A metal oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three…
A metal oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by metal ions. The formula of metal oxide is
$M O$
$M_3 \mathrm{O}_4$
$M_2 \mathrm{O}_5$
$M_2 \mathrm{O}_3$
Solution
Unit cell of hcp structure $=6$ number of oxide ions per unit cell $=6$ number of octahedral voids $=6$. Since, ferric ions occupy only two out of every three octahedral voids, number of octahedral holes occupied by ferric ions $\frac{2}{3} \times 6=4$ stoichiometric ratio of metal and ion $=4: 6=2: 3$.
Hence, formula of metal oxide is $M_2 \mathrm{O}_3$.