How to Find the n-Factor in a Disproportionation Reaction | CHEMCA

How to Find the n-Factor in a Disproportionation Reaction

Published by Abhishek Sengar | CHEMCA India

In the Redox Reactions chapter, finding the n-factor (valency factor) is crucial for calculating Equivalent Weight. For a normal molecule that just oxidizes or reduces, finding the n-factor is easy: simply multiply the change in oxidation state by the number of atoms.

But what do you do in a Disproportionation Reaction, where the exact same molecule acts as BOTH the oxidizing agent and the reducing agent simultaneously? Let's decode the secret mathematical formula.

Video Tutorial: The Net n-Factor Formula

Watch Abhishek Sengar sir from CHEMCA break down the classic disproportionation of Chlorine gas and apply the harmonic sum formula to find the net n-factor.

Step-by-Step Mathematical Breakdown

The Reaction:

Cl₂ → Cl^- + ClO₃^-

The Special Formula:
When a single reactant disproportionates, its overall n-factor (n_net) is calculated using the harmonic sum of its individual oxidation and reduction n-factors:

1 / n_net = 1 / n_ox + 1 / n_red
(Or simplified: n_net = (n_ox × n_red) / (n_ox + n_red))

Identify Oxidation States:
Reactant: Cl₂ (Oxidation State = 0)
Reduction Product: Cl^- (Oxidation State = -1)
Oxidation Product: ClO₃^- (Oxidation State = +5)
Calculate n-factor for Reduction (n_red):
Look only at the reduction pathway: Cl₂ → 2Cl^-.
The OS changes from 0 to -1. Change per atom = |0 - (-1)| = 1.

Crucial Mistake to Avoid!
You MUST multiply the change by the number of atoms in the original reactant molecule! Since we start with Cl₂, there are 2 atoms.
n_red = (Change per atom) × (Number of atoms) = 1 × 2 = 2.
Calculate n-factor for Oxidation (n_ox):
Look only at the oxidation pathway: Cl₂ → 2ClO₃^-.
The OS changes from 0 to +5. Change per atom = |0 - 5| = 5.
n_ox = (Change per atom) × (Number of atoms) = 5 × 2 = 10.

Fig: Visually tracking the oxidation and reduction pathways to calculate the harmonic sum.

Apply the Formula:
1 / n_net = 1/2 + 1/10
1 / n_net = 5/10 + 1/10 = 6/10 = 3/5
Reverse the fraction: n_net = 5/3 (or approx 1.67).

Final Answer: The n-factor for Cl₂ in this reaction is 5/3.

Practice Questions for JEE & NEET

Let's apply this exact mathematical trick to the two most common exam variations!

Question 1: Using the n-factor we just calculated, what is the Equivalent Weight (E) of Cl₂ in this reaction, assuming its Molar Mass is M?

Answer: Equivalent Weight = 3M / 5.

Reasoning:

The standard formula for Equivalent Weight is:
Equivalent Weight (E) = Molar Mass (M) / n-factor

Substitute the n-factor we calculated (5/3):
E = M / (5/3)
E = 3M / 5.

Note: In multiple-choice questions, the answer is sometimes written in the expanded form: E = M/2 + M/10.

Question 2: Phosphorus (P₄) undergoes disproportionation in basic medium: P₄ + OH^- → PH₃ + H₂PO₂^-.
Calculate the net n-factor for P₄.

Answer: Net n-factor = 3.

Step-by-Step Solution:

Step 1: Oxidation States. P₄ (0). PH₃ (-3). H₂PO₂^- (+1).
Step 2: Find n_red. P₄ → 4PH₃. Change = |0 - (-3)| = 3. Total change for 4 atoms = 3 × 4 = 12.
Step 3: Find n_ox. P₄ → 4H₂PO₂^-. Change = |0 - 1| = 1. Total change for 4 atoms = 1 × 4 = 4.
Step 4: Apply Formula. 1/n_net = 1/12 + 1/4 = 1/12 + 3/12 = 4/12 = 1/3. Therefore, n_net = 3.

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