Q 19: Ferric oxide crystallises in a hexagonal close–packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Q 19: Ferric oxide crystallises in a hexagonal close–packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Solution:

Suppose the number of oxide (O2–) ions = N.

Number of octahedral void = number of anions

So that the number of octahedral voids = N

Given that

Two out of every three octahedral holes are occupied by ferric ions.

So that the number of ferric (Fe3+) ions = 2N/3

The ratio of the number of Fe3+ ions to the number of O²⁻ ions,

                                    Fe³⁺: O²⁻ = 2N/3 : N

Multiply 3 and divide by N we get

                                    Fe3+ : O2− = 2 : 3

Hence, the formula of the ferric oxide is Fe2O3.