JEE Advanced Chemistry Question (2022) — Solution

Step-by-step solution

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 × 1 8 + 6 × 1 2 = 4 Atom X in a unit cell (in T.V.) = 1 2 × 8 = 4 Total atom X  in one unit cell = 8 For relation between a and r, since Tetrahedral void  forms at 1 4 th of body diagonal, a 3 4 = 2 r a = 8 r 3 Packing efficiency = 8 × 4 3 πr 3 a 3 × 100 = 8 × 4 3 πr 3 8 r 3 3 × 100 = 35 %