Substitution reactions (A-level only) — AQA A-Level Chemistry
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Substitution reactions (A-level only) explained
Haem is an octahedral iron(II) complex. The multidentate ligand is a porphyrin ring, which donates four lone pairs from four nitrogen atoms and so occupies four of the six co-ordination sites, all in one plane.
Read the full explanation
The fifth site is filled by a nitrogen atom from the protein globin, which anchors the haem group in place. That leaves exactly one site free, and this sixth position holds a water molecule when no oxygen is present and an oxygen molecule when oxygen is available. Haemoglobin is the whole assembly, four haem groups held within one globin protein, so a single haemoglobin can carry up to four O₂ molecules. The iron remains Fe²⁺ throughout and is not oxidised when oxygen binds.
Oxygen forms a co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported in the blood.
Oxygen transport works because the bond formed is a co-ordinate bond and, crucially, a reversible one. In the lungs, where the partial pressure of oxygen is high, an O₂ molecule donates a lone pair into the vacant sixth co-ordination site on the Fe²⁺ of haem, forming oxyhaemoglobin. Blood carries this to respiring tissue, where the partial pressure of oxygen is low, the equilibrium shifts back, O₂ is released and a water molecule takes the site instead. The iron stays as Fe²⁺ throughout. The bond must be strong enough to pick oxygen up but weak enough to let it go again; a ligand that bound irreversibly would load the haemoglobin once and then stop it working, which is exactly what carbon monoxide does.
Carbon monoxide is toxic because it replaces oxygen co- ordinately bonded to Fe(II) in haemoglobin.
Carbon monoxide is a ligand: the carbon atom carries a lone pair and donates it to Fe²⁺ exactly as oxygen does, forming carboxyhaemoglobin. The problem is strength. Carbon monoxide binds to haem around two hundred times more strongly than O₂ does, so the substitution is effectively irreversible at ordinary concentrations. Once carbon monoxide occupies the sixth site, that haemoglobin molecule can no longer pick up oxygen in the lungs or release it to tissue, so the oxygen carrying capacity of the blood falls and cells are starved of oxygen. The toxicity is therefore not a poison attacking a cell directly but a ligand substitution reaction with a far more favourable position of equilibrium.
Bidentate and multidentate ligands replace monodentate ligands from complexes. This is called the chelate effect.
Put a bidentate or multidentate ligand into a solution containing a complex of monodentate ligands and the substitution runs almost to completion. [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O is the standard case: one EDTA⁴⁻ ion displaces all six water ligands because it has six donor atoms of its own. The co-ordination number does not change; only the number of separate particles does, as two reactant particles become seven product particles. The chelate effect is the name for this strong tendency, and its cause is that increase in the number of particles, not any great difference in bond strength, since the bonds broken and made are all co-ordinate bonds of similar strength.
Students should be able to: explain the chelate effect, in terms of the balance between the entropy and enthalpy change in these reactions.
This is a Gibbs free energy argument and the marks are usually split between the two terms. The enthalpy change is close to zero because the bonds broken, metal to water co-ordinate bonds, and the bonds made, metal to chelate co-ordinate bonds, are of very similar strength, so ΔH barely contributes. The entropy change is clearly positive: [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O turns two particles into seven, so the system has many more ways of arranging itself. Substituting into ΔG = ΔH − TΔS, a near zero ΔH with a positive ΔS makes −TΔS the dominant term, so ΔG is negative and the reaction is feasible. The chelate effect is entropy driven.
Your focus
- State the oxidation state of the iron in haem and the co-ordination number of the complex.
- Account for all six co-ordination sites in haem, naming the ligand that occupies each one.
- Distinguish haem from haemoglobin in terms of the globin protein and the number of haem groups present.
Show all 15 objectives
- Explain oxygen transport as reversible co-ordinate bonding of O₂ at the sixth site on Fe(II).
- Predict whether oxygen will bind or be released from the partial pressure of oxygen in a given tissue.
- State what occupies the sixth co-ordination site when oxygen is not bound.
- Explain the toxicity of carbon monoxide as a ligand substitution at the Fe(II) site of haemoglobin.
- Compare the strength and reversibility of the bonds to carbon monoxide and to oxygen, and deduce the consequence for oxygen transport.
- Identify the donor atom in carbon monoxide and draw the co-ordinate bond that it forms.
- Write balanced equations for the displacement of monodentate ligands by bidentate or multidentate ligands.
- Show that the co-ordination number is unchanged while the number of free particles increases.
- Name the chelate effect and identify it as the reason such reactions go almost to completion.
- Explain why ΔH is close to zero for a chelate substitution by comparing the bonds broken and the bonds made.
- Deduce the sign of ΔS from the number of particles on each side of a given equation.
