Question

A solid has three types of atoms $X,Y$ and $Z$. $X$ forms an $FCC$ lattice with $Y$ atoms occupying all the tetrahedral voids and $Z$ atoms occupying half of the octahedral voids. The formula of the solid is

• A. $X_2{Y}_{4}Z$
• B. $X{Y}_2{Z}_4$
• C. $X_4{Y}_{2}Z$
• D. $X_4{Y}_2{Z}_3$

Answer 1

The correct option is A $X_2{Y}_{4}Z$

Given that $X$ forms an fcc lattice. Thus, the number of $X$ atoms $=4$ (number of atoms per unit cell).
Now, since $Y$ occupies all the tetrahedral voids, thus, the number of atoms of $Y=4\times 2=8$ (number of tetrahedral voids $=$ twice the number of octahedral voids).
Finally, we have $Z$ atoms occupying half of the number of octahedral voids. Hence, $Z=\frac{1}{2}\times 4=2$
Thus, the formula of the solid is $X_4{Y}_8{Z}_2$ or $X_2{Y}_{4}Z$.