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n-Factor in Intramolecular Redox Reactions | Chemca

n-Factor in Intramolecular Redox Reactions | Chemca
Advanced Redox Concepts

n-Factor in Intramolecular Redox Reactions

What do you do when one element in a compound gets oxidized and another element in the same compound gets reduced? Master this advanced equivalent weight concept for JEE and NEET.

By Abhishek Sengar 10 Min Read

Standard redox reactions usually involve two distinct reactants: an oxidizing agent and a reducing agent. But sometimes, a single chemical compound acts as both. When different elements within the same molecule undergo simultaneous oxidation and reduction, the process is called an Intramolecular Redox Reaction.

Calculating the equivalent weight of such compounds in JEE Main, JEE Advanced, and NEET frequently causes panic because students don't know which element to use for the n-factor calculation.


The Golden Rule for n-Factor

In any balanced redox reaction, the total number of electrons lost must equal the total number of electrons gained. Therefore, for an intramolecular redox compound, the n-factor is:

n-factor = | Total e- lost by oxidized element |
OR
n-factor = | Total e- gained by reduced element |

CRUCIAL: You calculate one OR the other. You NEVER add them together!

Case Study 1: Decomposition of Potassium Chlorate (KClO3)

This is a classic NTA favorite. When KClO3 is heated, it decomposes to form potassium chloride and oxygen gas.

Step-by-Step Calculation

KClO3KCl + 3/2 O2

The Reduction Part
  • Element: Chlorine (Cl)
  • Initial State: +5 (in KClO3)
  • Final State: -1 (in KCl)
  • Change per atom: Gains 6 e-
  • Atoms per molecule: 1
  • Total e- gained = 6
The Oxidation Part
  • Element: Oxygen (O)
  • Initial State: -2 (in KClO3)
  • Final State: 0 (in O2)
  • Change per atom: Loses 2 e-
  • Atoms per molecule: 3
  • Total e- lost = 3 × 2 = 6

n-factor of KClO3 = 6

Case Study 2: Decomposition of Ammonium Dichromate ((NH4)2Cr2O7)

This reaction (the "chemical volcano") is the ultimate test of your understanding. Here, Nitrogen is oxidized, and Chromium is reduced.

Step-by-Step Calculation

(NH4)2Cr2O7N2 + Cr2O3 + 4 H2O

The Oxidation Part
  • Element: Nitrogen (N)
  • Initial State: -3 (in NH4+)
  • Final State: 0 (in N2)
  • Change per atom: Loses 3 e-
  • Atoms per molecule: 2
  • Total e- lost = 2 × 3 = 6
The Reduction Part
  • Element: Chromium (Cr)
  • Initial State: +6 (in Cr2O72-)
  • Final State: +3 (in Cr2O3)
  • Change per atom: Gains 3 e-
  • Atoms per molecule: 2
  • Total e- gained = 2 × 3 = 6

n-factor of (NH4)2Cr2O7 = 6

The Deadly Addition Trap

In both examples above, the electrons lost (6) perfectly match the electrons gained (6), confirming the reaction is balanced internally.

Equivalent Weight Calculation

Equivalent Weight = Molecular Weight (M)

                    n-factor

WRONG Answer

Eq. Wt = M / 12

(Student added 6 lost + 6 gained)

CORRECT Answer

Eq. Wt = M / 6

(Student took the magnitude of transfer)

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