QJEEAdv22Chdcee: JEE Advanced 2022 Chemistry - Class 12 JEEAdv22Chdcee Solution: Atom occupies the fcc lattice sites as well as alternate…

Atom $X$ occupies FCC lattice sites as well as alternate tetrahedral voids of FCC.

In FCC, tetrahedral voids are 8 (in a unit cell)

Hence

Atom $X$ in a unit cell (FCC lattice sites) $=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$

Atom $X$ in a unit cell (in T.V.) $=\frac{1}{2} \times 8=4$

Total atom $\mathrm{X}$ in one unit cell $=8$

For relation between a and r, since T.V. forms at $\frac{1}{4}$ th of body diagonal,

$\begin{aligned}& \frac{a \sqrt{3}}{4}=2 r \\\\&\Rightarrow a=\frac{8 r}{\sqrt{3}}\end{aligned}$

Packing efficiency

$=\frac{8 \times \frac{4}{3} \pi r^3}{a^3} \times 100=\frac{8 \times \frac{4}{3} \pi r^3}{\left(\frac{8 r}{\sqrt{3}}\right)^3} \times 100 \simeq 35 \%$