In ionic solid, anions are arranged in ccp array and cations occupy 1 / 3 tetrahedral voids. What is the formula of ionic compound? [Consider A = cation ; B = anion]

Step-by-step solution

In a cubic close-packed (ccp) array of anions (B), the number of atoms per unit cell is 4. The number of tetrahedral voids is twice this number, so there are 2 4 = 8 tetrahedral voids. Cations (A) occupy 1 3 of these voids, giving 1 3 8 = 8 3 A ions per unit cell. The ratio of A to B ions is 8 3 : 4 , which simplifies to 2 : 3 by multiplying both sides by 3 and then dividing by 4. The formula of the compound is therefore A ₂ B ₃ , corresponding to option C .