- Use ΔG = ΔH − TΔS to justify why chelate substitutions are feasible and describe them as entropy driven.
Substitution reactions (A-level only) exam tips
Marking Points
- one mark for identifying the oxidation state of the iron in haem as +2
- one mark for stating that the porphyrin ring is multidentate and donates four lone pairs from four nitrogen atoms
- one mark for a co-ordination number of 6 and an octahedral shape
- one mark for naming globin as the fifth ligand and oxygen or water as the sixth
- one mark for stating that O₂ donates a lone pair to Fe(II) to form a co-ordinate bond
- one mark for identifying the sixth co-ordination site, previously occupied by water, as the binding site
- one mark for describing the binding as reversible, with uptake in the lungs and release in the tissues
- one mark for linking release of oxygen to the lower partial pressure of oxygen in respiring tissue
- one mark for stating that carbon monoxide forms a co-ordinate bond to Fe(II) by donating a lone pair from the carbon atom
- one mark for identifying the process as ligand substitution, with carbon monoxide replacing O₂ or H₂O
- one mark for stating that carbon monoxide binds more strongly than oxygen, so the substitution is effectively irreversible
- one mark for the consequence, that the haemoglobin can no longer transport oxygen
- one mark for a balanced equation showing the multidentate ligand displacing all the monodentate ligands, such as [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O
- one mark for stating that the co-ordination number is unchanged by the substitution
- one mark for counting the particles on each side and noting the increase
- one mark for naming the chelate effect as the reason such substitutions go almost to completion
- one mark for stating that ΔH is approximately zero because similar numbers of co-ordinate bonds of similar strength are broken and made
- one mark for a positive ΔS justified by the increase in the number of particles in solution
- one mark for using ΔG = ΔH − TΔS to show that ΔG is negative
- one mark for concluding that the reaction is entropy driven rather than enthalpy driven
Examiner Tips
- 💡Account for all six co-ordination sites when describing haem: four ring nitrogens, one globin, one O₂ or H₂O.
- 💡Keep the iron as Fe²⁺ in every statement you write about oxygen transport.
- 💡Use the word multidentate or tetradentate about the ring rather than describing it as large.
- 💡Use the word reversible, because it is usually the mark that separates a transport answer from a bonding answer.
- 💡Name the site that oxygen occupies rather than saying oxygen attaches to the haemoglobin.
- 💡Link the direction of the shift to partial pressure whenever a question mentions lungs and tissues.
- 💡Use the phrase ligand substitution and name the ligand that leaves as well as the one that arrives.
- 💡Say the bond is stronger than the bond to oxygen, because carbon monoxide is poisonous gains nothing on its own.
- 💡Keep the iron in the +2 oxidation state throughout your answer.
- 💡Balance chelate equations by donor atoms: six water ligands out means six lone pairs in.
- 💡Check the overall charge on both sides of the equation before moving on.
- 💡Name the effect when asked for the reason, then give the particle count as your evidence.
- 💡Count the particles in the equation you have written and quote both numbers in your answer.
- 💡Say ΔH is close to zero rather than there is no enthalpy change, which examiners read as careless.
- 💡Finish with the sign of ΔG, because the conclusion usually carries a mark of its own.
Common Mistakes
- describing haem as an iron(III) complex, or saying the iron is oxidised when oxygen binds
- calling the porphyrin ring bidentate because it looks like one flat ligand
- confusing haem with haemoglobin, which also contains the globin protein
- giving the co-ordination number as 4 by counting only the nitrogen atoms of the ring
- saying the oxygen is reduced or that Fe²⁺ is oxidised to Fe³⁺ when oxygen binds
- describing the bond as ionic, or as a normal covalent bond rather than a co-ordinate bond
- claiming oxygen displaces the globin rather than the water at the sixth site
- treating the binding as irreversible, which would leave no mechanism for delivering oxygen
- saying carbon monoxide destroys or denatures the haemoglobin rather than bonding to it
- drawing the donation from the oxygen atom of carbon monoxide instead of from the carbon atom
- describing the binding of carbon monoxide as reversible in the same way as oxygen, which removes the reason for the toxicity
- saying carbon monoxide reacts with oxygen in the blood to form carbon dioxide
- changing the co-ordination number when the chelating ligand replaces the water molecules
- forgetting to include the six displaced water molecules on the product side of the equation
- giving the wrong overall charge on the product because the charge on EDTA⁴⁻ or C₂O₄²⁻ was ignored
- explaining the chelate effect by claiming the new bonds are stronger, which is not what the mark scheme rewards
- claiming ΔH is large and negative because the chelate forms a stronger complex
- counting the wrong species, for example including solvent water, so getting the sign of ΔS wrong
- writing ΔG = ΔH + TΔS and concluding the reaction is not feasible
- stating that entropy increases without saying which side of the equation has more particles