Ligand exchange and the Stability of Complexes
A concise revision guide to ligand exchange: replacing water ligands by ammonia and chloride in copper(II) and cobalt(II) complexes, changes in coordination number and colour, and stability constants, Kstab.
- 28.2.1-a
- 28.2.1-b
- 28.2.1-c
- 28.2.7-a
- 28.2.7-b
- 28.3.5-a
- 28.3.5-b
- 28.5.1
- 28.5.2
- 28.5.3
- 28.5.4-a
- 28.5.4-b
What these spec points say
- 28.2.1-a describe and explain the reactions of transition elements with ligands to form complexes
- 28.2.1-b describe and explain the complexes of copper(II) ions with water and ammonia molecules and hydroxide and chloride ions
- 28.2.1-c describe and explain the complexes of cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions
- 28.2.7-a explain qualitatively that ligand exchange can occur, including the complexes of copper(II) ions with water and ammonia molecules and hydroxide and chloride ions
- 28.2.7-b explain qualitatively that ligand exchange can occur, including the complexes of cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions
- 28.3.5-a use the complexes of copper(II) ions with water and ammonia molecules and hydroxide and chloride ions as examples of ligand exchange affecting the colour observed
- 28.3.5-b use the complexes of cobalt(II) ions with water and ammonia molecules and hydroxide and chloride ions as examples of ligand exchange affecting the colour observed
- 28.5.1 define the stability constant, Kstab, of a complex as the equilibrium constant for the formation of the complex ion in a solvent (from its constituent ions or molecules)
- 28.5.2 write an expression for a Kstab of a complex ([H2O] should not be included)
- 28.5.3 use Kstab expressions to perform calculations
- 28.5.4-a describe and explain ligand exchanges in terms of Kstab values
- 28.5.4-b understand that a large Kstab is due to the formation of a stable complex ion
Before You Start
This page builds on Ligands, Complex Ions and Their Shapes and Colour of Transition element Ions.
What you need to remember
- A ligand donates a lone pair of electrons to a central metal ion to form a dative covalent bond.
- The coordination number is the number of dative covalent bonds to the central metal ion.
- In water, a metal ion exists as an aqua ion such as [Cu(H₂O)₆]²⁺, which is octahedral.
- Small ligands such as H₂O and NH₃ give octahedral complexes. The larger Cl⁻ ion gives tetrahedral complexes.
- The colour of a complex depends on the metal, its oxidation state, the ligands and the coordination number.
- An equilibrium constant is written as products over reactants, with each concentration raised to the power of its coefficient.
What Ligand Exchange Means
In a ligand exchange reaction, one ligand in a complex is replaced by another. Most examples start from an aqua ion, in which water ligands are replaced.
- The incoming ligand uses a lone pair to form a new dative covalent bond to the metal ion.
- The oxidation state of the metal does not change, so the reaction is not a redox reaction.
- All of the ligands may be replaced, or only some of them.
- The coordination number and the shape may stay the same or may change.
- The colour changes, because different ligands split the d orbitals by a different amount.
| Incoming ligand | Compared with H₂O | Effect on the coordination number |
|---|---|---|
| NH₃ | similar size, no charge | stays at 6: the complex stays octahedral |
| Cl⁻ | larger, and negatively charged | falls from 6 to 4: the complex becomes tetrahedral |
Ligand exchange in copper(II) complexes: ammonia gives a deep blue solution and concentrated hydrochloric acid gives a yellow solution.
Key idea: Ligand exchange swaps one ligand for another. The oxidation state of the metal stays the same, but the colour changes, and the coordination number changes when the new ligand is larger.
Copper(II) with Ammonia
Aqueous copper(II) ions are blue: [Cu(H₂O)₆]²⁺. Aqueous ammonia is added drop by drop, and then in excess.
| Ammonia added | Observation | Copper species formed |
|---|---|---|
| a few drops | pale blue precipitate | Cu(OH)₂(H₂O)₄, copper(II) hydroxide |
| excess | precipitate dissolves to give a deep blue solution | [Cu(NH₃)₄(H₂O)₂]²⁺ |
The overall change is a ligand exchange reaction:
[Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l)
The two stages
- A few drops: ammonia acts as a base. [Cu(H₂O)₆]²⁺ + 2NH₃ → Cu(OH)₂(H₂O)₄ + 2NH₄⁺
- Excess: ammonia acts as a ligand and the precipitate dissolves. Cu(OH)₂(H₂O)₄ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2H₂O + 2OH⁻
The precipitate stage is explained on Reactions of Metal Ions with Sodium Hydroxide and Ammonia.
