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

Step-by-step solution

The hexagonal close-packed (hcp) array of anions B contains 6 atoms per unit cell. In close-packed structures, the number of octahedral voids equals the number of atoms, giving 6 voids. Since cations A occupy 2 3 of these voids, the number of A cations is 2 3 6 = 4 . The A:B ratio becomes 4:6, which simplifies to 2:3. The formula is therefore A ₂ B ₃ . Final answer: B