AQA A-Level Chemistry Paper 3, June 2018: Question 26
1 mark · Medium difficulty · Multiple Choice
Identify the correct thermodynamic statement regarding a ligand substitution reaction involving the chelate effect in a cobalt(II) complex.
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Question text
26 Which statement is correct about this reaction?
[Co(NH ) ]2+ + 3H NCH CH NH [Co(H NCH CH NH ) ]2+ + 6NH
36 2 2 2 2 2 2 2 2 3 3
[1 mark]
A The co-ordination number of cobalt decreases.
B The enthalpy change is large and positive.
C The entropy change is large and positive.
D The shape of the complex changes from octahedral.
Mark scheme
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26 C
How to answer it
The Chelate Effect & Ligand Substitution
This question assesses understanding of ligand exchange reactions involving transition metal complexes:
- The Chelate Effect: How substitution of monodentate ligands by multidentate ligands impacts entropy (ΔS).
- Denticity & Co-ordination Number: Distinguishing monodentate (NH₃) from bidentate (ethane-1,2-diamine) ligands.
- Enthalpy Changes (ΔH): Why breaking and forming similar coordinate bonds (Co–N) results in near-zero enthalpy changes.
- Complex Geometry: Deducing the 3D shape based on the total number of coordinate bonds formed.
Ligand Substitution of Hexaamminecobalt(II)
Reaction: [Co(NH₃)₆]²⁺ + 3H₂NCH₂CH₂NH₂ → [Co(H₂NCH₂CH₂NH₂)₃]²⁺ + 6NH₃
✅ Correct Option: C
"The entropy change is large and positive."
On the reactant side, there are 4 particles in total (1 complex ion + 3 ligand molecules). On the product side, there are 7 particles (1 complex ion + 6 ligand molecules). An increase in the total number of particles in solution leads to a significant increase in disorder, making ΔS large and positive.
💡 Key Knowledge: The Chelate Effect
- Bidentate Ligand: Ethane-1,2-diamine (often abbreviated as en) has two nitrogen lone pairs and forms 2 coordinate bonds per molecule.
- Enthalpy Change (ΔH ≈ 0): Six Co–N coordinate bonds are broken and six very similar Co–N coordinate bonds are formed. Hence, ΔH is very small (near zero), neither large and positive nor large and negative.
- Feasibility: Since ΔG = ΔH - TΔS, a large positive ΔS makes ΔG significantly negative, driving the reaction forward and favouring the chelated complex.
📐 Step-by-Step Option Elimination
- Check Option A (Coordination Number):
Reactant has 6 monodentate NH₃ ligands → CN = 6.
Product has 3 bidentate ligands → 3 × 2 = 6 coordinate bonds → CN = 6.
Result: Incorrect (coordination number stays 6). - Check Option B (Enthalpy Change):
Six Co–N bonds broken; six Co–N bonds made. Bond enthalpies are almost identical.
Result: Incorrect (ΔH is close to 0, not large and positive). - Check Option C (Entropy Change):
Particles before = 1 + 3 = 4.
Particles after = 1 + 6 = 7.
Result: Correct (ΔS > 0 and large). - Check Option D (Shape):
Both reactant and product complexes have a coordination number of 6, which corresponds to an octahedral geometry.
Result: Incorrect (the shape remains octahedral).
❌ Common Errors & Misconceptions
- Confusing Ligand Count with Coordination Number: Students often see "3" ligands in [Co(en)₃]²⁺ and assume the coordination number dropped to 3 or changed to planar/trigonal. Remember: Coordination number = total number of coordinate bonds , not total number of ligands.
- Assuming Endothermic Breakdown: Thinking that because 6 NH₃ ligands leave, substantial energy is absorbed (Option B). This overlooks that 6 new coordinate bonds are simultaneously formed.
🧠 Exam Technique: 5-Second Multiple Choice Rule
Whenever you see monodentate ligands being replaced by bidentate or multidentate ligands (like EDTA⁴⁻):
- Count particles immediately: 4 → 7 .
- Particle increase = large positive ΔS.
- This is the textbook definition of the chelate effect. Look directly for the entropy statement.
Topics
Inorganic Chemistry · Physical Chemistry · 3.2.5 Transition Metals · 3.1.8 Thermodynamics
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.