Question

In ionic solid, anions are arranged in hcp array and cations occupy \(\frac{1}{2}\) tetrahedral voids. What is the formula of ionic compound ?
[Consider A = Cation ; B = anion]

• A. \(\text{AB}_2\)
• B. \(\text{AB}\)
• C. \(\text{A}_2\text{B}\)
• D. \(\text{AB}_3\)

Hint:

In hcp, N anions give 2N tetrahedral voids; half of them occupied means N cations.

Answer 1

The Correct Option is B

Step 1: Understand the concept
In a hexagonal close packed (hcp) array, if there are \(N\) atoms (here anions B), there are \(N\) octahedral voids and \(2N\) tetrahedral voids.

Step 2: Count the anions and voids
Take \(N\) anions of B. The number of tetrahedral voids is \(2N\).

Step 3: Count the cations
Cations A occupy one half of the tetrahedral voids, so the number of A is \(\frac{1}{2} \times 2N = N\).

Step 4: Write the formula
The ratio is A : B = \(N : N = 1 : 1\), so the formula is AB. Option (A) \(\text{AB}_2\) would need all tetrahedral voids to be half empty with fewer cations, option (C) \(\text{A}_2\text{B}\) needs all tetrahedral voids filled, and (D) does not fit a void count of 2N.

Final Answer:
The cation to anion ratio is 1 : 1. This is option (B). \[ \boxed{\text{(B) }\text{AB}} \]