What to notice
- Only four of the six water ligands are replaced: the exchange is incomplete.
- The coordination number stays at 6 and the shape stays octahedral, because NH₃ and H₂O are similar in size and both are uncharged.
- The charge stays at 2+, because both ligands are neutral molecules.
- The two water ligands that remain are opposite each other, above and below the square of four ammonia ligands.
![Wedge-and-dash drawings of two octahedral copper(II) complexes joined by an arrow labelled + 4NH₃ and − 4H₂O: the blue hexaaquacopper(II) ion, [Cu(H₂O)₆]²⁺, becomes the deep blue [Cu(NH₃)₄(H₂O)₂]²⁺ ion, whose two remaining water ligands are opposite each other.](https://www.onlinelearningsystem.net/wp-content/uploads/2026/09/transition-t17-cuexchange-v2.svg)
The hexaaquacopper(II) ion and the tetraamminediaquacopper(II) ion. Both are octahedral, and the two remaining water ligands are opposite each other.
Check: Ligand Exchange with Ammonia
Decide what changes and what stays the same when ammonia replaces water ligands: the number of ligands replaced, the coordination number, the shape, the charge and the colour.
Copper(II) and Cobalt(II) with Chloride Ions
Concentrated hydrochloric acid supplies a high concentration of chloride ions. With aqueous copper(II) ions, four chloride ligands replace all six water ligands.
[Cu(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6H₂O(l)
- Colour: blue → green → yellow. The green solution is a mixture of the blue and yellow complexes.
- Coordination number: 6 → 4. The shape changes from octahedral to tetrahedral.
- Reason: Cl⁻ ions are larger than water molecules and are charged, so only four fit round the Cu²⁺ ion.
- Charge: (2+) + 4 × (1−) = 2−.
A reversible reaction
- Adding more concentrated hydrochloric acid moves the equilibrium to the right: the solution turns yellow.
- Adding water lowers the chloride concentration and moves the equilibrium to the left: the solution turns blue again.
Cobalt(II)
[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l)
- The pink solution turns blue as concentrated hydrochloric acid is added.
- The coordination number again falls from 6 to 4, and the shape changes from octahedral to tetrahedral.
- Adding water turns the blue solution pink again.
Ammonia and water are similar in size, so six ligands still fit round the metal ion. Chloride ions are larger, so only four fit.
The cobalt(II) equilibrium: concentrated hydrochloric acid turns the pink solution blue, and adding water turns it pink again.
Common error: Dissolving solid copper(II) chloride in water gives blue [Cu(H₂O)₆]²⁺, not [CuCl₄]²⁻. The chloride complex forms only when the concentration of chloride ions is very high.
Check: Chloride Ions and a Change in Coordination Number
Work with chloride complexes: the change in the number of ligands, the shape, the overall charge and what dilution does to the equilibrium.
Cobalt(II) with Ammonia
With excess ammonia, all six water ligands of the cobalt(II) aqua ion are replaced.
[Co(H₂O)₆]²⁺(aq) + 6NH₃(aq) → [Co(NH₃)₆]²⁺(aq) + 6H₂O(l)
- The pink solution becomes a pale yellow-brown (straw coloured) solution.
- A blue precipitate of cobalt(II) hydroxide, Co(OH)₂(H₂O)₄, forms first with a few drops of ammonia. It dissolves in excess.
- The exchange is complete, and the coordination number stays at 6.
- The solution darkens on standing in air: oxygen oxidises [Co(NH₃)₆]²⁺ to [Co(NH₃)₆]³⁺.
Copper(II) and cobalt(II) complexes compared
| Ligand | Copper(II) | Cobalt(II) |
|---|---|---|
| H₂O | [Cu(H₂O)₆]²⁺, blue solution | [Co(H₂O)₆]²⁺, pink solution |
| OH⁻ | Cu(OH)₂(H₂O)₄, pale blue precipitate | Co(OH)₂(H₂O)₄, blue precipitate |
| NH₃ (excess) | [Cu(NH₃)₄(H₂O)₂]²⁺, deep blue solution | [Co(NH₃)₆]²⁺, yellow-brown solution |
| Cl⁻ (concentrated) | [CuCl₄]²⁻, yellow solution | [CoCl₄]²⁻, blue solution |
- Each ligand splits the d orbitals by a different energy gap, ΔE.
- A different ΔE means that light of a different frequency is absorbed.
- The colour seen is the complementary colour, so each complex has its own colour: see Colour of Transition element Ions.
- The reactions with hydroxide ions are covered on Reactions of Metal Ions with Sodium Hydroxide and Ammonia.
Summary and Unfamiliar Reactions
| Reaction | Complex formed | Colour change | Coordination number | Shape |
|---|---|---|---|---|
| [Cu(H₂O)₆]²⁺ + 4NH₃ | [Cu(NH₃)₄(H₂O)₂]²⁺ | blue → deep blue | 6 → 6 | octahedral → octahedral |
| [Cu(H₂O)₆]²⁺ + 4Cl⁻ | [CuCl₄]²⁻ | blue → yellow | 6 → 4 | octahedral → tetrahedral |
| [Co(H₂O)₆]²⁺ + 4Cl⁻ | [CoCl₄]²⁻ | pink → blue | 6 → 4 | octahedral → tetrahedral |
| [Co(H₂O)₆]²⁺ + 6NH₃ | [Co(NH₃)₆]²⁺ | pink → yellow-brown | 6 → 6 | octahedral → octahedral |
Writing an equation for an unfamiliar complex
Questions often give the formula of a new complex. The equation is built in the same way each time.
- Write the aqua ion and the incoming ligand on the left.
- Write the new complex on the right, with its charge.
- Release one water molecule for every water ligand replaced.
- Check that the total charge is the same on both sides.
Worked example 1
Excess ammonia converts [Ni(H₂O)₆]²⁺ into [Ni(NH₃)₆]²⁺. Write the equation.
Answer. [Ni(H₂O)₆]²⁺ + 6NH₃ → [Ni(NH₃)₆]²⁺ + 6H₂O
Worked example 2
One thiocyanate ion, SCN⁻, replaces one water ligand in [Fe(H₂O)₆]³⁺ to give a blood-red complex. Deduce its formula and write the equation.
Answer. [Fe(H₂O)₆]³⁺ + SCN⁻ → [Fe(SCN)(H₂O)₅]²⁺ + H₂O
Quick Check: Equation for a New Complex
Use the ligand count and the charges to choose the correct equation for an unfamiliar complex.
Stability Constants, Kstab
Ligand exchange reactions are equilibria. The stability constant, Kstab, measures how far the formation of a complex goes.
Definition: Kstab is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.
Writing the expression
[Cu(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6H₂O(l)
Kstab = ([[CuCl₄]²⁻]) / ([[Cu(H₂O)₆]²⁺] × [Cl⁻]⁴)
- The complex formed goes on the top. The aqua ion and the free ligand go on the bottom.
- The ligand concentration is raised to the power of its coefficient.
- [H₂O] is not included: water is the solvent, and its concentration is effectively constant.
- The aqua ion is often written as Cu²⁺(aq), which gives Kstab = ([CuCl₄²⁻]) / ([Cu²⁺][Cl⁻]⁴).
Units
Units = (mol dm⁻³) / ((mol dm⁻³) × (mol dm⁻³)⁴) = (mol dm⁻³)⁻⁴ = mol⁻⁴ dm¹². The units depend on the number of ligands in the expression.
What the value means
| Complex | Ligand | Kstab | Units |
|---|---|---|---|
| [CuCl₄]²⁻ | Cl⁻ | 4.2 × 10⁵ | mol⁻⁴ dm¹² |
| [Cu(NH₃)₄(H₂O)₂]²⁺ | NH₃ | 1.4 × 10¹³ | mol⁻⁴ dm¹² |
| [Cu(EDTA)]²⁻ | EDTA⁴⁻ | 6.3 × 10¹⁸ | mol⁻¹ dm³ |
These are typical values. A value needed in a question is always given.
- A large Kstab means the equilibrium lies well to the right: a stable complex ion is formed.
- The larger the Kstab, the more stable the complex.
The stability constant expression for the tetrachlorocuprate(II) ion, and three copper(II) complexes ranked by their stability constants.
Kstab Calculations and Ligand Exchange
Worked example 1: a ratio of concentrations
Kstab for [CuCl₄]²⁻ is 4.2 × 10⁵ mol⁻⁴ dm¹². Calculate the ratio ([CuCl₄²⁻]) / ([Cu²⁺]) when the chloride concentration at equilibrium is 2.0 mol dm⁻³.
Step 2. Ratio = 4.2 × 10⁵ × 2.0⁴ = 4.2 × 10⁵ × 16
With [Cl⁻] = 0.10 mol dm⁻³ the ratio is 4.2 × 10⁵ × 0.10⁴ = 42. This is why concentrated acid is needed, and why dilution reverses the reaction.
Worked example 2: finding Kstab
For Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq), an equilibrium mixture contains 2.0 × 10⁻⁷ mol dm⁻³ Ag⁺, 0.50 mol dm⁻³ NH₃ and 0.85 mol dm⁻³ [Ag(NH₃)₂]⁺. Calculate Kstab.
