Oxidation Number – IIT‑JEE (Main & Advanced) Study Notes
Chemistry – Inorganic/Physical
1. Simple Explanation
Oxidation Number (ON) (also called Oxidation State) is a conceptual charge assigned to an atom in a compound, representing the number of electrons that an atom would lose, gain, or share if the compound were completely ionic.
- Purpose:
- Track electron transfer in redox reactions.
- Predict the direction of oxidation/reduction.
- Determine the empirical/empirical formula of compounds.
Important: ON is a formal number; the real electronic distribution may differ (especially in covalent compounds).
2. Key Points & Rules
| # | Rule | Example |
|---|---|---|
| 1 | Element in its free state has ON = 0. | ( \text{O}_2, \text{N}_2, \text{Fe} ) → ON = 0 |
| 2 | Alkali metals (group 1) have ON = +1; Alkaline earth metals (group 2) have ON = +2. | ( \text{NaCl} ) → Na = +1, Cl = –1 |
| 3 | Oxygen (except in peroxides and superoxides) has ON = –2. | ( \text{CO}_2 ): O = –2 |
| 4 | Fluorine is the most electronegative element → ON = –1. | ( \text{SF}_6 ): F = –1 |
| 5 | Halogens (F–Cl–Br–I) usually have ON = –1 (except when bonded to F). | ( \text{Cl}_2\text{O}_7 ): Cl = +7 |
| 6 | Hydrogen bonded to non‑metal (or halogen) has ON = +1; bonded to metal has ON = –1. | ( \text{H}_2\text{O} ): H = +1; ( \text{NaH} ): H = –1 |
| 7 | Sum of ONs in a neutral compound = 0. | ( \text{H}_2\text{SO}_4 ): 2(+1) + 1(+6) + 4(–2) = 0 |
| 8 | Sum of ONs in a polyatomic ion = its charge. | ( \text{NO}_3^- ): N + 3(–2) = –1 |
| 9 | P and S in oxyanions: ON = +2, +4, +6 (O = –2). | ( \text{H}_2\text{S}_2\text{O}_7 ): S = +5 (each) |
| 10 | Exceptions: |
- Oxygen in peroxides: –1.
- Oxygen in superoxides: –½.
- S in ( \text{S}_2\text{O}_7^{2-} ): +5 each. | ( \text{H}_2\text{O}_2 ): O = –1 | | 11 | Use of “effective valence”: In covalent molecules, electrons are shared equally to satisfy ON rules. | ( \text{C}_6\text{H}_6 ): C = +4, H = +1 |
Redox Reaction Check
- Oxidation: ON of atom increases.
- Reduction: ON of atom decreases.
3. Worked Example
Problem
Find the oxidation numbers of all atoms in the compound ( \text{K}_2\text{SO}_4 ) and identify the oxidised/reduced species when ( \text{K}_2\text{SO}_4 ) reacts with ( \text{H}_2\text{O}_2 ) to form ( \text{K}_2\text{S}_2\text{O}_7 + \text{O}_2 ).
Step‑by‑Step
| Atom | Rule used | Oxidation Number |
|---|---|---|
| K | Group‑1 metal | +1 |
| S | In an oxyanion, S + 4(–2) = charge (0) → S = +4 | |
| O | In a neutral compound, O = –2 |
Check
(2(+1) + (+4) + 4(–2) = 0 ) ✔️
Reaction
( \text{K}_2\text{SO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{K}_2\text{S}_2\text{O}_7 + \text{O}_2 )
- In ( \text{K}_2\text{S}_2\text{O}_7 ): S has +5 (since 2S + 7(–2) = –1 → 2S = +13 → S = +6.5? Wait, let's calculate properly.
Let x = oxidation number of S.
(2x + 7(-2) = -1) (charge of ( \text{S}_2\text{O}_7^{2-}) is –2, but overall product is neutral because 2K⁺).
(2x - 14 = -1) → (2x = 13) → (x = +6.5).
But S cannot have fractional ON → Actually, ( \text{K}_2\text{S}_2\text{O}_7 ) is a disulfate ion: each S +5 and one S +6? No, check: Each S in ( \text{S}_2\text{O}_7^{2-}) is +5.**
Let's redo: In ( \text{S}_2\text{O}_7^{2-}), each S is +5.**
(Because 2S + 7(–2) = –2 → 2S – 14 = –2 → 2S = 12 → S