Question

In an ionic solid anions are arranged in hcp array and cations occupy $\frac{2}{3}$ of octahedral voids. What is the formula of ionic compound? [consider A = cation; B = anion]}

• A. AB
• B. $A_2B_3$
• C. $A_3B_2$
• D. $AB_3$

Hint:

Number of Octahedral Voids = $N$; Number of Tetrahedral Voids = $2N$.

Answer 1

The Correct Option is B

Step 1: Void Analysis
In hcp or ccp, if the number of anions (B) is $N$, then the number of octahedral voids is also $N$.

Step 2: Calculation
- Number of Anions (B) = $N$. - Number of Cations (A) = $\frac{2}{3} \times (\text{Octahedral voids}) = \frac{2}{3}N$.

Step 3: Ratio
Ratio A : B = $\frac{2}{3}N : N = \frac{2}{3} : 1 = 2 : 3$.

Step 4: Conclusion
The formula is $A_2B_3$.
Final Answer: (B)