Monodentate Ligand Exchange Revision Notes | A Level AQA Chemistry

1. What Makes a Ligand? The Lone Pair Requirement

Picture a transition metal ion sitting in solution — it carries a positive charge and has empty orbitals available. Any species that can donate a lone pair of electrons directly into those empty orbitals becomes a ligand, forming a co-ordinate (dative covalent) bond.

Three species you must know for AQA are water ($\text{H}_2\text{O}$), ammonia ($\text{NH}_3$), and the chloride ion ($\text{Cl}^-$). Each carries at least one lone pair and can therefore act as a monodentate ligand — donating exactly one lone pair per ligand molecule or ion to the central metal.

Notice that $\text{H}_2\text{O}$ and $\text{NH}_3$ are both uncharged molecules of similar size, whereas $\text{Cl}^-$ carries a negative charge and is significantly larger. This distinction in size and charge drives the differences in ligand exchange behaviour you need to explain in the exam.

2. NH3 and H2O Exchange — Same Size, Same Co-ordination Number

Because $\text{NH}_3$ and $\text{H}_2\text{O}$ are similar in size and both uncharged, one can replace the other in a complex without altering the geometry or the co-ordination number. The metal ion swaps one ligand for another of comparable dimensions.

A classic example is the reaction of the hexaaquacobalt(II) ion with excess ammonia. The six water ligands are progressively replaced by ammonia molecules, but the co-ordination number stays at 6 throughout because each incoming $\text{NH}_3$ occupies exactly the same space as the departing $\text{H}_2\text{O}$.

The same pattern holds for copper(II). Adding excess ammonia to the hexaaquacopper(II) ion produces a deep blue solution — but here the substitution stops after four ammonia molecules have replaced four water ligands. This is an example of incomplete substitution, and the product is $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$, the tetraamminecopper(II) ion.

The equilibrium does not favour replacement of the remaining two water ligands under normal conditions — which is why the substitution is described as incomplete.

3. Why Cl- Exchange Changes the Co-ordination Number

Here is where the chemistry becomes more interesting. The chloride ion is significantly larger than either $\text{H}_2\text{O}$ or $\text{NH}_3$. When $\text{Cl}^-$ replaces water ligands, fewer chloride ions can fit around the central metal ion — the steric bulk forces a change in co-ordination number, typically from 6 down to 4. AQA wording is that ligand exchange can involve a change in coordination number, typically from 6 to 4, rather than that it always does.

The charge also changes. Chloride carries a $-1$ charge, so replacing uncharged water ligands with chloride ions alters the overall charge on the complex. This is why the colour and properties of the solution change so dramatically when concentrated hydrochloric acid is added to a transition metal aqua complex.

The equations for these reactions follow the same pattern: concentrated $\text{HCl}$ provides a high concentration of $\text{Cl}^-$ ions, which drives the equilibrium towards the chloro complex. The reaction is reversible — adding water shifts it back towards the aqua complex.

4. Comparing the Three Ligands — A Structured Analysis

Now that you have seen each ligand in action, it is worth drawing the comparisons together explicitly, because AQA 'compare' and 'explain' questions will ask you to do exactly this.

The key principle to carry into any exam question is this: when two ligands are similar in size and charge, exchange occurs without a change in co-ordination number. When one ligand is significantly larger (as $\text{Cl}^-$ is relative to $\text{H}_2\text{O}$), the co-ordination number typically decreases to accommodate the bulkier ligands.

5. Worked Example — Writing and Explaining Ligand Exchange Equations

Exam questions on ligand exchange typically ask you to write an equation, state the change in co-ordination number, and explain why that change occurs (or does not occur). The step-by-step below models the full mark-winning response for a copper(II) question.

Worked example: Writing a Ligand Exchange Equation and Explanation for Cu2+

Write an equation for the reaction of [Cu(H2O)6]2+(aq) with excess concentrated hydrochloric acid. State the co-ordination number of the product and explain why it differs from that of the reactant complex. [4 marks]

1. Reactant ligand: $\text{H}_2\text{O}$ (neutral, small) Incoming ligand: $\text{Cl}^-$ (charged, large) — Recognising the size and charge difference is the foundation of the explanation. [B1 — identifying Cl- as larger]
2. Reactant: $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ — co-ordination number 6 Product: $[\text{CuCl}_4]^{2-}$ — co-ordination number 4 — Because Cl- is larger, only 4 can fit around Cu2+, reducing the co-ordination number from 6 to 4. [B1 — correct co-ordination number of product]
3. $$[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CuCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)$$ — Check: 1 Cu on each side; 4 Cl on each side; 6 O (as H2O) on each side; charge: 2+ + 4(1-) = 2- on left, 2- on right. Balanced. [M1 — correct equation; A1 — balanced with state symbols]
4. Cl- is larger than H2O. Fewer Cl- ligands can fit around the central Cu2+ ion. The co-ordination number therefore decreases from 6 to 4. — This is the mark-winning explanation. Refer explicitly to the relative sizes of the ligands and link this directly to the change in co-ordination number.
5. Left side: $2+ + 4 \times (1-) = 2+ - 4 = 2-$ Right side: $[\text{CuCl}_4]^{2-} = 2-$ Charges match. — Always verify charge balance in complex ion equations — examiners check this and it is a common source of lost marks.

Answer: $$[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CuCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)$$ Co-ordination number changes from 6 to 4 because Cl- is larger than H2O, so fewer Cl- ligands can fit around the central Cu2+ ion.

Notice how the charge balance check in Step 5 is non-negotiable. A common error is writing $[\text{CuCl}_4]^{2+}$ instead of $[\text{CuCl}_4]^{2-}$ — forgetting that four $\text{Cl}^-$ ligands contribute $-4$ to the overall charge of the complex.

6. Incomplete Substitution — Why [Cu(NH3)4(H2O)2]2+ Forms

Incomplete substitution is a concept that trips up many students because it seems to contradict the idea that excess reagent drives a reaction to completion. The key is that the equilibrium position for replacing the final two water ligands in copper(II) does not favour full substitution — the axial water molecules are held in a different electronic environment and are not displaced by ammonia under standard conditions.

The product $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ is the AQA-specified example of incomplete substitution. Four equatorial positions are occupied by $\text{NH}_3$, and two axial positions retain $\text{H}_2\text{O}$. The co-ordination number remains 6 throughout — this is not a change in geometry, only a partial replacement of ligands.

When writing this formula in an exam, take care with the charge: $\text{Cu}^{2+}$ contributes $+2$; four $\text{NH}_3$ and two $\text{H}_2\text{O}$ are all neutral, so the overall charge on the complex is $2+$. The formula is $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$.

7. Exam Technique — Scoring Full Marks on Ligand Exchange Questions

Ligand exchange questions appear regularly on AQA Paper 2 and can carry up to 6 marks. The marks are distributed across the equation, the co-ordination number, the explanation of size, and the charge balance. Missing any one of these is the difference between 4 and 6 marks.

Do not overstate the geometry rule. Ligand exchange can involve a change in coordination number, but that depends on the specific complex and ligand set involved.