# Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions.

**Question:**

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:**

Let the number of oxide (O2−) ions be *x*.

So, number of octahedral voids = *x*

It is given that two out of every three octahedral holes are occupied by ferric ions.

So, number of ferric $\left(\mathrm{Fe}^{3+}\right)$ ions $=\frac{2}{3} x$

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

$\mathrm{Fe}^{3+}: \mathrm{O}^{2-}=\frac{2}{3} x: x$

$=\frac{2}{3}: 1$

= 2 : 3

Hence, the formula of the ferric oxide is Fe2O3.

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