AQA A-Level Inorganic Chemistry 3.2.5 Transition Metals Ligand Substitution Reactions
3.2.5

Ligand Substitution Reactions

Learn the mechanisms, equations, and colour changes for ligand exchange, and master the thermodynamic principles of the chelate effect.

In a aqueous solution, transition metal ions exist as hexaaqua complexes, where six water molecules act as monodentate ligands. These water ligands can be replaced by other molecules or ions in a process known as ligand substitution. These reactions are often accompanied by changes in coordination number, shape, and colour.

📖 Definition: Ligand Substitution

A ligand substitution reaction is a chemical process in which one or more ligands bonded to a central metal ion in a complex are replaced by other ligands.

Substitution by Neutral Ligands of Similar Size

When ligands are of a similar size and carry no charge, such as water (\(\text{H}_2\text{O}\)) and ammonia (\(\text{NH}_3\)), substitution can occur without a change in the coordination number or the overall shape of the complex.

The Reaction of Copper(II) Complexes with Ammonia

When excess ammonia is added to a solution containing hexaaquacopper(II) ions, a ligand substitution reaction takes place. However, only four of the six water ligands are replaced:

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

During this reaction, the following changes occur:

Substitution by Larger, Charged Ligands

When small, neutral ligands (like \(\text{H}_2\text{O}\)) are replaced by larger, negatively charged ligands (like chloride ions, \(\text{Cl}^-\)), both the coordination number and the shape of the complex typically change. This is due to electrostatic repulsion and steric crowding around the central metal ion.

🔑 Key Principle: Coordination Number Changes with Chloride

Chloride ions (\(\text{Cl}^-\)) are larger than water molecules and carry a negative charge. Their larger size means fewer can pack around the metal ion, and their negative charges repel one another. Consequently, only four chloride ligands can coordinate to the metal ion, reducing the coordination number from 6 to 4 and changing the shape from octahedral to tetrahedral.

The Reaction of Copper(II) Complexes with Chloride

Adding concentrated hydrochloric acid (a source of \(\text{Cl}^-\) ions) to a solution of hexaaquacopper(II) ions results in the following reversible reaction:

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

This reaction displays distinct characteristics:

The Reaction of Cobalt(II) Complexes with Chloride

Similarly, adding concentrated hydrochloric acid to a solution containing pink hexaaquacobalt(II) ions drives a substitution reaction:

\[ [\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CoCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \]
Copper(II) Ligand Substitution Flowchart OCTAHEDRAL [Cu(H₂O)₆]²⁺ Pale Blue Solution TETRAHEDRAL [CuCl₄]²⁻ Green-Yellow Sol. + 4Cl⁻ (conc. HCl) - 6H₂O OCTAHEDRAL [Cu(NH₃)₄(H₂O)₂]²⁺ Deep Blue Solution + 4NH₃ (excess) - 4H₂O

The Chelate Effect and Entropy

Complexes containing multidentate ligands (such as bidentate 1,2-diaminoethane or hexadentate \(\text{EDTA}^{4-}\)) are significantly more stable than complexes containing monodentate ligands. This phenomenon is known as the chelate effect.

📖 Definition: Chelate Effect

The chelate effect refers to the thermodynamic stabilization of transition metal complexes containing multidentate ligands compared to those containing only monodentate ligands, driven primarily by a positive change in entropy.

The chelate effect is explained by the thermodynamic equation for Gibbs free energy change:

\[ \Delta G = \Delta H - T\Delta S \]

When a multidentate ligand replaces monodentate ligands, the reaction proceeds with a very small enthalpy change (\(\Delta H \approx 0\)) because the coordinate bonds being broken are very similar in strength to the new coordinate bonds being formed (e.g. oxygen-metal bonds replaced by nitrogen-metal bonds).

However, there is a large, positive entropy change (\(\Delta S > 0\)) because the number of free particles in solution increases:

🔑 Key Principle: Entropy and the Chelate Effect

Compare these two substitution reactions:

  • Bidentate Substitution: \[ [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 3\text{en}(\text{aq}) \rightarrow [\text{Cu}(\text{en})_3]^{2+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \] There are 4 reactant particles (1 complex + 3 ligands) forming 7 product particles (1 complex + 6 waters). The increase in particles increases disorder, meaning \(\Delta S\) is positive.
  • Hexadentate Substitution: \[ [\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + \text{EDTA}^{4-}(\text{aq}) \rightarrow [\text{Cu}(\text{EDTA})]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \] Here, 2 reactant particles form 7 product particles. This results in an even larger increase in particles and a highly positive \(\Delta S\).
A positive \(\Delta S\) value makes the term \(-T\Delta S\) negative. Since \(\Delta H \approx 0\), the overall Gibbs free energy change \(\Delta G\) becomes highly negative, making the reaction extremely favourable and virtually irreversible.

