(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