Oxidation NumbersFoundation

The bookkeeping of redox — interactive Chemistry simulation for IIT-JEE.

Concept

The oxidation number is the charge an atom would carry if every bond were fully ionic. Rules in priority order: elements 0; F always −1; Group 1/2 metals +1/+2; H +1 (−1 in metal hydrides); O −2 (−1 in peroxides, −½ in superoxides, +2 in OF₂). Everything must sum to the species' charge — solve for the unknown.

Key formula

\sum \text{(oxidation numbers)} = \text{overall charge}

Derivation

Assign known atoms by priority (higher rules win conflicts), then solve the one-unknown equation.

KMnO₄: +1 + x + 4(−2) = 0 → x = +7. Cr₂O₇²⁻: 2x + 7(−2) = −2 → x = +6. Fe₃O₄: 3x − 8 = 0 → x = 8/3 — a fractional AVERAGE hiding one Fe²⁺ and two Fe³⁺.

Scenarios to explore

Oxidation Numbers — Solve KMnO₄, dichromate, peroxides & mixed oxides step by step.

Real-world applications

Identifying oxidised/reduced species (ON up = oxidised).
Balancing redox equations by electron count.
Naming: iron(III) chloride, manganate(VII).

JEE exam tips

Maximum ON = group number (S: +6, Cl: +7, Mn: +7); minimum = group − 8 for non-metals.
Disproportionation: same element goes BOTH up and down (Cl₂ → Cl⁻ + ClO⁻).
In S₄O₆²⁻ (tetrathionate) the two middle S are 0, the end ones +5 — average 2.5.

Common mistakes

O = −2 in peroxides (it's −1) and OF₂ (it's +2 — F outranks O).
H = +1 in NaH (it's −1: metal hydride).
Rejecting fractional answers — they're valid averages (Fe₃O₄, C₃O₂, S₄O₆²⁻).

Exam traps to avoid

CrO₅ has two peroxide linkages: Cr is +6, not +10!
Oxidation number ≠ formal charge ≠ actual charge — three different bookkeepings.