Ferric oxide crystallizes 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

In hexagonal close packed arrangement of oxide ions, oxide ions are present at every corner of the cube
`:.` No. of oxide ions per unit cell = `8 xx 1/8` = 1
For each oxide ion, there is one octahedral hole and two out of three are occupied by `Fe^(3+)` ions.
`:.` No. of `Fe^(3+)` ions present per octahedral void = `2/3 Fe^(3+)`
Thus, formula of ferric oxide should be `Fe_(2//3)O_1` or `Fe_2O_3` (by taking the number of atoms in formula as whole number).