# n Factor

(a) In Non Redox Change :

(i) n-factor for element : Valency of the element

(ii) For Acids : Acids will be treated as species which furnish H+ ions when dissolved in a solvent.

The n factor of an acid is the no. of acidic H+ ions that a molecule of the acid would give when dissolved in a solvent (Basicity).

e.g. for HCl (n = 1) , HNO3 (n = 1), H2SO4 (n = 2), H3PO4 (n = 3) and H3PO3 ( n= 2)

(iii) For Bases : Bases will be treated as species which furnish OH ions when dissolved in a solvent. The n factor of a base is the no. of OH ions that a molecule of the base would give when dissolved in a solvent (Acidity).

e.g. NaOH (n = 1), Ba(OH)2 (n = 2), Al(OH)3 (n = 3), etc.

(iv) For Salts: A salt reacting such that no atom of the salt undergoes any change in oxidation state.

e.g. 2AgNO3 + MgCl2 –> Mg(NO3)2 + 2AgCl

In this reaction it can be seen that the oxidation state of Ag, N, O, Mg and Cl remains

the same even in the product. The n factor for such a salt is the total charge on cation or anion.

(b) In Redox Change: For oxidising agent or reducing agent n   factor is the change in oxidation number per mole of the substance.

Illustration : Calculate the n   factor in the following chemical changes.

(i) KMnO4 H+—> Mn2+

(ii) KMnO4 H20 —> Mn4+

(iii) KMnO4 oH—> Mn6+

(iv) K2Cr2O7 H+–> Cr3+

(v) C2O42- –> CO2

(vi) FeSO4 –> Fe2O3

(vii) Fe2O3 —> FeSO4

Solution:

(i) In this reaction, KMnO4 which is an oxidising agent, itself gets reduced to Mn2+ under acidic conditions.

n = |1×(+7) – 1×(+2) | = 5

(ii) In this reaction, KMnO4 gets reduced to Mn4+ under neutral or slightly (weakly) basic conditions.

n = |1×(+7) – 1×(+4) | = 3

(iii) Here KMnO4 gets reduced to Mn6+ under basic conditions.

n = |1×(+7)- 1×(+6) | = 1

(iv) Here K2Cr2O7 which acts as an oxidising agent gets reduced to Cr+3 under acidic conditions. (It does not react under basic conditions).

n = |2×(+6)- 2×(+3) | = 6

(v) C2O42- (oxalate ion) gets oxidised to CO2 when it is reacted with an oxidising agent.

n = |2×(+3)- 2×(+4) | = 2

(vi) Ferrous ions get oxidised to ferric ions

n = |1×(+2)- 1×(+3) | = 1

(vii) Here ferric ions are getting reduced to ferrous ions

n = |2×(+3)- 2×(+2) | = 2