Question

The anions (A) of \( \text{C}_p\text{A}_q \) molecule form an fcc lattice. Cations (C) are positioned at the body center and half of the edge centers. The formula of the molecule is:

• A. CA
• B. \( \text{CA}_2 \)
• C. \( \text{C}_3\text{A}_4 \)
• D. \( \text{C}_5\text{A}_8 \)

Hint:

Use effective atoms per unit cell for lattice positions: corners (1/8), edges (1/4), faces (1/2), body center (1).

Answer 1

The Correct Option is D

- Anions (A) form FCC lattice → Total A per unit cell: \[ \text{8 corners} \Rightarrow \frac{1}{8} \times 8 = 1, \quad \text{6 faces} \Rightarrow \frac{1}{2} \times 6 = 3 \Rightarrow \text{Total = 4 A atoms} \] - Cations (C) occupy:
- 1 body center → 1 atom
- 12 edges, with half occupied → \( \frac{1}{4} \times 12 = 3 \) atoms
(each edge contributes \( \frac{1}{4} \) if only half edges occupied)
\[ \text{Total C atoms} = 1 + 3 = 4 \] So, in one unit cell:
- A = 4
- C = 4
But the formula asks for smallest whole number ratio. Multiplying by 2: \[ \Rightarrow \text{C}_5\text{A}_8 \]