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Reactions of transition metal ions with bases and ligandsEdexcel A-Level Chemistry: Revision notes

Section 1

Precipitates with sodium hydroxide and ammonia

Aqueous hexaaqua ions react with aqueous hydroxide (NaOH, or the OH⁻ in NH₃(aq)). OH⁻ removes H⁺ from water ligands, giving neutral metal hydroxide precipitates:

[Cu(H2O)6]2++2OH−→Cu(OH)2(H2O)4+2H2O\mathrm{[Cu(H_2O)_6]^{2+} + 2OH^- \rightarrow Cu(OH)_2(H_2O)_4 + 2H_2O}

Observations (dropwise, then excess):

  • Cr³⁺: green precipitate; dissolves in excess NaOH to dark green [Cr(OH)6]3−[\mathrm{Cr(OH)_6}]^{3-}.
  • Fe²⁺: green precipitate, insoluble in excess; darkens to brown in air (oxidised).
  • Fe³⁺: orange-brown precipitate, insoluble in excess.
  • Co²⁺: blue precipitate, insoluble in excess NaOH; dissolves in excess ammonia to a pale brown solution.
  • Cu²⁺: pale blue precipitate, insoluble in excess NaOH; dissolves in excess ammonia to a deep blue solution.
Key termsprecipitatehydroxide
Exam tip

Learn two columns: colour of precipitate, and what happens in excess NaOH and in excess NH₃.

Section 2

Amphoteric behaviour

An amphoteric hydroxide can react with both an acid and a base. Cr(OH)3(H2O)3\mathrm{Cr(OH)_3(H_2O)_3} is amphoteric:

  • with acid it accepts H⁺: Cr(OH)3(H2O)3+3H+→[Cr(H2O)6]3+\mathrm{Cr(OH)_3(H_2O)_3 + 3H^+ \rightarrow [Cr(H_2O)_6]^{3+}}
  • with excess hydroxide it donates H⁺: Cr(OH)3(H2O)3+3OH−→[Cr(OH)6]3−+3H2O\mathrm{Cr(OH)_3(H_2O)_3 + 3OH^- \rightarrow [Cr(OH)_6]^{3-} + 3H_2O}

This is an acid–base (proton transfer) reaction, and the precipitate dissolves to a dark green solution.

Key termsamphoteric

Section 3

Ligand exchange with ammonia

In ligand exchange one ligand is replaced by another with no change in oxidation number and no proton transfer. Excess ammonia replaces water and hydroxide ligands on copper(II):

Cu(OH)2(H2O)4+4NH3→[Cu(NH3)4(H2O)2]2++2OH−+2H2O\mathrm{Cu(OH)_2(H_2O)_4 + 4NH_3 \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+} + 2OH^- + 2H_2O}

The pale blue precipitate becomes a deep blue solution, still with coordination number 6 and four NH₃ and two H₂O ligands. Overall: [Cu(H2O)6]2++4NH3→[Cu(NH3)4(H2O)2]2++4H2O\mathrm{[Cu(H_2O)_6]^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O}.

Key termsligand exchange
Common mistake

Calling the dissolving of Cu(OH)₂ in ammonia 'amphoteric'. It is ligand exchange; Cu(OH)₂ does not dissolve in excess NaOH.

Section 4

Ligand exchange with chloride ions

Adding concentrated HCl replaces water by chloride, which is larger and charged, so fewer ligands fit and the coordination number changes from 6 to 4:

[Cu(H2O)6]2++4Cl−⇌[CuCl4]2−+6H2O\mathrm{[Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O} (pale blue → yellow-green)

[Co(H2O)6]2++4Cl−⇌[CoCl4]2−+6H2O\mathrm{[Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O} (pink → blue)

The complexes change from octahedral to tetrahedral. Adding water reverses the change.

Key termscoordination number change

Section 5

Stability of complexes and entropy

Replacing monodentate ligands by a bidentate or multidentate ligand gives a more stable complex:

[Ni(H2O)6]2++3en→[Ni(en)3]2++6H2O\mathrm{[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O}

  • Particles increase from 4 to 7, so there is a large positive increase in ΔS system.
  • ΔH\Delta H is about zero: the same number and type of bonds (six Ni–N for six Ni–O).
  • ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S becomes negative, so the reaction is feasible and the equilibrium lies well to the right.

The same applies to EDTA⁴⁻ replacing six water ligands: 2 particles become 7.

Key termsΔS systemstability
Exam tip

State three things: particles increase, ΔS system positive, ΔH approximately zero, therefore ΔG negative.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Reactions of transition metal ions with bases and ligands

  1. A student has four separate solutions containing the hexaaqua ions of chromium(III), iron(III), cobalt(II) and copper(II). She adds aqueous sodium hydroxide dropwise to a sample of each solution and then adds an excess.
    Write the ionic equation for the formation of the precipitate when aqueous sodium hydroxide is added to the copper(II) solution, and state the colour of the precipitate.2 marks
  2. A student adds aqueous ammonia dropwise, and then in excess, to a pale blue solution of copper(II) sulfate containing [Cu(H₂O)₆]²⁺.
    Write the ionic equation for the ligand exchange reaction that forms the deep blue complex ion from [Cu(H₂O)₆]²⁺ and ammonia, and state the coordination number of copper before and after the reaction.2 marks
  3. Hexaaqua complexes of copper(II) and nickel(II) take part in ligand exchange reactions. Concentrated hydrochloric acid is added to a solution of [Cu(H₂O)₆]²⁺, and, separately, excess ethane-1,2-diamine (en), NH₂CH₂CH₂NH₂, is added to a solution of [Ni(H₂O)₆]²⁺.
    Write the equation for the reaction of [Cu(H₂O)₆]²⁺ with concentrated hydrochloric acid and state the colour change. Explain why the coordination number of copper changes.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).