JEE Advanced 2022 Paper 2 Q15 Chemistry Physical Chemistry Atomic Structure Hard

JEE Advanced 2022 Paper 2 · Q15 · Atomic Structure

Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to

  1. A. 25
  2. B. 35
  3. C. 55
  4. D. 75
Reveal answer + step-by-step solution

Correct answer:B

Solution

Step 1 — effective number of atoms per unit cell. FCC lattice: 4 atoms/unit cell (8 corners × 1/8 + 6 faces × 1/2 = 1 + 3 = 4). Alternate tetrahedral voids: there are 8 tetrahedral voids in an fcc unit cell; alternate ⇒ 4 are occupied by X. Total X per unit cell = 4 + 4 = 8.

Step 2 — radius constraint. Along the body diagonal of one octant (where the tetrahedral void sits), atom-to-atom contact constrains $r$: the corner atom and the tetrahedral-void atom touch along the (1/4, 1/4, 1/4) body diagonal segment of length $a\sqrt{3}/4$. Setting $2r = a\sqrt{3}/4$ gives $r = a\sqrt{3}/8$.

Step 3 — packing fraction. Volume of 8 atoms $= 8 \times (4/3)\pi r^3 = 8 \times (4/3)\pi (a\sqrt{3}/8)^3 = 8 \times (4/3)\pi \times (3\sqrt{3}\,a^3/512) = (4/3)\pi \times (3\sqrt{3}\,a^3/64)$. Volume of unit cell = $a^3$. Packing fraction $= \frac{(4/3)\pi (3\sqrt{3}/64)}{1} = \frac{\pi \sqrt{3}}{16} \approx \frac{3.1416 \times 1.732}{16} \approx \frac{5.441}{16} \approx 0.34 = 34\% \approx 35\%$.

Closest option = (B) 35%.

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