# 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.

$\begin{array}{1 1}Fe_2O_3\\Fe_3O_2\\FeO\\Fe_4O_3\end{array}$

Suppose the number of oxide ions ($O^{2-})$ in the packing = N
$\therefore$ As 2/3$^{rd}$ of the octahedral voids are occupied by ferric ions,therefore number of ferric ions present.
$\Rightarrow \large\frac{2}{3}$$\times N=\large\frac{2N}{3} Ratio of Fe^{3+} : O^{2-}=\large\frac{2N}{3}$$ : N$
$\Rightarrow 2 : 3$
Hence,the formula of ferric oxide =$Fe_2O_3$