Step 1. Kstab = ([[Ag(NH₃)₂]⁺]) / ([Ag⁺] × [NH₃]²)
Step 2. Kstab = (0.85) / (2.0 × 10⁻⁷ × 0.50²) = (0.85) / (5.0 × 10⁻⁸)
Answer. Kstab = 1.7 × 10⁷ mol⁻² dm⁶
Worked example 3: the concentration of the aqua ion
Kstab for [Cu(NH₃)₄(H₂O)₂]²⁺ is 1.4 × 10¹³ mol⁻⁴ dm¹². At equilibrium the complex is 0.050 mol dm⁻³ and the ammonia is 0.20 mol dm⁻³. Calculate [Cu²⁺(aq)].
Step 1. [Cu²⁺] = ([complex]) / (Kstab × [NH₃]⁴)
Step 2. [Cu²⁺] = (0.050) / (1.4 × 10¹³ × 0.20⁴) = (0.050) / (2.24 × 10¹⁰)
Explaining ligand exchange with Kstab
- Ligand exchange goes towards the complex with the larger Kstab.
- Adding ammonia to yellow [CuCl₄]²⁻ gives deep blue [Cu(NH₃)₄(H₂O)₂]²⁺, because 1.4 × 10¹³ is far larger than 4.2 × 10⁵.
- Adding EDTA⁴⁻ to either complex gives [Cu(EDTA)]²⁻, which has the largest Kstab of the three.
- A ligand does not displace another from a complex that has a much larger Kstab.
Exam focus: Leave [H₂O] out of every Kstab expression, raise the ligand concentration to the correct power, and work out the units each time: they change with the number of ligands.
Quick Check: Working With New Kstab Values
Work out each answer on paper, then flip the card to check it.
Common Exam Points
Say
- “The coordination number changes from 6 to 4 because chloride ions are larger than water molecules.”
- “The exchange is incomplete: only four of the six water ligands are replaced by ammonia.”
- “The complex with the larger Kstab is more stable, so it is the one that forms.”
Do not say
- “[Cu(NH₃)₆]²⁺ forms in excess ammonia” (the product is [Cu(NH₃)₄(H₂O)₂]²⁺).
- “[CuCl₄]²⁻ is square planar” (it is tetrahedral).
- “The oxidation state of copper changes” (it stays at +2: only the ligands change).
- “Kstab includes the concentration of water” (it never does).
Watch for
- Charges: neutral ligands leave the charge unchanged, and each Cl⁻ lowers it by one.
- Six water molecules are released when four chloride ions replace six water ligands.
- State both colours in an observation: the colour before and the colour after.
- The power in the Kstab expression and the power in its units must match.
Check: Ligand Exchange, Colour and Coordination Number
Bring the copper(II) reactions together: predict observations, deduce formulae and charges, and explain each change in colour and coordination number.
FAQs
Use these quick answers to check ligand exchange reactions, their colour changes and stability constants.
What happens when ammonia is added to copper(II) sulfate solution?
A little ammonia gives a pale blue precipitate of copper(II) hydroxide. Excess ammonia dissolves it to give a deep blue solution: [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O. Only four of the six water ligands are replaced.
Why does the coordination number change when chloride ions replace water?
Chloride ions are larger than water molecules, so only four fit round the metal ion: [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O. The coordination number falls from 6 to 4 and the shape changes from octahedral to tetrahedral.
Why does copper(II) solution turn green before yellow with concentrated hydrochloric acid?
The reaction is an equilibrium. Part of the way, the solution holds both blue [Cu(H₂O)₆]²⁺ and yellow [CuCl₄]²⁻, so the mixture looks green. More chloride ions shift it to yellow. Adding water reverses the change, back to blue.
What is the colour change when cobalt(II) ions react with chloride ions?
Pink [Co(H₂O)₆]²⁺ turns to blue [CoCl₄]²⁻ with concentrated hydrochloric acid: [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O. The shape changes from octahedral to tetrahedral. Adding water turns the solution pink again.
What does a large Kstab value mean?
Kstab is the equilibrium constant for forming a complex from its ions or molecules in solution. A larger value means a more stable complex. Ligand exchange goes towards the complex with the larger Kstab, so ammonia displaces chloride from [CuCl₄]²⁻.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
Keep this note — free
Save your progress across every Cambridge topic. A free account remembers which topics you have covered, saves your question scores, and syncs across your phone and laptop.
- Track every topic you have finished
- Keep your practice-question scores
- No payment, no card, free forever
Already registered? Log in