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Learn7405Paper 1 · Paper 1: Physical and inorganic chemistryTransition metalsTransition element properties
Transition metals · Subtopic 1 of 2

Transition Element Properties

A transition element forms one or more stable ions with an incomplete d subshell (so Sc and Zn are d-block but not transition elements). Configurations: 4s fills before 3d but is emptied first on ionisation; Cr is [Ar]3d⁵4s¹ and Cu is [Ar]3d¹⁰4s¹. Properties: variable oxidation states (similar 4s and 3d energies), coloured ions, complex formation and catalysis (Fe in the Haber process, V₂O₅ in the Contact process, Fe²⁺/Fe³⁺ in I⁻/S₂O₈²⁻). Redox titrations: MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\mathrm{MnO_4^-+5Fe^{2+}+8H^+\rightarrow Mn^{2+}+5Fe^{3+}+4H_2O} (self-indicating: purple to colourless); Cr2O72−\mathrm{Cr_2O_7^{2-}} reacts 1 : 6 with Fe²⁺.

Cambridge 9701 Paper 4Edexcel YCH11 WCH15AQA 7405 Paper 1≈ 50 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Define a transition element.
  • Write electron configurations of atoms and ions.
  • Describe variable oxidation states and catalytic behaviour.
  • Carry out redox titration calculations.
Pause and think

Zinc is in the d-block. Why isn't it a transition element?

Section 2 of 14

Core idea

Start hereMetals with many faces

Manganese can be +2 in a pale pink solution or +7 in deep purple permanganate. Few elements outside the d-block can do anything like that.

The big idea

Transition elements form ions with partly filled d subshells. Close 4s and 3d energies give variable oxidation states, which underpin colour, catalysis and redox titrations.

1. Electron configurations

Configuration of Fe³⁺
  1. Fe:[Ar]3d64s2\mathrm{Fe}:[\mathrm{Ar}]3d^64s^2
    26 electrons; 4s fills before 3d.
  2. Fe2+:[Ar]3d6\mathrm{Fe^{2+}}:[\mathrm{Ar}]3d^6
    Both 4s electrons lost first.
  3. Fe3+:[Ar]3d5\mathrm{Fe^{3+}}:[\mathrm{Ar}]3d^5
    Then one 3d electron.

2. Redox titrations

Worked exampleIron(II) by permanganate

25.0 cm³ of Fe²⁺(aq) needs 20.0 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find [Fe²⁺].

Goal
Concentration.
Strategy
MnO₄⁻ : Fe²⁺ = 1 : 5.
  1. n(MnO4−)=0.0200×0.0200=4.00×10−4n(\mathrm{MnO_4^-})=0.0200\times0.0200=4.00\times10^{-4}
    Moles of titrant.
  2. n(Fe2+)=5×4.00×10−4=2.00×10−3n(\mathrm{Fe^{2+}})=5\times4.00\times10^{-4}=2.00\times10^{-3}
    Ratio from half-equations.
  3. [Fe2+]=2.00×10−30.0250=0.0800[\mathrm{Fe^{2+}}]=\frac{2.00\times10^{-3}}{0.0250}=0.0800
    Divide by volume.
Answer0.0800 mol dm⁻³.
✓ CheckMnO₄⁻ gains 5 e⁻; each Fe²⁺ loses 1 ✓.
How transition metals catalyse

Two routes: surfaces and changing oxidation states.

Summary

  • Partly filled d subshell in an ion.
  • 4s lost before 3d.
  • Variable oxidation states, colour, catalysis.
  • MnO₄⁻ 1 : 5 Fe²⁺.
Key words
Transition element
d-block element forming at least one stable ion with an incomplete d subshell.
Heterogeneous catalyst
Catalyst in a different phase from the reactants.
Autocatalysis
A product of the reaction acts as its catalyst.
HistoryHaber's catalyst. Alwin Mittasch tested over 2500 catalysts in 1909–1912 to find the iron-based catalyst still used in ammonia production today.
Fun factStained glass. Medieval glassmakers coloured windows with transition metal oxides: cobalt for blue, copper for green, manganese for purple.
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