Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB6C4_tP22_137_a_df_g-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/ZQRP
or https://aflow.org/p/AB6C4_tP22_137_a_df_g-001
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Li$_{6}$CoO$_{4}$ Structure: AB6C4_tP22_137_a_df_g-001

Picture of Structure; Click for Big Picture
Prototype CoLi$_{6}$O$_{4}$
AFLOW prototype label AB6C4_tP22_137_a_df_g-001
ICSD 62688
Pearson symbol tP22
Space group number 137
Space group symbol $P4_2/nmc$
AFLOW prototype command aflow --proto=AB6C4_tP22_137_a_df_g-001
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{4}, \allowbreak z_{4}$

Other compounds with this structure

F$_{6}$OCd$_{4}$,  F$_{6}$OMn$_{4}$,  Li$_{6}$CoO$_{4}$,  Li$_{6}$CoZn$_{4}$,  Li$_{6}$FeO$_{4}$,  Li$_{6}$IrN$_{4}$,  Li$_{6}$MnO$_{4}$,  Li$_{6}$NiO$_{4}$,  Li$_{6}$RuN$_{4}$,  Li$_{6}$WN$_{4}$,  Na$_{6}$MgO$_{4}$


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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (2a) Co I
$\mathbf{B_{2}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (2a) Co I
$\mathbf{B_{3}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4d) Li I
$\mathbf{B_{4}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4d) Li I
$\mathbf{B_{5}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (4d) Li I
$\mathbf{B_{6}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4d) Li I
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{8}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{9}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{12}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{13}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{14}}$ = $x_{3} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8f) Li II
$\mathbf{B_{15}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+a y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{16}}$ = $\frac{1}{4} \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{17}}$ = $- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{18}}$ = $y_{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a y_{4} \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{19}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{20}}$ = $\frac{3}{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}- a y_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{21}}$ = $\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I
$\mathbf{B_{22}}$ = $- y_{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a y_{4} \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (8g) O I

References

  • R. Luge and R. Hoppe, Ein neues Cobaltat mit Inselstruktur: Li$_{6}$[CoO$_{4}$], Z. Anorganische und Allgemeine Chemie 534, 61–68 (1986), doi:10.1002/zaac.19865340308.

Found in

  • A. Jain, S. P. Ong, G. Hautier, W. Chen, W. D. Richards, S. Dacek, S. Cholia, D. Gunter, G. D. Ceder, and K. A. Persson, {Commentary: The Materials Project: A materials genome approach to accelerating materials innovation}, APL Materials 1, 011002 (2013), doi:10.1063/1.4812323.

Prototype Generator

aflow --proto=AB6C4_tP22_137_a_df_g --params=$a,c/a,z_{2},x_{3},y_{4},z_{4}$

Species:

Running:

Output: