Question

A crystal is made up of particles $X,Y$ and $Z$. $X$ forms FCC packing. $Y$ occupies all the octahedral voids of $X$ and $Z$ occupies all the tetrahedral voids of $X$. If all the particles along one body diagonal are removed, then the formula of the crystal would be:

• A. $XY{Z}_2$
• B. $X_{2}Y{Z}_2$
• C. $X_8{Y}_4{Z}_5$
• D. $X_5{Y}_4{Z}_8$

Answer 1

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

For fcc, number of $X$ atoms = 4/unit cell
Number of tetrahedral voids(TVs) $=8Z$
Number of octahedral voids (OVs) $=4Y$
Number of atoms removed along one body diagonal
$=2X$ (corner) and $2Z$ (TVs) and $1Y$ (OV at body centre)
$\therefore$ Number of $X$ atoms left $=4-(2\times \frac{1}{8})=\frac{15}{4}$
Number of $Y$ atoms left $=4-(1\times 1)=3$
Number of $Z$ atoms left $=8-(2\times 1)=6$
The simplest formula $=X_{15/4}{Y}_3{Z}_6\Rightarrow X_{15}{Y}_{12}{Z}_{24}$
$\Rightarrow X_5{Y}_4{Z}_8$