Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A18B2C3_cF184_227_fg_d_ac-001

This structure originally had the label A18B2C3_cF184_227_fg_d_ac. Calls to that address will be redirected here.

If you are using this page, please cite:
D. Hicks, M.J. Mehl, M. Esters, C. Oses, O. Levy, G.L.W. Hart, C. Toher, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 3, Comp. Mat. Sci. 199, 110450 (2021). (doi=10.1016/j.commatsci.2021.110450)

Links to this page

https://aflow.org/p/3DEV
or https://aflow.org/p/A18B2C3_cF184_227_fg_d_ac-001
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Mg$_{3}$Cr$_{2}$Al$_{18}$ Structure: A18B2C3_cF184_227_fg_d_ac-001

Picture of Structure; Click for Big Picture
Prototype Al$_{18}$Cr$_{2}$Mg$_{3}$
AFLOW prototype label A18B2C3_cF184_227_fg_d_ac-001
ICSD 57659
Pearson symbol cF184
Space group number 227
Space group symbol $Fd\overline{3}m$
AFLOW prototype command aflow --proto=A18B2C3_cF184_227_fg_d_ac-001
--params=$a, \allowbreak x_{4}, \allowbreak x_{5}, \allowbreak z_{5}$

Other compounds with this structure

LaCr$_{2}$Al$_{20}$,  CeCr$_{2}$Al$_{20}$,  PrCr$_{2}$Al$_{20}$,  SmCr$_{2}$Al$_{20}$,  YbCr$_{2}$Al$_{20}$,  CeTi$_{2}$Al$_{20}$,  PrTi$_{2}$Al$_{20}$,  SmTi$_{2}$Al$_{20}$,  YbTi$_{2}$Al$_{20}$,  CeV$_{2}$Al$_{20}$,  GdV$_{2}$Al$_{20}$,  LaV$_{2}$Al$_{20}$,  PrV$_{2}$Al$_{20}$,  SmV$_{2}$Al$_{20}$,  CeNi$_{2}$Cd$_{20}$,  GdNi$_{2}$Cd$_{20}$,  LaNi$_{2}$Cd$_{20}$,  NdNi$_{2}$Cd$_{20}$,  PrNi$_{2}$Cd$_{20}$,  SmNi$_{2}$Cd$_{20}$,  YNi$_{2}$Cd$_{20}$,  CePd$_{2}$Cd$_{20}$,  PrPd$_{2}$Cd$_{20}$,  SmPd$_{2}$Cd$_{20}$,  UOs$_{2}$Zn$_{20}$


