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3.1.7Additional lessons

Oxidation States — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 54 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Part 1 of 5Rules for Assigning Oxidation States

3.1.7 Oxidation States
Free elements
  • An atom in an uncombined element, such as Fe, O₂ or S₈, has an oxidation state of zero.
Total charge
  • The oxidation states in a neutral compound add to zero, and in a polyatomic ion they add to the ion's charge.
Oxygen
  • Oxygen is usually −2, with exceptions such as −1 in peroxides like H₂O₂ and positive values bonded to fluorine.
Hydrogen
  • Hydrogen is usually +1, but it is −1 in metal hydrides such as NaH.
Groups 1, 2 and halogens
  • Group 1 metals are +1, Group 2 metals are +2, and halogens are −1 unless combined with a more electronegative element.

Part 2 of 5Using Oxidation States

1
Write the known oxidation states first, then calculate the unknown one so that the total equals the overall charge.
2
Oxidation is an increase in oxidation state and reduction is a decrease, so changes reveal which species were oxidised.
3
A species that is oxidised is the reducing agent, while the species reduced is the oxidising agent.
4
Roman numerals in names show the oxidation state, so iron(III) oxide contains Fe³⁺ and copper(I) oxide contains Cu⁺.
5
Reactions where no element changes oxidation state, such as neutralisation, are not redox reactions.
SpeciesElementOxidation state
MnO₄⁻Mn+7
SO₄²⁻S+6
NO₃⁻N+5
H₂O₂O−1

Part 3 of 5Redox Examples

Redox Examples
Disproportionation
  • In 2H₂O₂ → 2H₂O + O₂ the same element, oxygen, is both oxidised and reduced.
Displacement
  • In Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), zinc goes from 0 to +2 and copper goes from +2 to 0.
Combustion
  • In CH₄ + 2O₂ → CO₂ + 2H₂O carbon changes from −4 to +4, so it is oxidised.
Non-integer
  • Fractional values such as Fe in Fe₃O₄ represent an average over different sites in the structure.

Part 4 of 5Worked Example

Question Calculate the oxidation state of chromium in the dichromate ion, Cr₂O₇²⁻.

  1. Let the oxidation state of Cr be x.
  2. Oxygen is −2, so seven oxygen atoms contribute −14.
  3. The ion has an overall charge of −2, so 2x + (−14) = −2.
  4. 2x = +12, so x = +6.

Answer Cr is +6 in Cr₂O₇²⁻

Part 5 of 5Exam Checklist & Tips

3.1.7 – Key Points for the Exam

  • Oxidation state The charge an atom would have if all its bonds were treated as fully ionic.
  • Oxidising agent The species that gains electrons and is itself reduced.
  • Reducing agent The species that loses electrons and is itself oxidised.
  • Disproportionation A reaction in which one element is simultaneously oxidised and reduced.

Common Mistakes

  • Confusing oxidation state with the number of bonds or the charge of the whole molecule.
  • Writing the oxidation state as 2+ instead of +2, which is the convention for charge on ions.
  • Saying the oxidising agent is oxidised; it is the species that gets reduced.
  • Assuming oxygen is always −2, including in H₂O₂ and OF₂.

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