbf{a}_{1}- \left(x_{7} - y_{7}\right) \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{7} - y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{7} \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{42}}$ = $- y_{7} \, \mathbf{a}_{1}- x_{7} \, \mathbf{a}_{2}- \left(z_{7} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{7} + y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{7} - y_{7}\right) \,\mathbf{\hat{y}}- c \left(z_{7} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{43}}$ = $- \left(x_{7} - y_{7}\right) \, \mathbf{a}_{1}+y_{7} \, \mathbf{a}_{2}- \left(z_{7} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{7} + 2 y_{7}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{7} \,\mathbf{\hat{y}}- c \left(z_{7} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) O I
$\mathbf{B_{44}}$ = $x_{7} \, \mathbf{a}_{1}+\left(x_{7} - y_{7}\right) \, \mathbf{a}_{2}- \left(z_{7} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{7} - y_{7}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{7} \,\mathbf{\hat{y}}- c \left(z_{7} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (12i) O I

References

  • L. Viciu, Q. Huang, E. Morosan, H. W. Zandbergen, N. I. Greenbaum, T. McQueen, and R.J.Cav, Structure and basic magnetic properties of the honeycomb lattice compounds Na$_{2}$Co$_{2}$TeO$_{6}$ and Na$_{3}$Co$_{2}$SbO$_{6}$, J. Solid State Chem. 180, 1060–1067 (2007), doi:10.1016/j.jssc.2007.01.002.
  • G. Xiao, Z. Xia, W. Zhang, X. Yue, S. Huang, X. Zhang, F. Yang, Y. Song, M. Wei, H. Deng, and D. Jiang, Crystal Growth and the Magnetic Properties of Na$_{2}$Co$_{2}$TeO$_{6}$ with Quasi-Two-Dimensional Honeycomb Lattice, Crystal Growth and Design 19, 2658–2662 (2019), doi:10.1021/acs.cgd.8b01770.

Prototype Generator

aflow --proto=A2B13C6D_hP44_182_bc_a2i_i_d --params=$a,c/a,x_{5},y_{5},z_{5},x_{6},y_{6},z_{6},x_{7},y_{7},z_{7}$

Species:

Running:

Output: