Question

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.

Answer 1

Ferric oxide crystalizes in a hexagonal close packed array of oxide ions.
Two out of every three octahedral holes are occupied by ferric ions.
Let the number of oxide ions = x
Since, two out of every three octahedral holes are fitted by ferric ions.
Thus, voids filled by ferric ions ,Fe^3+ = 2x/3
therefore, number of ferric ions , Fe^3+ = 2x/3
now, Now, the ratio of ferric ions to the oxide ions
i.e. Fe^3+ : O^2- = 2x/3 : x = 2/3 : 1 = 2 : 3
$\bf{hence,formula\:of\:ferric\:oxide\:is \:Fe_2O_3}$

Answer 2

Let the number of oxide (O2−) ions be x.

So, number of octahedral voids = x

It is given that two out of every three octahedral holes are occupied by ferric ions.

So, number of ferric (Fe3+) ions

2x/3

Therefore, ratio of the number of Fe3+ ions to the number of O2− ions,
Fe3+ : O2−

2x/3:x

2/3:1


= 2 : 3

Hence, the formula of the ferric oxide is Fe2O3.


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