Monodentate Ligand Exchange Revision Notes | A Level AQA Chemistry
When large chloride ions replace water ligands, the co-ordination number falls from 6 to 4, as in [CoCl₄]²⁻. Free AQA notes and 12 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Monodentate Ligand Exchange is part of Transition metals (A-level only) in A Level AQA Chemistry.
6 min
Reading time
26
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A Level AQA
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Content review
What this lesson covers
When large chloride ions replace water ligands, the co-ordination number falls from 6 to 4, as in [CoCl₄]²⁻. Free AQA notes and 12 questions answered.
Explain why H2O, NH3, and Cl- can act as monodentate ligands, with reference to lone pairs and charge
Describe and explain ligand exchange reactions involving NH3 and H2O, including why co-ordination number is retained
Explain why substitution of H2O by Cl- can involve a change in co-ordination number, with reference to ligand size
Apply knowledge of ligand exchange to specific transition metal ions including Co2+, Cu2+, and Fe3+
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Use this lesson to stay inside Transition metals (A-level only) while connecting the detail back to the wider Inorganic chemistry topic.
Monodentate ligand: A ligand that donates one lone pair of electrons to a central metal ion, forming a single co-ordinate (dative covalent) bond. Examples include H2O, NH3, and Cl-.
Ligand exchange: A reaction in which one or more ligands in a complex ion are replaced by other ligand species. The co-ordination number may or may not change depending on the relative sizes of the ligands involved.
Co-ordination number: The total number of co-ordinate bonds formed between ligands and the central metal ion in a complex. Common values at A-Level are 4 and 6.
Co-ordinate (dative covalent) bond: A covalent bond in which both electrons in the shared pair are donated by one atom or ion — in transition metal chemistry, always donated by the ligand to the metal ion.
Revision notes
Read the core explanations from Monodentate Ligand Exchange before testing yourself from memory.
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.
Feature
H2O
NH3
Cl-
Charge
0 (neutral)
0 (neutral)
-1
Relative size
Small
Small (similar to H2O)
Large
Donor atom
Oxygen
Nitrogen
Chlorine
Lone pairs available
2
1
4
Typical co-ordination number retained?
Yes (when swapped with NH3)
Yes (when swapped with H2O)
No — changes from 6 to 4
Geometry of resulting complex
Octahedral (CN = 6)
Octahedral (CN = 6)
Tetrahedral (CN = 4)
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]
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]
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]
$$[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CuCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)$$ — 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]
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.
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.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
A monodentate ligand forms one co-ordinate bond to the central metal ion by donating a single lone pair. Water donates via oxygen, ammonia via nitrogen and the chloride ion via one of its four lone pairs.
Why does the coordination number stay the same when ammonia replaces water?
When ammonia replaces water ligands, the co-ordination number stays at 6 because NH₃ and H₂O are similar in size and both uncharged. Each incoming NH₃ occupies the same space as the departing H₂O, so the octahedral geometry is kept, as when [Co(H₂O)₆]²⁺ reacts with excess ammonia.
Why does the coordination number change when chloride replaces water?
The co-ordination number typically falls from 6 to 4 when Cl⁻ replaces water because chloride is much larger than H₂O, so fewer chloride ligands fit around the metal ion: [Cu(H₂O)₆]²⁺ + 4 Cl⁻ ⇌ [CuCl₄]²⁻ + 6 H₂O. The charge on the complex also changes because each chloride carries −1.
Why does cobalt(II) solution turn from pink to blue with concentrated HCl?
Adding concentrated hydrochloric acid to pink [Co(H₂O)₆]²⁺ forms blue [CoCl₄]²⁻, as the large chloride ligands replace water and the co-ordination number falls from 6 to 4. The reaction is reversible, so adding water shifts it back towards the pink aqua complex.
Why is the substitution of water by ammonia in copper(II) incomplete?
Adding excess ammonia to [Cu(H₂O)₆]²⁺ replaces only four water ligands, giving the deep blue [Cu(NH₃)₄(H₂O)₂]²⁺ ion. The equilibrium does not favour replacing the two axial water molecules, so substitution is incomplete and the co-ordination number stays at 6.
What happens when concentrated HCl is added to iron(III) ions?
When concentrated hydrochloric acid is added to [Fe(H₂O)₆]³⁺, the pale yellow octahedral aqua complex is replaced by the yellow tetrahedral [FeCl₄]⁻ ion. The coordination number falls from 6 to 4 because Cl⁻ is larger than H₂O, and the overall charge changes from 3+ to 1−.
What happens when excess ammonia is added to cobalt(II) ions?
When excess ammonia is added to [Co(H₂O)₆]²⁺, the six water ligands are progressively replaced by ammonia molecules. The coordination number stays at 6 throughout because NH₃ and H₂O are similar in size and both uncharged, so each NH₃ occupies the same space as the H₂O it replaces.
Why does the charge on a complex change when Cl⁻ replaces water?
The charge on a complex changes when Cl⁻ replaces water because each chloride carries a −1 charge, while water is uncharged. For example, [Cu(H₂O)₆]²⁺ becomes [CuCl₄]²⁻: Cu²⁺ contributes +2 and four Cl⁻ contribute −4, so the complex is 2−.
How many lone pairs do water, ammonia and chloride ligands have?
Water has 2 lone pairs and donates through oxygen, ammonia has 1 lone pair and donates through nitrogen, and the chloride ion has 4 lone pairs and donates through chlorine. Each acts as a monodentate ligand, donating just one lone pair to the central metal ion.
What is the charge on the CuCl₄ complex ion?
The charge on the [CuCl₄]²⁻ complex ion is 2−. Cu²⁺ contributes +2 and four Cl⁻ ligands contribute −4. Writing [CuCl₄]²⁺ is a common error that comes from forgetting the charge of the chloride ligands.
Is ligand substitution reversible?
Ligand substitution with chloride is reversible. Concentrated hydrochloric acid provides a high Cl⁻ concentration that drives the equilibrium towards the chloro complex, and adding water shifts it back towards the aqua complex, reversing the colour change.
What is a coordinate bond in a complex ion?
A co-ordinate bond, or dative covalent bond, in a complex ion forms when a ligand donates a lone pair of electrons directly into an empty orbital of the central transition metal ion. Water donates via oxygen, ammonia via nitrogen and chloride via one of its lone pairs.