Question

A crystalline solid is of $\text{X, Y}$ and $\text{Z}$ elements. Atoms of $\text{X}$ form $\text{fcc}$ packing; atoms of $\text{Y}$ occupy octahedral voids while atoms of $\text{Z}$ occupy tetrahedral voids. What will be the simplest formula of a solid if atoms along one body diagonal are removed.

• A. $X_5{Y}_4{Z}_8$
• B. $XYZ$
• C. $X_8{Y}_4{Z}_5$
• D. $X_{2}YZ$

Answer 1

The correct option is A $X_5{Y}_4{Z}_8$

Number of atoms of $\text{X}$ (at packing site, i.e., at corners and face-centres
$=8\times \frac{1}{8}+6\times \frac{1}{2}=4$

Number of atoms of $\text{Y}=4$
Number of atoms of $\text{Z}=8$

Along one body diagonal there will be two $\text{X}$ atoms, one $\text{Y}$ atom and two $\text{Z}$ atoms are found and are removed.

Number of atoms of $\text{X}$ will be $=4-\frac{1}{8}\times 2=\frac{15}{4}$
Number of atoms of $\text{Y}$ will be $4-1=3$
Number of atoms of $\text{Z}$ will be $8-2=6$

$\text{X:Y:Z}$
$\frac{15}{4}:3:65:4:8$
$\therefore$ Simplest formula will be $X_5{Y}_4{Z}_8$