Ferric oxide crystalliizes 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.

Explanation

Suppose the number of oxide ions `(O^(2-))` in the packing =n
`therefore` No. of octahedral voids =n
As 2/3rd of the octahedral voids are occupied by ferric ions, therefore , number of ferric ions present `=2/3xxn="2n"/3`
`therefore` Ratio of `Fe^(3+) : O^(2-)="2n"/3 : n=2:3`
Hence, the formula of ferric oxide is `Fe_2O_3`