Substitution reactions (A-level only)

What you'll learn:

How the size of a ligand dictates whether the coordination number changes during a substitution reaction.
Why substitution is sometimes incomplete, resulting in a mixed-ligand complex.
The role of multidentate ligands in biological molecules like haemoglobin.
How to explain the exceptional stability of complexes with multidentate ligands using the 'chelate effect' and entropy.

Transition metal complexes aren't permanently fixed. Ligands can swap places with other ligands in a process called ligand substitution (or ligand exchange). Whether the complex changes shape, charge, or coordination number during this swap depends almost entirely on the size and charge of the incoming ligands.

1. Monodentate ligands of similar size

Water ( $H2O\text{H}_2\text{O}$ ) and ammonia ( $NH3\text{NH}_3$ ) are both small, uncharged, monodentate ligands. Because they are practically identical in size, swapping one for the other does not cause any overcrowding around the central metal ion.

When exchange occurs between these similarly sized ligands, there is no change in coordination number and no change in the shape of the complex. An octahedral complex will remain octahedral.

Complete substitution

In some cases, all six water ligands can be replaced by ammonia. For example, with cobalt(II) ions:

[\text{Co}(\text{H}_2\text{O})_6]^{2+} + 6\text{NH}_3 \rightarrow [\text{Co}(\text{NH}_3)_6]^{2+} + 6\text{H}_2\text{O}

Here, both the reactant and the product are octahedral complexes with a coordination number of $6$ .

Incomplete substitution

Sometimes, the substitution doesn't go all the way to completion. Copper(II) ions are the classic A-level example of this. When excess concentrated ammonia is added to a solution containing hexaaquacopper(II) ions, only four of the six water molecules are replaced.

[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}

This forms a distinctively deep blue solution. The coordination number remains $6$ , and the shape remains octahedral.

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2. Monodentate ligands of different sizes

The chloride ion ( $Cl−\text{Cl}^-$ ) is much larger than both $H2O\text{H}_2\text{O}$ and $NH3\text{NH}_3$ . It is also negatively charged, so the ligands repel each other more strongly.

Because of this increased size and repulsion, you cannot fit six chloride ions around a central transition metal ion. Only four will fit. Therefore, exchanging water ligands for chloride ligands involves a change of coordination number (from $6$ to $4$ ) and a change of shape (from octahedral to tetrahedral).

This happens with metal ions like $Co2+\text{Co}^{2+}$ , $Cu2+\text{Cu}^{2+}$ , and $Fe3+\text{Fe}^{3+}$ when concentrated hydrochloric acid is added.

[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}

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Common Mistake

Forgetting the change in overall charge

When neutral ligands (like $H2O\text{H}_2\text{O}$ ) are replaced by charged ligands (like $Cl−\text{Cl}^-$ ), the overall charge of the complex ion must change. If a $Cu2+\text{Cu}^{2+}$ ion binds to four $Cl−\text{Cl}^-$ ions, the overall charge is $\times -1) = -2$ .

3. Bidentate and multidentate ligands

Ligands don't just have to form one coordinate bond.

Definition

Bidentate and Multidentate Ligands

Bidentate ligand: A molecule or ion that can form two coordinate bonds to a central metal ion.
Multidentate ligand: A molecule or ion that can form more than two coordinate bonds to a central metal ion.

You need to know specific examples of these:

Ethane-1,2-diamine ( $H2NCH2CH2NH2\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ , often abbreviated to 'en'): A neutral bidentate ligand. The lone pairs are on the two nitrogen atoms.
Ethanedioate ( $C2O42−\text{C}_2\text{O}_4^{2-}$ ): A bidentate ligand with a $- 2$ charge. The lone pairs are on the negatively charged oxygen atoms.
EDTA $^{4-}$ : A hexadentate ligand. It can form six coordinate bonds to a single central metal ion, entirely wrapping around it to form an exceptionally stable octahedral complex.

