Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B3_tP20_136_j_cfg-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/T3N4
or https://aflow.org/p/A2B3_tP20_136_j_cfg-001
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Na$_{3}$Hg$_{2}$ Structure: A2B3_tP20_136_j_cfg-001

Picture of Structure; Click for Big Picture
Prototype Hg$_{2}$Na$_{3}$
AFLOW prototype label A2B3_tP20_136_j_cfg-001
ICSD 104328
Pearson symbol tP20
Space group number 136
Space group symbol $P4_2/mnm$
AFLOW prototype command aflow --proto=A2B3_tP20_136_j_cfg-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak z_{4}$

Other compounds with this structure

Al$_{2}$Zr$_{3}$


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

Basis vectors

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

References

  • J. W. Nielsen and N. C. Baenzizer, The Crystal Structure of NaHg$_3$, NaHg and Na$_3$Hg, Acta Cryst. 7, 277–282 (1954), doi:10.1107/S0365110X54000783.

Prototype Generator

aflow --proto=A2B3_tP20_136_j_cfg --params=$a,c/a,x_{2},x_{3},x_{4},z_{4}$

Species:

Running:

Output: