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: Ratios of voids
For each close packed anion there are 2 tetrahedral voids and 1 octahedral void.

Step 2: Use per-anion counting
Per anion B, tetrahedral voids = 2. Half occupied means 1 void filled, so 1 cation A per anion B.

Step 3: Formula
$\text{A}_1\text{B}_1$, written AB. This is the zinc blende / wurtzite type of arrangement.

Step 4: Contrast
If all tetrahedral voids were filled, we would get $\text{A}_2\text{B}$, which is option (C), but the question says only one half are filled.

Final Answer:
Half of the 2N tetrahedral voids gives N cations. This is option (B). \[ \boxed{\text{(B) }\text{AB}} \]