Haem and carbon monoxide poisoning

Haem is a naturally occurring multidentate complex found in blood. It consists of an iron(II) ( $Fe2+\text{Fe}^{2+}$ ) central ion coordinated to a multidentate ligand called a porphyrin ring (which forms four coordinate bonds to the iron). A protein chain (globin) provides a fifth coordinate bond.

This leaves exactly one coordinate bond position free. Oxygen ( $O2\text{O}_2$ ) forms a coordinate bond to this vacant site on the $Fe2+\text{Fe}^{2+}$ ion in haemoglobin, allowing oxygen to be transported around the body in the blood.

Carbon monoxide ( $CO\text{CO}$ ) is incredibly toxic because it can undergo a ligand substitution reaction with the oxygen in haemoglobin. The $CO\text{CO}$ molecule is a much better ligand than oxygen and binds to the $Fe2+\text{Fe}^{2+}$ ion irreversibly (or at least, much more strongly). This permanently replaces the oxygen, preventing the haemoglobin from transporting any more oxygen to your tissues.

4. The Chelate Effect

When a bidentate or multidentate ligand replaces a monodentate ligand, the resulting complex is remarkably stable. This is known as the chelate effect.

To explain why this happens, we must look at the thermodynamics of the substitution reaction, specifically focusing on the balance between enthalpy ( $ΔH\Delta H$ ) and entropy ( $ΔS\Delta S$ ) changes.

Consider the substitution of water ligands by $EDTA4−\text{EDTA}^{4-}$ :

[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + \text{EDTA}^{4-}(\text{aq}) \rightarrow [\text{Cu}(\text{EDTA})]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})

Example

Evaluating the feasibility of a chelation reaction

We can use the Gibbs free energy equation, $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , to explain why this reaction goes essentially to completion.

Evaluate the enthalpy change ( $ΔH\Delta H$ ): We are breaking six $Cu-O\text{Cu-O}$ bonds (in the water complex) and making six new bonds (a mix of $Cu-O\text{Cu-O}$ and $Cu-N\text{Cu-N}$ bonds in the EDTA complex). Because the bonds being broken and the bonds being made have similar strengths, the overall enthalpy change is very small ( $ΔH≈0\Delta H \approx 0$ ).
Determine the sign of the entropy change ( $ΔS\Delta S$ ): Count the particles in the equation. We start with $2$ species on the left (the aqua complex and the EDTA ion) and produce $7$ species on the right (the new complex and $6$ water molecules). An increase from $2$ particles to $7$ particles causes a massive increase in disorder, so $ΔS\Delta S$ is large and positive.
Substitute into the Gibbs equation: Since $ΔH\Delta H$ is roughly zero, the equation simplifies to $ΔG≈−TΔS\Delta G \approx -T\Delta S$ .
Determine feasibility: Temperature ( $T$ ) is always positive (in Kelvin), and $ΔS\Delta S$ is large and positive. Therefore, $−TΔS-T\Delta S$ will be a large negative number. This means $ΔG\Delta G$ is highly negative ( $ΔG<0\Delta G < 0$ ), making the substitution reaction highly feasible and practically irreversible.

Key Idea

Thermodynamics of the Chelate Effect

The chelate effect is primarily driven by a large, positive entropy change ( $ΔS\Delta S$ ), because replacing monodentate ligands with multidentate ligands dramatically increases the total number of particles in the solution.

Tip

Counting particles for entropy

Whenever an exam question asks you to explain the chelate effect, explicitly write down the number of moles of particles on the left versus the right side of the chemical equation. Stating " $\text{ moles} \rightarrow 7 \text{ moles}$ " is the clearest way to justify your claim that entropy increases.

Exam technique

In the exam

When writing equations for ligand substitution, always double-check the overall charge of the complex ion on the right-hand side. Sum the charge of the central metal ion and all the newly attached ligands.
If asked why a complex changes shape upon substitution (e.g. from $H2O\text{H}_2\text{O}$ to $Cl−\text{Cl}^-$ ), state clearly that the incoming ligand is larger and repels more strongly, so fewer can fit around the metal ion.
For chelate effect explanations, hit three key phrases: "enthalpy change is negligible as similar bonds are broken and made", "large increase in the number of particles causes a large increase in entropy ( $ΔS\Delta S$ is positive)", and " $ΔG\Delta G$ is therefore negative".
Remember that carbon monoxide poisoning is a ligand substitution reaction where $CO\text{CO}$ replaces $O2\text{O}_2$ by forming a coordinate bond to the $Fe2+\text{Fe}^{2+}$ central ion in haem.

