Question

In ionic solid, anions are arranged in ccp array and cations occupy all the octahedral voids. What is the formula of ionic compound? [consider \(\mathrm{A}=\) cation; \(\mathrm{B}=\) anion]

• A. \(\mathrm{AB}_{3}\)
• B. \(\mathrm{A}_{2}\text{ }\mathrm{B}_{3}\)
• C. \(\mathrm{AB}_{4}\)
• D. AB

Hint:

In ccp, number of octahedral voids equals the number of atoms in the array.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In a ccp (fcc) lattice, if there are \(N\) anions, there are \(N\) octahedral voids and \(2N\) tetrahedral voids.

Step 2: Count particles
Let anions \(\text{B} = 4\) per unit cell. Octahedral voids \(= 4\), all occupied by cations \(\text{A}\), so \(\text{A} = 4\).

Step 3: Formula
Ratio \(\text{A} : \text{B} = 4 : 4 = 1 : 1\), so the formula is AB.

Final Answer:
The formula is AB. \[ \boxed{\text{(D) }\text{AB}} \]