Question

Find the n-factor in the following chemical changes.
$FeS{{O}_{4}}\to F{{e}_{2}}{{O}_{3}}$
(A) 2
(B) 1
(C) 3
(D) none of these

Answer 1

Write down the oxidation state of each atom of the reactant and product as well. Find the change in oxidation state per mole of atom to find the n-factor. Add all the changes in oxidation state to obtain the final n-factor.

Complete step by step answer:
We will define the term n-factor for better understanding
-N-factor is the valency factor or conversion factor which indicates the number of moles of electrons lost or gained per mole of reactant.

-For acids, n-factor is defined as the number of ${{H}^{+}}$ replaced in one mole of acid.
Let us write the oxidation state of atoms in the reactant as well as the product.
Reactant: $FeS{{O}_{4}}$
Oxidation state of Fe = +2
Oxidation state of S = +6
Oxidation state of O = -2
Product: $F{{e}_{2}}{{O}_{3}}$
Oxidation state of Fe = +3
Oxidation state of S = +6
Oxidation state of O= -2

We observe that the oxidation states of S and O do not change. Hence n factor of the reaction will be the change in oxidation state of the Fe atom.
$2F{{e}^{2+}}\to 2F{{e}^{3+}}$
The change in oxidation state is 2, however n-factor is defined for 1 mole of reactant. Hence the n-factor becomes 1.
So, the correct answer is “Option B”.

Note: The question does not mention the product of sulfur atoms after the reaction. This is the reason why we do not consider the change in oxidation state of the S atom while calculating n-factor.