Explain ligand exchange using stability constants
Compare complex stability and Kstab values to explain ligand-exchange equilibria in transition-metal complexes.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 28. Use the first-row transition elements titanium to copper and the species explicitly named in the syllabus. Where redox potentials or stability constants are needed, use the values supplied in the question. Colour descriptions can depend on conditions and concentration. Written practice is self-checked
Before you begin
- Use electron configurations, oxidation numbers, ionic equations, equilibrium expressions and the AS redox and bonding models.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens to the ligand set in ligand exchange?
What you will learn
- Describe ligand exchange as substitution around a central metal
- Use Kstab to explain which complex is favoured
- Relate complex stability to the equilibrium position without overgeneralising across equations
Ligand exchange replaces coordinated species
In a ligand-exchange reaction, one ligand set around a central metal is replaced by another. The metal can retain the same oxidation state while coordination number, geometry or colour changes.
Write a balanced equilibrium including any ligands released to solution. Check charges and the number of each donor ligand on both sides.
Use Kstab values with the reaction written clearly
A larger stability constant for a defined formation reaction indicates a greater tendency to form that complex from its constituents under the stated conditions. For exchange, this can help explain which complex is favoured when the relevant formation constants are supplied.
Do not compare raw Kstab values for differently written reactions without relating the expressions. If an exchange equation is built from two formation equilibria, the exchange equilibrium constant depends on the ratio of their formation constants, with the direction specified.
Pause and explain
When can two Kstab values be used to calculate an exchange constant?
Worked example
Two complexes of the same metal are formed from the same aqua ion. Supplied data give Kstab(A) = 2.0 × 10⁴ and Kstab(B) = 5.0 × 10⁸ for their stated formation equations. Which formation equilibrium is more product-favoured?
Show the worked solution
- Check that each K is for the equation as written.
- Compare the magnitudes under the supplied conditions.
- Do not infer an exchange ratio without matching stoichiometry.
Answer Formation of complex B is more product-favoured for its stated equation because its supplied Kstab is larger. To calculate a ligand-exchange equilibrium constant, account for both formation reactions and their stoichiometries.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A metal aqua ion reacts with 4L to form ML₄. What must a formation expression include, and what happens to water if it is the solvent?
Give me a hint
Use the coefficient of L and omit solvent water.
Compare my reasoning
- Put [ML₄] in the numerator.
- Use [M(aqua)][L]⁴ in the denominator.
- Omit liquid water from the expression.
Kstab = [ML₄]/([M(aqua)][L]⁴), with solvent water omitted.
Look for these in your work
- Product identified
- Ligand power correct
- Water convention applied
For M(aqua) + 4L ⇌ ML₄ + water, Kstab(ML₄) is much larger than the supplied Kstab for M(aqua). What qualitative conclusion can be drawn if both values use comparable conventions?
Give me a hint
Compare the stated formation preferences and keep the condition explicit.
Compare my reasoning
- The ML₄ formation equilibrium is more product-favoured.
- The metal may be found more often in the ML₄ complex when L is available.
- The constants still do not state the exchange rate.
The ML₄ complex is more favoured relative to its stated constituents under comparable conditions, but the exchange rate cannot be read from Kstab.
Look for these in your work
- Equilibrium direction inferred
- Conditions qualified
- Rate separated
An exchange reaction is the reverse of formation of complex A plus formation of complex B. Describe how to construct its equilibrium constant from the two supplied Kstab values.
Give me a hint
Reverse the formation constant for the complex being broken up, and combine with formation of the new complex.
Compare my reasoning
- Write both balanced formation equilibria.
- Reverse the one corresponding to the complex that dissociates.
- Multiply equilibrium constants in the chosen direction, including stoichiometric powers if needed.
Write the component equilibria in the exchange direction, invert Kstab for any reversed formation step, and multiply the constants (raising them for any scaled equation). The exact ratio depends on the balanced exchange equation.
Look for these in your work
- Equations written
- Reverse step inverted
- Stoichiometry accounted for
Common mistakes
- Treating ligand exchange as necessarily redox
- Using Kstab to predict rate
- Comparing constants with different stoichiometries as if they were the same reaction
- Forgetting that exchange equilibria depend on ligand concentrations
What can you explain now?
What happens to the ligand set in ligand exchange?
Compare with the explanation
One or more coordinated ligands are replaced
The central metal's coordination environment changes.
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Make it stick
How do you use formation equilibria and Kstab values to explain the direction of ligand exchange?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.