📋 Structured Questions
Complete each question on paper, then check your answers against the mark scheme.
Question 1: Ligand Substitution and the Chelate Effect
9 marks(a) Describe what is observed when concentrated hydrochloric acid is added dropwise to an aqueous solution of hexaaquacopper(II) ions, and write a balanced equation for the reaction. Explain why the coordination number changes. [4]
(b) When a solution containing [Cu(H2O)6]2+ is mixed with EDTA4-, the hexaaqua complex is converted into [Cu(EDTA)]2-. Explain why this reaction is thermodynamically favourable. Refer to entropy and Gibbs free energy in your answer. [5]
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(a)
- Observation: Pale blue solution turns yellow / green [1]
- [Cu(H2O)6]2+ + 4Cl- → [CuCl4]2- + 6H2O (or reversible arrow) [1]
- The chloride ligand (Cl-) is larger than the water ligand (H2O) [1]
- Only four Cl- ligands can fit around the central Cu2+ ion due to steric hindrance / electrostatic repulsion, changing the coordination number from 6 to 4 [1]
(b)
- Equation: [Cu(H2O)6]2+ + EDTA4- → [Cu(EDTA)]2- + 6H2O [1]
- There are 2 reactant particles yielding 7 product particles in solution, which represents a large increase in disorder / positive entropy change (ΔSθ is positive) [1]
- The enthalpy change (ΔHθ) is very close to zero because the same number and type of coordinate bonds (metal-oxygen and metal-nitrogen) are broken and formed [1]
- In the Gibbs free energy equation ΔGθ = ΔHθ - TΔSθ, because ΔSθ is positive and ΔHθ is small, ΔGθ will be highly negative [1]
- A negative ΔGθ indicates the reaction is thermodynamically feasible / stable complex is formed [1]
Question 2: Why are Transition Metal Complexes Coloured?
7 marks(a) Explain how colour arises in transition metal complexes. [5]
(b) State three factors that can cause the colour of a transition metal complex to change. [2]
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(a)
- Ligands donate lone pairs to the central transition metal ion, causing the five 3d orbitals to split into two different energy levels [1]
- An electron in the lower energy d orbital absorbs a photon of visible light and is promoted to a higher energy d orbital (this is a d-d transition) [1]
- The energy of the absorbed photon is related to the frequency of light by ΔE = hν (or ΔE = hc / λ) [1]
- This absorption removes certain wavelengths of visible light from the spectrum [1]
- The remaining wavelengths of light are transmitted / reflected, and the complementary colour is observed [1]
(b)
- Award [2] for any three of:
- Identity of the transition metal
- Oxidation state of the metal ion
- Identity of the ligands
- Coordination number / shape of the complex
- (Award [1] if only two factors are listed).
Question 3: Transition Metals as Homogeneous Catalysts
6 marks(a) Distinguish between homogeneous and heterogeneous catalysts. [2]
(b) The reaction between peroxodisulfate (S2O82-) and iodide (I-) ions is catalysed by Fe2+ ions. Explain, using equations, how Fe2+ acts as a catalyst, and explain why the uncatalysed reaction is very slow. [4]
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(a)
- A homogeneous catalyst is in the same phase / state as the reactants [1]
- A heterogeneous catalyst is in a different phase / state from the reactants [1]
(b)
- The uncatalysed reaction is very slow because both peroxodisulfate (S2O82-) and iodide (I-) ions are negatively charged [1]
- They repel each other, resulting in a very high activation energy for the reaction [1]
- Fe2+ acts as a catalyst by providing an alternative pathway with lower activation energy using two redox steps:
- S2O82-(aq) + 2Fe2+(aq) → 2SO42-(aq) + 2Fe3+(aq) [1]
- 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq) [1]