Self review

Check yourself

Write the balanced equation for the incomplete substitution of hexaaquacopper(II) with ammonia.
Why does the coordination number change when $Cl−\text{Cl}^-$ replaces $H2O\text{H}_2\text{O}$ in a transition metal complex?
What are the charges on the bidentate ligands ethane-1,2-diamine and ethanedioate?
Explain the chelate effect in terms of $ΔH\Delta H$ and $ΔS\Delta S$ .

What you'll learn:

How the size of a ligand dictates whether the coordination number changes during a substitution reaction.
Why substitution is sometimes incomplete, resulting in a mixed-ligand complex.
The role of multidentate ligands in biological molecules like haemoglobin.
How to explain the exceptional stability of complexes with multidentate ligands using the 'chelate effect' and entropy.

1. Monodentate ligands of similar size

When exchange occurs between these similarly sized ligands, there is no change in coordination number and no change in the shape of the complex. An octahedral complex will remain octahedral.

Complete substitution

In some cases, all six water ligands can be replaced by ammonia. For example, with cobalt(II) ions:

[\text{Co}(\text{H}_2\text{O})_6]^{2+} + 6\text{NH}_3 \rightarrow [\text{Co}(\text{NH}_3)_6]^{2+} + 6\text{H}_2\text{O}

Here, both the reactant and the product are octahedral complexes with a coordination number of $6$ .

Incomplete substitution

[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}

This forms a distinctively deep blue solution. The coordination number remains $6$ , and the shape remains octahedral.

alt text

2. Monodentate ligands of different sizes

The chloride ion ( $Cl−\text{Cl}^-$ ) is much larger than both $H2O\text{H}_2\text{O}$ and $NH3\text{NH}_3$ . It is also negatively charged, so the ligands repel each other more strongly.

This happens with metal ions like $Co2+\text{Co}^{2+}$ , $Cu2+\text{Cu}^{2+}$ , and $Fe3+\text{Fe}^{3+}$ when concentrated hydrochloric acid is added.

[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}

alt text

Common Mistake

Forgetting the change in overall charge

3. Bidentate and multidentate ligands

Ligands don't just have to form one coordinate bond.

Definition

Bidentate and Multidentate Ligands

Bidentate ligand: A molecule or ion that can form two coordinate bonds to a central metal ion.
Multidentate ligand: A molecule or ion that can form more than two coordinate bonds to a central metal ion.

You need to know specific examples of these:

Ethane-1,2-diamine ( $H2NCH2CH2NH2\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ , often abbreviated to 'en'): A neutral bidentate ligand. The lone pairs are on the two nitrogen atoms.
Ethanedioate ( $C2O42−\text{C}_2\text{O}_4^{2-}$ ): A bidentate ligand with a $- 2$ charge. The lone pairs are on the negatively charged oxygen atoms.
EDTA $^{4-}$ : A hexadentate ligand. It can form six coordinate bonds to a single central metal ion, entirely wrapping around it to form an exceptionally stable octahedral complex.

Haem and carbon monoxide poisoning

4. The Chelate Effect

When a bidentate or multidentate ligand replaces a monodentate ligand, the resulting complex is remarkably stable. This is known as the chelate effect.

Consider the substitution of water ligands by $EDTA4−\text{EDTA}^{4-}$ :

[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + \text{EDTA}^{4-}(\text{aq}) \rightarrow [\text{Cu}(\text{EDTA})]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})

Example

Evaluating the feasibility of a chelation reaction

We can use the Gibbs free energy equation, $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , to explain why this reaction goes essentially to completion.

