Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: ABC3_cP20_198_a_a_b-001

This structure originally had the label ABC3_cP20_198_a_a_b. 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/QEXW
or https://aflow.org/p/ABC3_cP20_198_a_a_b-001
or PDF Version

Sodium Chlorate (NaClO$_{3}$, $G0_{3}$) Structure: ABC3_cP20_198_a_a_b-001

Picture of Structure; Click for Big Picture
Prototype NaClO$_{3}$
AFLOW prototype label ABC3_cP20_198_a_a_b-001
Strukturbericht designation $G0_{3}$
Mineral name sodium chlorate
ICSD 16714
Pearson symbol cP20
Space group number 198
Space group symbol $P2_13$
AFLOW prototype command aflow --proto=ABC3_cP20_198_a_a_b-001
--params=$a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak y_{3}, \allowbreak z_{3}$

Other compounds with this structure

NaBrO$_{3}$


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

Basis vectors

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

References

  • G. N. Ramachandran and K. S. Chandrasekaran, The absolute configuration of sodium chlorate 10, 671–675 (1957), doi:10.1107/S0365110X57002327.

Found in

  • A. A. Kaminskii, S. N. Bagayev, J. Hulliger, H. Eichler, J. Findeisen, and R. Macdonald, Acentric cubic NaClO$_{3}$–a new crystal for Raman lasers, Applied Physics B 67, 157–162 (1998), doi:10.1007/s003400050487.

Prototype Generator

aflow --proto=ABC3_cP20_198_a_a_b --params=$a,x_{1},x_{2},x_{3},y_{3},z_{3}$

Species:

Running:

Output: