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 .

Answer 1

There is one octahedral hole for each atom in hexagonal close packed arrangement. If the number of oxide ions $\left(O^{2-}\right)$ per unit cell is 1, then the number of $F{e}^{3+}$ ions $=\frac{2}{3}\times$ octahedral holes $=\frac{2}{3}\times 1=\frac{2}{3}$ Thus, the formula of the compound $=F{e}_{\frac{2}{3}}{O}_{1}orF{e}_2{O}_3$