How to Find the n-Factor in a Disproportionation Reaction
In Redox Reactions, calculating the n-factor (or valence factor) can sometimes get tricky—especially when dealing with a disproportionation reaction. A disproportionation reaction is a type of redox reaction where the same element simultaneously undergoes both oxidation and reduction from a single initial oxidation state into two distinct final oxidation states.
To help Indian engineering and medical aspirants cracking JEE Main, JEE Advanced, and NEET, this guide breaks down the ultimate mathematical shortcut formula to find the net n-factor instantly without balancing the whole chemical equation.
Video Tutorial: Step-by-Step Calculation
Watch this quick classroom breakdown by Abhishek Sengar sir from CHEMCA explaining the exact methodology using the conversion of Chlorine gas (Cl₂) in an alkaline medium:
The Master Shortcut Formula
When an element undergoes simultaneous oxidation and reduction, you calculate the individual n-factors for both pathways separately based on one mole of the reactant molecule. Once you have both values, apply the reciprocal formula below:
Solved Example: Chlorine Reaction
Consider the classical disproportionation of Chlorine:
Cl₂ → Cl⁻ + ClO₃⁻
- Assign Oxidation Numbers:
- Reactant Cl₂ = 0
- Reduced Product Cl⁻ = -1
- Oxidized Product ClO₃⁻ = +5
- Calculate n-factor for Reduction (nr):
Path: Cl₂ (0) → 2Cl⁻ (-1)
Total change in oxidation state for one molecule of Cl₂ = |0 - (-1 × 2)| = 2. Therefore, nreduction = 2. - Calculate n-factor for Oxidation (no):
Path: Cl₂ (0) → 2ClO₃⁻ (+5)
Total change in oxidation state for one molecule of Cl₂ = |0 - (5 × 2)| = 10. Therefore, noxidation = 10. - Apply the Net n-factor Formula:
1 / nnet = 1/2 + 1/10
1 / nnet = (5 + 1) / 10 = 6 / 10 = 3 / 5
nnet = 5 / 3 ≈ 1.67
Practice Questions for JEE & NEET
Test your understanding of the concepts with these exam-targeted questions. Click to reveal the complete calculation steps!
Question 1: Find the n-factor of Phosphorus (P₄) in the following basic medium disproportionation reaction:
P₄ → PH₃ + H₂PO₂⁻
Solution:
- Oxidation states: P₄ (0), PH₃ (-3), H₂PO₂⁻ (+1).
- Reduction path (P₄ → 4PH₃): Change per P₄ molecule = |0 - (-3 × 4)| = 12. So, nr = 12.
- Oxidation path (P₄ → 4H₂PO₂⁻): Change per P₄ molecule = |0 - (+1 × 4)| = 4. So, no = 4.
- Using the formula: 1/nnet = 1/12 + 1/4 = (1 + 3)/12 = 4/12 = 1/3.
- Answer: nnet = 3.
Question 2: Calculate the n-factor of Sulphur (S₈) when it undergoes disproportionation in an aqueous alkaline medium to form Thiosulphate (S₂O₃²⁻) and Sulphide (S²⁻) ions.
Solution:
- Oxidation states: S₈ (0), S²⁻ (-2), S₂O₃²⁻ (+2).
- Reduction path (S₈ → 8S²⁻): Change = |0 - (-2 × 8)| = 16. So, nr = 16.
- Oxidation path (S₈ → 4S₂O₃²⁻): Change = |0 - (+2 × 8)| = 16. So, no = 16.
- Using the formula: 1/nnet = 1/16 + 1/16 = 2/16 = 1/8.
- Answer: nnet = 8.
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