Entropy Increase in Chelation Reactions [M(H₂O)₆]²⁺ EDTA⁴⁻ 2 Particles (Lower Entropy) ΔS > 0 [M(EDTA)]²⁻ W W W W W W 7 Particles (Higher Entropy)

Haemoglobin and Carbon Monoxide Poisoning

Haemoglobin is an iron-containing protein found in red blood cells that is responsible for transporting oxygen around the body. The structure of haemoglobin features a central iron(II) ion (\(\text{Fe}^{2+}\)) situated in an octahedral complex:

Oxygen Transport Mechanism

In the lungs, where the oxygen concentration is high, water at the sixth coordinate site is replaced by an oxygen molecule via ligand substitution:

\[ \text{Haemoglobin} + \text{O}_2 \rightleftharpoons \text{Oxyhaemoglobin} \]

This oxyhaemoglobin is carried through the bloodstream to tissues that require oxygen. In the tissues, where oxygen concentration is low, the equilibrium shifts to the left, and the oxygen ligand is released, returning to haemoglobin.

Carbon Monoxide Poisoning

Carbon monoxide (\(\text{CO}\)) can also act as a ligand, coordinating to the \(\text{Fe}^{2+}\) ion at the sixth site. However, carbon monoxide forms a far stronger coordinate bond with the iron(II) ion than oxygen does. This results in a ligand substitution reaction that is essentially irreversible under normal physiological conditions:

\[ \text{Oxyhaemoglobin} + \text{CO} \rightarrow \text{Carboxyhaemoglobin} + \text{O}_2 \]

Because the coordinate bond in carboxyhaemoglobin is so strong, the carbon monoxide ligand does not readily dissociate. This permanently blocks the oxygen-binding site, preventing haemoglobin from transporting oxygen, which leads to oxygen starvation in vital organs (carbon monoxide poisoning).

📝 AQA Examiner Tip

When discussing carbon monoxide poisoning in the exam, always state that carbon monoxide forms a stronger coordinate bond with the iron(II) ion in haemoglobin than oxygen. Avoid saying that carbon monoxide binds "irreversibly" without explaining the difference in bond strength, as it is the stronger dative covalent bond that drives this substitution and prevents oxygen transport.

✏️ Worked Example: Copper and Hydrochloric Acid
Write the equation for the reaction that occurs when excess concentrated hydrochloric acid is added to an aqueous solution of copper(II) sulfate. State the shape, coordination number, and colour of both the reactant and product complexes.

Step 1: Write the balanced chemical equation

Concentrated hydrochloric acid provides chloride ions (\(\text{Cl}^-\)) which replace water ligands in the hexaaquacopper(II) complex:

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

Step 2: Identify the properties of the reactant complex

  • Formula: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\)
  • Shape: Octahedral
  • Coordination number: 6
  • Colour: Pale blue

Step 3: Identify the properties of the product complex

  • Formula: \([\text{CuCl}_4]^{2-}\)
  • Shape: Tetrahedral
  • Coordination number: 4
  • Colour: Yellow (though the mixture will appear green-yellow)
✏️ Worked Example: Chelate Effect Thermodynamics
The reaction of a cobalt(II) aqua complex with EDTA⁴⁻ is represented by the equation below: \[ [\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + \text{EDTA}^{4-}(\text{aq}) \rightleftharpoons [\text{Co}(\text{EDTA})]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \] The enthalpy change (\(\Delta H\)) for this reaction is close to zero. Explain, in terms of entropy, why this reaction has a very large equilibrium constant.

Step 1: Count the number of species on each side of the equation

  • Reactant side: 1 complex ion + 1 EDTA⁴⁻ ion = 2 species
  • Product side: 1 metal-EDTA complex ion + 6 water molecules = 7 species

Step 2: Relate the change in species count to entropy (\(\Delta S\))

Since the number of particles in solution increases from 2 to 7, there is a significant increase in the disorder of the system. Therefore, the entropy change (\(\Delta S\)) is positive.

Step 3: Link entropy and enthalpy to Gibbs free energy (\(\Delta G\))

Using \(\Delta G = \Delta H - T\Delta S\):

  • Since \(\Delta H \approx 0\), the equation simplifies to \(\Delta G \approx -T\Delta S\).
  • Since \(\Delta S\) is positive, \(-T\Delta S\) is negative, which makes \(\Delta G\) highly negative.

A highly negative \(\Delta G\) indicates that the reaction is highly thermodynamically feasible, meaning the equilibrium lies far to the right and the equilibrium constant is very large.

📝 AQA Examiner Tip: Ammonia acting as a base

Be careful when adding ammonia (\(\text{NH}_3\)) to iron(II), iron(III), or aluminium(III) aqua complexes. Under these conditions, ammonia acts as a Bronsted-Lowry base by accepting protons from the water ligands, rather than acting as a nucleophile to perform ligand substitution. This results in the formation of neutral, insoluble metal hydroxides like \(\text{Fe}(\text{OH})_2\), \(\text{Fe}(\text{OH})_3\), and \(\text{Al}(\text{OH})_3\) as precipitates. We will cover this in detail in Topic 3.2.6 Aqueous Ions.

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