Question

A crystalline solid is made of $X,Y$ and $Z$ elements. Atoms of $X$ form fcc packing; atoms of $Y$ occupy octahedral voids while atoms of $Z$ occupy tetrahedral voids. What will be the simplest formula of solid is 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 $X$ (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 $Y=4$

Number of atoms of $Z=8$

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

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

Number of atoms of $Z$ will be $=8-2=6$

$X:Y:Z$

$\frac{15}{4}:3:6$

$5:4:8$

$\therefore$ Simplest formula will be $X_5{Y}_4{Z}_8$.

Hence, the correct option is $\text{A}$