  • (Samson, 1958) gives the atomic coordinates in terms of Setting 1 of space group $F4dm$ #227. We have shifted this to the standard Setting 2, where the inversion site of the lattice is at the origin.
  • If the (8a), (16c) and (16d) sites are occupied by the same type of atom this becomes the Zn$_{22}$Zr structure.
  • In the ternary compounds LnMX$_{20}$, the rare earth (Ln) metal occupies the (8a) site, the transition metal (M) the (16d) site, and X=Al,Cd,Zn occupies the (16c), (48gf) and (96g) sites. These compounds are sometimes listed under the CeCr$_{2}$Al$_{20}$ prototype.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{1}{8} \, \mathbf{a}_{1}+\frac{1}{8} \, \mathbf{a}_{2}+\frac{1}{8} \, \mathbf{a}_{3}$ = $\frac{1}{8}a \,\mathbf{\hat{x}}+\frac{1}{8}a \,\mathbf{\hat{y}}+\frac{1}{8}a \,\mathbf{\hat{z}}$ (8a) Mg I
$\mathbf{B_{2}}$ = $\frac{7}{8} \, \mathbf{a}_{1}+\frac{7}{8} \, \mathbf{a}_{2}+\frac{7}{8} \, \mathbf{a}_{3}$ = $\frac{7}{8}a \,\mathbf{\hat{x}}+\frac{7}{8}a \,\mathbf{\hat{y}}+\frac{7}{8}a \,\mathbf{\hat{z}}$ (8a) Mg I
$\mathbf{B_{3}}$ = $0$ = $0$ (16c) Mg II
$\mathbf{B_{4}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}$ (16c) Mg II
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (16c) Mg II
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{1}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (16c) Mg II
$\mathbf{B_{7}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ (16d) Cr I
$\mathbf{B_{8}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ (16d) Cr I
$\mathbf{B_{9}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (16d) Cr I
$\mathbf{B_{10}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (16d) Cr I
$\mathbf{B_{11}}$ = $- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+\frac{1}{8}a \,\mathbf{\hat{y}}+\frac{1}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{12}}$ = $x_{4} \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{2}- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{4}\right) \,\mathbf{\hat{x}}+\frac{1}{8}a \,\mathbf{\hat{y}}+\frac{1}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ = $\frac{1}{8}a \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}+\frac{1}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{14}}$ = $- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $\frac{1}{8}a \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{15}}$ = $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ = $\frac{1}{8}a \,\mathbf{\hat{x}}+\frac{1}{8}a \,\mathbf{\hat{y}}+a x_{4} \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{16}}$ = $- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{4}\right) \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ = $\frac{1}{8}a \,\mathbf{\hat{x}}+\frac{1}{8}a \,\mathbf{\hat{y}}- a \left(x_{4} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{17}}$ = $\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $\frac{3}{8}a \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{3}{4}\right) \,\mathbf{\hat{y}}+\frac{3}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{18}}$ = $- x_{4} \, \mathbf{a}_{1}+\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ = $\frac{3}{8}a \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}+\frac{3}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{19}}$ = $- x_{4} \, \mathbf{a}_{1}+\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{2}+\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{3}{4}\right) \,\mathbf{\hat{x}}+\frac{3}{8}a \,\mathbf{\hat{y}}+\frac{3}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{20}}$ = $\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{3}{8}a \,\mathbf{\hat{y}}+\frac{3}{8}a \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{21}}$ = $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ = $\frac{3}{8}a \,\mathbf{\hat{x}}+\frac{3}{8}a \,\mathbf{\hat{y}}- a x_{4} \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{22}}$ = $\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{3}{4}\right) \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ = $\frac{3}{8}a \,\mathbf{\hat{x}}+\frac{3}{8}a \,\mathbf{\hat{y}}+a \left(x_{4} + \frac{3}{4}\right) \,\mathbf{\hat{z}}$ (48f) Al I
$\mathbf{B_{23}}$ = $z_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}+a z_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{24}}$ = $z_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+a z_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{25}}$ = $\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{1}- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}- a \left(z_{5} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{26}}$ = $- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{y}}- a \left(z_{5} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{27}}$ = $\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a z_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}+a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{28}}$ = $- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a z_{5} \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{y}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{29}}$ = $z_{5} \, \mathbf{a}_{1}+\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{2}- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(z_{5} - \frac{1}{4}\right) \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{30}}$ = $z_{5} \, \mathbf{a}_{1}- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{3}$ = $- a \left(z_{5} - \frac{1}{4}\right) \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{31}}$ = $z_{5} \, \mathbf{a}_{1}+\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+a z_{5} \,\mathbf{\hat{y}}+a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{32}}$ = $z_{5} \, \mathbf{a}_{1}- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{x}}+a z_{5} \,\mathbf{\hat{y}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{33}}$ = $- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}- a \left(z_{5} - \frac{1}{4}\right) \,\mathbf{\hat{y}}- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{34}}$ = $\left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}- \left(2 x_{5} + z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{4}\right) \,\mathbf{\hat{x}}- a \left(z_{5} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{35}}$ = $- z_{5} \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}+\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{y}}- a z_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{36}}$ = $- z_{5} \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}- a z_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{37}}$ = $- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{1}+\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}+a \left(z_{5} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{38}}$ = $\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+a \left(z_{5} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{39}}$ = $- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}+\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{x}}+a \left(z_{5} + \frac{1}{4}\right) \,\mathbf{\hat{y}}- a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{40}}$ = $\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+a \left(z_{5} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{41}}$ = $- z_{5} \, \mathbf{a}_{1}- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- a z_{5} \,\mathbf{\hat{y}}- a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{42}}$ = $- z_{5} \, \mathbf{a}_{1}+\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{x}}- a z_{5} \,\mathbf{\hat{y}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{43}}$ = $- z_{5} \, \mathbf{a}_{1}- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{2}+\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(z_{5} + \frac{1}{4}\right) \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{y}}- a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{44}}$ = $- z_{5} \, \mathbf{a}_{1}+\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{3}$ = $a \left(z_{5} + \frac{1}{4}\right) \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{45}}$ = $\left(2 x_{5} + z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a z_{5} \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+a \left(x_{5} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ (96g) Al II
$\mathbf{B_{46}}$ = $- \left(2 x_{5} - z_{5}\right) \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a z_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}- a x_{5} \,\mathbf{\hat{z}}$ (96g) Al II

References

  • S. Samson, The Crystal Structure of the Intermetallic Compound Mg$_{3}$Cr$_{2}$Al$_{18}$, Acta Cryst. 11 (1958), doi:10.1107/S0365110X58002425.

Found in


Prototype Generator

aflow --proto=A18B2C3_cF184_227_fg_d_ac --params=$a,x_{4},x_{5},z_{5}$

Species:

Running:

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