In a normal spinel type structure, the oxide ions are arranged in ccp, whereas 1 / 8 tetrahedral holes are occupied by Zn ²⁺ ions and 50 % of octahedral holes are occupied by Fe ³⁺ ions. The formula of compound is

Step-by-step solution

In a ccp arrangement of oxide ions, let the number of oxide ions ( O²⁻ ) be N . The number of tetrahedral voids is 2N and the number of octahedral voids is N . According to the problem, Zn ²⁺ ions occupy 1/8 of the tetrahedral voids. Therefore, the number of Zn ²⁺ ions is 1 8 2N = 1 4 N . The Fe ³⁺ ions occupy 50 % of the octahedral voids. Therefore, the number of Fe ³⁺ ions is 50 100 N = 1 2 N . The ratio of Zn ²⁺ : Fe ³⁺ : O²⁻ is 1 4 N : 1 2 N : N . Multiplying by 4 to get the simplest integer ratio, we get 1 : 2