What is oxidation number?
In a covalent bond, electrons are shared, not fully given. To track them, we use a pretend charge called the oxidation number (or oxidation state).
Rule of pretending: give all shared electrons to the atom that is more electronegative (pulls harder). The charge each atom would then have is its oxidation number.
In H₂O, O pulls harder. So O gets both electron pairs → O = −2, each H = +1.
Try it: at step 3 of the 3D, watch the two yellow electrons slide towards O.
Rules for finding oxidation number
- Free element (Na, O₂, P₄, S₈): 0.
- Single-atom ion: equal to its charge (Na⁺ = +1, Mg²⁺ = +2, Cl⁻ = −1).
- Oxygen is usually −2. Exceptions: peroxides like H₂O₂ and Na₂O₂ (−1), superoxides like KO₂ (−½), and OF₂ (+2) because F pulls harder.
- Hydrogen is +1 with non-metals, but −1 in metal hydrides like NaH and CaH₂.
- Fluorine is always −1. Other halogens are −1 in halides, but can be positive with O (as in ClO₄⁻).
- Group 1 metals are +1, group 2 metals are +2 in compounds.
- Sum of all oxidation numbers = 0 for a neutral compound, and = the charge for an ion.
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Call the unknown atom x. Write known values. Make the total equal to the charge. Solve.
H₂SO₄: 2(+1) + x + 4(−2) = 0 → x = +6.
Cr₂O₇²⁻: 2x + 7(−2) = −2 → 2x = 12 → x = +6.
Fe₃O₄: 3x + 4(−2) = 0 → x = +8/3. A fraction means an average: really two Fe are +3 and one Fe is +2.
Try it: at step 6 pick KMnO₄, NH₄⁺ and Fe₃O₄. Solve first, then check.
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When a metal can have more than one oxidation number, we write it in Roman numerals in brackets after the name. This is Stock notation.
- FeO → iron(II) oxide; Fe₂O₃ → iron(III) oxide
- Cu₂O → copper(I) oxide; CuO → copper(II) oxide
- MnO₂ → manganese(IV) oxide; SnCl₄ → tin(IV) chloride
Redox in terms of oxidation number
Oxidation = increase in oxidation number. Reduction = decrease in oxidation number.
Oxidising agent: its oxidation number goes down. Reducing agent: its oxidation number goes up.
Example: 2Cu₂O + Cu₂S → 6Cu + SO₂. Cu goes +1 → 0 (reduced). S goes −2 → +4 (oxidised). So Cu₂S acts as reducing agent and also gets its own Cu reduced.
Types of redox reactions
- Combination: two things join. C + O₂ → CO₂ (C: 0 → +4). At least one reactant is an element.
- Decomposition: one compound breaks. 2H₂O → 2H₂ + O₂ (H: +1 → 0, O: −2 → 0). Not every decomposition is redox: CaCO₃ → CaO + CO₂ has no change.
- Displacement: an atom or ion in a compound is replaced. Metal displacement: Zn + CuSO₄ → ZnSO₄ + Cu. Non-metal displacement: 2Na + 2H₂O → 2NaOH + H₂; Cl₂ + 2KBr → 2KCl + Br₂ (F₂ > Cl₂ > Br₂ > I₂ in pushing out).
- Disproportionation: the same element in one substance is both oxidised and reduced. 2H₂O₂ → 2H₂O + O₂ (O: −1 → −2 and −1 → 0). Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O (Cl: 0 → +1 and 0 → −1). The element must be at a middle oxidation number, so F₂ (only −1 possible in compounds) cannot disproportionate.
Try it: at step 5 of the 3D watch O atoms split into a "down" group and an "up" group.
Key formulas and definitions
- Sum of oxidation numbers = 0 (neutral) or = charge (ion)
- O = −2 (peroxide −1, superoxide −½, OF₂ +2)
- H = +1 (metal hydride −1); F = −1 always
- Oxidation = increase in oxidation number; reduction = decrease
- Disproportionation: one element goes both up and down
Worked examples
1. Find the oxidation number of N in NH₃.
x + 3(+1) = 0 → x = −3.
2. Find the oxidation number of Mn in KMnO₄.
+1 + x + 4(−2) = 0 → x = +7.
3. Find the oxidation number of Cr in K₂Cr₂O₇.
2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6. Stock name: potassium dichromate(VI).
4. Find the oxidation number of N in NH₄⁺.
x + 4(+1) = +1 → x = −3.
5. Find the oxidation number of S in S₂O₃²⁻ (thiosulphate).
2x + 3(−2) = −2 → 2x = 4 → x = +2 (an average of the two S atoms).
6. Find the oxidation number of O in OF₂ and in H₂O₂.
OF₂: F is −1, so x + 2(−1) = 0 → x = +2. H₂O₂: 2(+1) + 2x = 0 → x = −1.
7. Is Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O a disproportionation? Show numbers.
Cl starts at 0. In ClO⁻, x − 2 = −1 → x = +1 (up). In Cl⁻ it is −1 (down). The same element goes up and down → yes, disproportionation.
8. Classify: (a) CaCO₃ → CaO + CO₂ (b) Zn + 2HCl → ZnCl₂ + H₂.
(a) Ca +2, C +4, O −2 stay the same → not redox. (b) Zn 0 → +2, H +1 → 0 → redox, a (non-metal) displacement reaction.
Common mistakes
- Taking O as −2 in every compound. In H₂O₂ it is −1 and in OF₂ it is +2.
- Forgetting to multiply by the number of atoms: in Cr₂O₇²⁻ oxygen gives 7 × (−2) = −14.
- Setting the total to 0 for an ion. For an ion the total equals its charge.
- Thinking a fractional oxidation number is wrong. It is an average over atoms that differ.