Question

In ionic solid, anions are arranged in ccp array and cations occupy $1/3$ tetrahedral voids. What is the formula of ionic compound?
[Consider $\text{A} = \text{cation}; \text{B} = \text{anion}$]

• A. $\text{AB}_3$
• B. $\text{A}_2\text{B}_3$
• C. $\text{A}_3\text{B}_2$
• D. $\text{AB}_4$

Hint:

Tet voids = $2 \times$ Atoms; Oct voids = Atoms.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a ccp (or fcc) lattice, the number of tetrahedral voids is twice the number of atoms in the lattice.
Step 2: Key Formula or Approach:
If number of anions is $\text{N}$, then number of tetrahedral voids $= 2\text{N}$.
Step 3: Detailed Explanation:
- Anions (B) form ccp lattice. Let the number of anions be $\text{N}$.
- Number of tetrahedral voids $= 2\text{N}$.
- Cations (A) occupy $1/3$ of these voids.
- Number of cations (A) $= 1/3 \times 2\text{N} = \frac{2\text{N}}{3}$.
Ratio of $\text{A} : \text{B} = \frac{2\text{N}}{3} : \text{N} = 2 : 3$.
The formula is $\text{A}_2\text{B}_3$.
Step 4: Final Answer:
The formula of the ionic compound is $\text{A}_2\text{B}_3$.