3.1.10Additional lessons
Variable Oxidation States in Transition Elements (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 57 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Why Several States Occur
Close energies
- Close energies The 4s and 3d sub-levels are close in energy, so electrons from both can be removed with a similar energy cost.
Gradual increase
- Gradual increase Successive ionisation energies rise gradually in a transition metal, so several electrons can be lost before a large jump.
Maximum state
- Maximum state The highest oxidation state of the first-row elements up to manganese equals the total number of 4s and 3d electrons, +7 for manganese.
Stability
- Stability Half-filled 3d⁵ and filled 3d¹⁰ arrangements make states such as Fe³⁺ and Cu⁺ relatively favourable.
Environment
- Environment Oxidation state stability depends on the ligands, the pH and the oxidising or reducing agent present.
| Element | Common states |
|---|---|
| Fe | +2, +3 |
| Cr | +3, +6 |
| Mn | +2, +4, +7 |
| Cu | +1, +2 |
Section 2 of 5Colour Changes in Redox
1
Dichromate(VI), Cr₂O₇²⁻, is orange and is reduced to green Cr³⁺ in acid by a suitable reducing agent.
2
Permanganate(VII), MnO₄⁻, is deep purple and is reduced to almost colourless Mn²⁺ in acid.
3
Vanadium shows a sequence of colours as it is reduced, from yellow VO₂⁺ through blue and green to violet V²⁺.
4
Iron(II) is pale green and iron(III) is yellow-brown in solution, so oxidation can be seen.
5
The disappearance of purple permanganate colour acts as its own end point in a titration.
Section 3 of 5Half-equations
Half-equations
Permanganate
- Permanganate MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O is the half-equation used in acidic titrations.
Dichromate
- Dichromate Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O reduces Cr(VI) to Cr(III).
Iron
- Iron Fe²⁺ → Fe³⁺ + e⁻ is oxidation, so Fe²⁺ acts as the reducing agent.
Combining
- Combining Balance electrons in the two half-equations before adding them to get the overall equation.
Section 4 of 5Worked Example
Question
- 25.0 cm³ of an Fe²⁺ solution requires 20.0 cm³ of 0.0200 mol dm⁻³ KMnO₄ in acid.
- Calculate [Fe²⁺].
- MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
- n(MnO₄⁻) = 0.0200 × 0.0200 = 4.00 × 10⁻⁴ mol.
- n(Fe²⁺) = 5 × 4.00 × 10⁻⁴ = 2.00 × 10⁻³ mol.
- [Fe²⁺] = 2.00 × 10⁻³ ÷ 0.0250 = 0.0800 mol dm⁻³.
Answer [Fe²⁺] = 0.0800 mol dm⁻³
Section 5 of 5Exam Checklist & Tips
3.1.10 – Key Points for the Exam
- Oxidation state The apparent charge on an atom when electrons are assigned to the more electronegative element.
- Redox titration A titration in which an oxidising agent reacts with a reducing agent.
- Half-equation An equation showing the electron transfer for oxidation or reduction alone.
- End point The point where the indicator or colour shows the reaction is complete.
Common Mistakes
- Stating that a transition metal has variable oxidation states because it has unfilled 3d orbitals only;
- the close energies of 4s and 3d matter.
- Giving oxidation states as 2+ rather than +2.
- Forgetting acidic conditions when using MnO₄⁻ or Cr₂O₇²⁻.
- Forgetting the 5:1 ratio between Fe²⁺ and MnO₄⁻.
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