Complex Ions, Ligands, Ligand Exchange
Explain ligand exchange in copper(II) complexes and in haemoglobin.
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This lesson follows the exact 9476 ligand-exchange sequence: define ligand and complex, track the copper(II) complexes formed with water, ammonia and chloride ions, and explain oxygen/carbon monoxide competition in haemoglobin.
Use the Transition Elements hub for the lesson sequence and d-Orbital Splitting and Colour for the orbital explanation behind an exchange colour change.
Definitions (Must Know)
A. Ligand
A ligand is an ion or molecule that donates a lone pair of electrons to a central metal ion to form a coordinate bond.
B. Complex
A complex contains a central metal atom or ion surrounded by ligands joined through coordinate bonds. A charged complex is written in square brackets with its overall charge outside, for example [Cu(H₂O)₆]²⁺.
Key Ideas (What Earns Marks)
- The assessed copper(II) sequence uses water, ammonia and chloride ions as ligands.
- In ligand exchange, one ligand replaces another around the same central ion; conserve every ligand, atom and charge in the equation.
- Aqueous Cu²⁺ is represented as the pale-blue hexaaquacopper(II) ion, [Cu(H₂O)₆]²⁺.
- Excess ammonia forms the deep-blue complex [Cu(NH₃)₄(H₂O)₂]²⁺ after the initial pale-blue precipitate dissolves.
- Concentrated chloride ions establish an equilibrium with [CuCl₄]²⁻; the observed colour depends on the mixture of complexes.
- Carbon monoxide and oxygen exchange as ligands at the metal centre in haemoglobin. This is ligand competition, not a redox explanation.
Learn the named copper(II) ligand exchanges and their colour evidence. A memorised spectrochemical series or a general catalogue of complex geometries is not required by 9476 Section 13.
Detailed Explanations
A. Why a complex forms
A ligand supplies both electrons in the new bond. Its lone pair is donated into an available orbital on the electron-deficient metal ion, producing a coordinate bond.
B. Water–ammonia sequence for copper(II)
Adding a little ammonia first supplies enough OH⁻ for a pale-blue copper(II) hydroxide precipitate: [Cu(H₂O)₆]²⁺ + 2OH⁻ → Cu(OH)₂(s) + 6H₂O
In excess ammonia, ligand exchange produces the deep-blue ammine complex: [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O
Describe the observations in order: pale-blue precipitate first, then dissolution to a deep-blue solution in excess ammonia.
C. Water–chloride ligand exchange
[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O
Increasing chloride concentration shifts the equilibrium towards the chloride complex. Dilution shifts it back towards the aqua complex. State that a different ligand environment changes the d-orbital splitting and hence the colour.
D. Oxygen–carbon monoxide exchange in haemoglobin
Oxygen binds reversibly as a ligand at the metal centre. Carbon monoxide competes for the same site and forms a more stable complex, displacing oxygen and reducing the number of sites available for oxygen transport. No detailed biological mechanism is required.
E. Equation audit
For every exchange equation, check the central metal, each ligand, total atoms and overall charge. Brackets identify the complete complex; the charge belongs outside them.
Worked Examples
Modelled example 1
Oxidation State in a Cyanide Complex
Problem
Study the worked solution
Total ligand charge
Method
Assign six CN⁻ ligands a total charge of -6.Reason
Each cyanide ligand carries charge -1.Working
6(-1) = -6.Match the complex charge
Method
Solve x-6 = -4.Reason
Metal plus ligand charges equal the overall complex charge.Working
x = +2.
Common misconception 2
Charge of a Cobalt Ammine Complex
Learner claim
Account for ligand charge
View solution step by step
Assign ligand charge
Method
Treat all six ammonia ligands as neutral.Reason
Ligand count does not imply ionic charge.Working
6(0) = 0.Write the complex
Method
Keep the cobalt(II) charge.Reason
+ 2 + 0 = +2.Working
[Co(NH₃)₆]²⁺.
Challenge 3
Copper(II) Hydroxide from a Hexaaqua Ion
Equation transfer
Balance charge, hydroxide and water
Hints
Hint 1: solid
Hint 2: ligands
View solution step by step
Form the neutral solid
Method
Combine copper(II) with two hydroxides.Reason
Cu(OH)₂ is charge neutral.Working
Cu²⁺ + 2OH⁻ → Cu(OH)₂(s).Account for aqua ligands
Method
Release six water molecules.Reason
The starting copper species is hexaaqua.Working
[Cu(H₂O)₆]²⁺ + 2OH⁻ → Cu(OH)₂(s) + 6H₂O.
Common Mistakes
- Describing water, ammonia or chloride as a catalyst rather than a ligand.
- Calling the initial copper(II) hydroxide precipitation ligand exchange.
- Omitting brackets or placing the overall charge on one ligand.
- Giving only a colour word without the species or stage of reagent addition.
- Claiming oxygen/carbon monoxide exchange is a change in the metal oxidation state.
Use the Transition Elements topic check to diagnose whether the error is equation balance, observation order or chemical classification.
Exam Tips
- For a definition, say donates a lone pair and central metal atom or ion.
- In an observation question, write “dropwise” and “in excess” stages separately.
- In an exchange equation, conserve the complete ligand count as well as total charge.
- To distinguish ligand exchange from redox, compare the metal oxidation state before and after.
- For haemoglobin, describe competing ligands and reversible site occupancy; detailed biology is not required.
Mind Stretchers
Mind stretcher 1: Following a reversible chloride exchangeExtension
Question. Concentrated chloride solution changes a pale-blue copper(II) solution to a different colour. Adding much water restores the pale-blue colour. Explain both changes and state what evidence would distinguish this process from a redox reaction.
Show Hint
Write the reversible exchange equation, then compare the copper oxidation state before and after the colour change.
Show Answer
Added chloride shifts [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O towards the chloride complex. Dilution lowers chloride concentration and shifts the equilibrium back towards pale-blue [Cu(H₂O)₆]²⁺. The copper oxidation state remains +2 on both sides; showing no oxidation-number change distinguishes ligand exchange from redox.
Mind stretcher 2: Competing ligands in haemoglobinExtension
Question. Carbon monoxide binds more strongly than oxygen to the same metal centre in haemoglobin. Explain, using ligand exchange and equilibrium, why even a modest carbon monoxide exposure can reduce oxygen transport.
Show Hint
Treat oxygen and carbon monoxide as competing ligands at the same binding site, not as redox reagents.
Show Answer
Oxygen and carbon monoxide act as ligands competing for the same metal centre. Stronger binding by carbon monoxide shifts ligand exchange towards the carbon monoxide complex, so fewer sites remain available for reversible oxygen binding. Oxygen transport therefore falls even though carbon monoxide is not itself consuming the oxygen in a redox reaction.
Syllabus and review details
- GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus
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