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

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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 =$Fe_2O_3$

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