Q.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.

Let oxide ions $= N$ (hcp). Number of octahedral holes $= N$. Ferric ions occupy $\tfrac23 N$ …

Step 1 – Count oxide ions and octahedral holes.

Let the number of $O^{2-}$ ions in the hcp array $= N$. The number of octahedral holes in a close-packed array equals the number of packing particles:

Step 2 – Number of ferric ions.

Two out of every three octahedral holes are filled by $Fe^{3+}$:

$\text{Number of } Fe^{3+} = \frac{2}{3}N$ …

Use the close-packing rule that octahedral holes equal the number of packing (oxide) ions; take the $2/3$ occupanc …

• Using tetrahedral count ($2N$) instead of octahedral ($N$). …