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]

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Let the number of anions (B) forming the ccp array be $N$ .

The number of tetrahedral voids generated in a ccp arrangement is $2 N$ .

Since cations (A) occupy $\frac{1}{3}$ of the tetrahedral voids, the number of cations (A) is $\frac{1}{3} \times 2 N = \frac{2 N}{3}$ .

The ratio of the number of cations (A) to the number of anions (B) is $\frac{2 N}{3} : N = 2 : 3$ .

Therefore, the formula of the ionic compound is $A_{2} B_{3}$ .

Answer: $A_{2} B_{3}$

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