n-Factor: When Multiple Elements are Oxidized in the Same Compound
What happens when an oxidizing agent attacks two different elements in the same molecule? Master the "Addition Rule" to conquer the hardest titration questions in JEE and NEET.
In standard redox reactions, usually, only one element in a compound undergoes a change in oxidation state. However, in advanced titration problems, you will encounter compounds where a strong oxidizing agent (like KMnO4 or K2Cr2O7) simultaneously oxidizes two or more elements within the same reactant molecule.
If you don't calculate the total electron transfer correctly, your equivalent weight calculation will be completely wrong.
The Addition Rule for n-Factor
Unlike intramolecular redox (where one element oxidizes and another reduces, and we DO NOT add them), here both elements are undergoing the same process (oxidation). Therefore, the molecule as a whole is acting as a massive electron donor.
Total n-factor = Σ (Moles of e- lost by Element 1) + (Moles of e- lost by Element 2) ...
Rule of Thumb: When elements suffer the same fate (both oxidized or both reduced), ADD their individual n-factors.
Case Study 1: Ferrous Oxalate (FeC2O4)
This is arguably the most frequently asked question in JEE physical chemistry. When Ferrous Oxalate reacts with an acidic solution of KMnO4, the permanganate oxidizes both the Ferrous ion and the Oxalate ion.
Step-by-Step Breakdown
FeC2O4 → Fe3+ + 2 CO2
Element 1: Iron (Fe)
- Process: Oxidation
- Initial State: +2 (Ferrous)
- Final State: +3 (Ferric)
- Change: Loses 1 e- per atom
- Atoms per molecule: 1
- e- lost by Fe = 1 × 1 = 1
Element 2: Carbon (C)
- Process: Oxidation
- Initial State: +3 (in Oxalate)
- Final State: +4 (in Carbon Dioxide)
- Change: Loses 1 e- per atom
- Atoms per molecule: 2
- e- lost by C = 2 × 1 = 2
Total n-factor = 1 (from Fe) + 2 (from C) = 3
Therefore, Equivalent Weight of FeC2O4 = M / 3
Case Study 2: Cuprous Sulfide (Cu2S)
This is a slightly more advanced example often seen in JEE Advanced. When Cuprous Sulfide is treated with a strong oxidizing agent, both Copper and Sulfur are oxidized to their higher stable states.
Step-by-Step Breakdown
Cu2S → 2 Cu2+ + SO2
Element 1: Copper (Cu)
- Initial State: +1 (Cuprous)
- Final State: +2 (Cupric)
- Change: Loses 1 e- per atom
- Atoms per molecule: 2
- e- lost by Cu = 2 × 1 = 2
Element 2: Sulfur (S)
- Initial State: -2 (Sulfide)
- Final State: +4 (in Sulfur Dioxide)
- Change: Loses 6 e- per atom
- Atoms per molecule: 1
- e- lost by S = 1 × 6 = 6
Total n-factor = 2 (from Cu) + 6 (from S) = 8
Therefore, Equivalent Weight of Cu2S = M / 8
The Subscript Trap
The number one mistake students make is calculating the change in oxidation state correctly but forgetting to multiply by the subscript of the element in the parent molecule.
Take Ferrous Oxalate (FeC2O4) as an example:
The Mistake
Fe loses 1.
Carbon loses 1.
n-factor = 1 + 1 = 2
(Forgot there are TWO carbons!)
The Correct Way
Fe loses 1.
Carbon loses 1 × 2 = 2.
n-factor = 1 + 2 = 3
(Multiplied by the subscript C2)
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