A crystalline solid is made of $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ elements. Atoms of $\mathrm{X}$ form $\mathrm{FCC}$packing, $\mathrm{Y}$ occupying octahedral voids, while atoms of $\mathrm{Z}$ occupying tetrahedral voids. What will be the simplest formula of the solid if atoms along any one body diagonal are removed?

Number of X's atoms (at packing site, i.e., at corners and face-centers)
$8\times \frac{1}{8}+6\times \frac{1}{2}=4$
Number of $\mathrm{Y}=4$ atoms
Number of $\mathrm{Z}=8$ atoms
Two $\mathrm{X}$ one $\mathrm{Y}$ and two $\mathrm{Z}$atoms will be located and eliminated along one body diagonal.
Atoms for $\mathrm{X}$will total $=4-\frac{1}{8}\times 2=\frac{15}{4}$
The atom count of $\mathrm{Y}$will be $4-1=3$.
$\mathrm{Z}$'s atom count will be $=8-2=6$
$\mathrm{X}:\mathrm{Y}:\mathrm{Z}$
$\frac{15}{4}:3:6$

$5:4:8$

∴ The easiest formula is ${\mathrm{X}}_5{\mathrm{Y}}_4{\mathrm{Z}}_8$.