The anions (A) of $\mathrm{C}_{\mathrm{p}} \mathrm{A}_{\mathrm{q}}$ molecule forms an fcc lattice. Cations…

The anions (A) of $\mathrm{C}_{\mathrm{p}} \mathrm{A}_{\mathrm{q}}$ molecule forms an fcc lattice. Cations (C) are positioned at the body center and half of the edge centers. The formula of the molecule is
  1. $\mathrm{CA}$
  2. $\mathrm{CA}_2$
  3. $\mathrm{C}_3 \mathrm{~A}_4$
  4. $\mathrm{C}_5 \mathrm{~A}_8$

Solution

In a unit cell of fcc lattice, $A$ is present at 8 corners and 6 faces. $\therefore \quad$ Effective no. of ${ }^{\prime} \mathrm{A}^{\prime}=\frac{1}{8} \times 8+\frac{1}{2} \times 6=4$ $\mathrm{C}$ is present at body centre and 6 edges. $\therefore \quad$ Effective no. of ' $\mathrm{C}^{\prime}=1+6 \times \frac{1}{4}=\frac{5}{2}$ $\therefore \quad \mathrm{C}_{\frac{5}{2}} \mathrm{~A}_4=\mathrm{C}_5 \mathrm{~A}_8$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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