Question

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 ferric oxide.

Answer 1

Hint: In hcp, the ratio of octahedral holes to the number of anions is 1:1. Assume ‘x’ to be the number of oxide ions present. Accordingly, the total number of octahedral holes will be ‘x’. Find out from the question of how many ferric ions occupy octahedral ions. Take the ratio of the voids occupied cations and anions to get the answer.

Complete step by step solution:
- In a hexagonal close-packed array or cubic close packing, the ratio of octahedral holes to the number of anions is 1:1.
- Let us assume ‘x’ to be the total number of oxide ions present in the hcp lattice.
- Therefore, the total number of octahedral voids occupying the octahedral voids will also be ‘x’.
- According to the question, two out of three total octahedral voids are occupied by ferric ions. That means, ferric ions occupy ${}^{2}/{}_{3}x$
- Therefore, the ratio of holes occupied by ferric ions to that of oxide ions will be ${}^{2}/{}_{3}x:x$
- So, we obtain the ratio 2:3 for ferric ions to oxide ions.

Therefore, the molecular formula of ferric oxide will be $F{{e}_{2}}{{O}_{3}}$.

Note: Remember in an hcp or ccp lattice, the number of anions is equal to the total number of octahedral voids. To find the molecular formula, just take the ratio of voids occupied by cations to that of anions. That value will be the subscript of atoms present in the compound to give molecular formula.