Evaluate the enthalpy change ( $ΔH\Delta H$ ): We are breaking six $Cu-O\text{Cu-O}$ bonds (in the water complex) and making six new bonds (a mix of $Cu-O\text{Cu-O}$ and $Cu-N\text{Cu-N}$ bonds in the EDTA complex). Because the bonds being broken and the bonds being made have similar strengths, the overall enthalpy change is very small ( $ΔH≈0\Delta H \approx 0$ ).
Determine the sign of the entropy change ( $ΔS\Delta S$ ): Count the particles in the equation. We start with $2$ species on the left (the aqua complex and the EDTA ion) and produce $7$ species on the right (the new complex and $6$ water molecules). An increase from $2$ particles to $7$ particles causes a massive increase in disorder, so $ΔS\Delta S$ is large and positive.
Substitute into the Gibbs equation: Since $ΔH\Delta H$ is roughly zero, the equation simplifies to $ΔG≈−TΔS\Delta G \approx -T\Delta S$ .
Determine feasibility: Temperature ( $T$ ) is always positive (in Kelvin), and $ΔS\Delta S$ is large and positive. Therefore, $−TΔS-T\Delta S$ will be a large negative number. This means $ΔG\Delta G$ is highly negative ( $ΔG<0\Delta G < 0$ ), making the substitution reaction highly feasible and practically irreversible.

Key Idea

Thermodynamics of the Chelate Effect

Tip

Counting particles for entropy

Exam technique

In the exam

When writing equations for ligand substitution, always double-check the overall charge of the complex ion on the right-hand side. Sum the charge of the central metal ion and all the newly attached ligands.
If asked why a complex changes shape upon substitution (e.g. from $H2O\text{H}_2\text{O}$ to $Cl−\text{Cl}^-$ ), state clearly that the incoming ligand is larger and repels more strongly, so fewer can fit around the metal ion.
For chelate effect explanations, hit three key phrases: "enthalpy change is negligible as similar bonds are broken and made", "large increase in the number of particles causes a large increase in entropy ( $ΔS\Delta S$ is positive)", and " $ΔG\Delta G$ is therefore negative".
Remember that carbon monoxide poisoning is a ligand substitution reaction where $CO\text{CO}$ replaces $O2\text{O}_2$ by forming a coordinate bond to the $Fe2+\text{Fe}^{2+}$ central ion in haem.

Self review

Check yourself

Write the balanced equation for the incomplete substitution of hexaaquacopper(II) with ammonia.
Why does the coordination number change when $Cl−\text{Cl}^-$ replaces $H2O\text{H}_2\text{O}$ in a transition metal complex?
What are the charges on the bidentate ligands ethane-1,2-diamine and ethanedioate?
Explain the chelate effect in terms of $ΔH\Delta H$ and $ΔS\Delta S$ .

1. Monodentate ligands of similar size

Complete substitution

Incomplete substitution

2. Monodentate ligands of different sizes

3. Bidentate and multidentate ligands

Haem and carbon monoxide poisoning

4. The Chelate Effect

Review Substitution reactions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Substitution reactions (A-level only)

1. Monodentate ligands of similar size

Complete substitution

Incomplete substitution

2. Monodentate ligands of different sizes

3. Bidentate and multidentate ligands

Haem and carbon monoxide poisoning

4. The Chelate Effect

Review Substitution reactions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Substitution reactions (A-level only)

1. Monodentate ligands of similar size

Complete substitution

Incomplete substitution

2. Monodentate ligands of different sizes

3. Bidentate and multidentate ligands

Haem and carbon monoxide poisoning

4. The Chelate Effect

Review Substitution reactions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Substitution reactions (A-level only) Revision Guide

Substitution reactions (A-level only)

1. Monodentate ligands of similar size

Complete substitution

Incomplete substitution

2. Monodentate ligands of different sizes

3. Bidentate and multidentate ligands

Haem and carbon monoxide poisoning

4. The Chelate Effect

Review Substitution reactions (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Substitution reactions (A-level only) Revision Guide