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.

(D) Let the number of oxide $(O^{2-})$ ions be $x$.
Since the oxide ions form a hexagonal close-packed $(HCP)$ array,the number of octahedral voids is equal to the number of oxide ions,which is $x$.
It is given that two out of every three octahedral holes are occupied by ferric ions.
So,the number of ferric $(Fe^{3+})$ ions $= \frac{2}{3}x$.
Therefore,the ratio of the number of $Fe^{3+}$ ions to the number of $O^{2-}$ ions is $Fe^{3+} : O^{2-} = \frac{2}{3}x : x = \frac{2}{3} : 1 = 2 : 3$.
Hence,the formula of the ferric oxide is $Fe_{2}O_{3}$.