AQA A-Level Chemistry Paper 1, 2022: Question 7

✅ Mark Scheme Requirements

• Mark 1: (Visible / white) light is absorbed (to promote / excite d electrons from ground to excited state).
• Mark 2: (Only) yellow light is transmitted or reflected.

🧠 Exam Technique & Precise Vocabulary

Always structure transition metal colour explanations in two clear, linked sentences:

1. State that visible light is absorbed to promote d electrons across the energy gap ΔE.
2. State that the observed colour (yellow) is the light that is transmitted (or reflected), representing the remaining unabsorbed frequencies.

❌ Common Errors & Misconceptions

• "Absorbs yellow light": Completely wrong physics. If it absorbs yellow light, it appears violet (complementary colour).
• "Emits yellow light when electrons drop back": Confuses transition metal coordination complex colour with atomic emission flame tests. Electron de-excitation in complexes releases energy thermally (vibrations), not via visible emission.
• Omitting the word "light" (e.g. stating simply "yellow is transmitted").

💡 Key Knowledge

In an isolated copper ion, all five 3d orbitals are degenerate (equal energy). When ligands approach, ligand lone-pair repulsion splits the 3d orbitals into two distinct energy levels separated by an energy gap ΔE.

📐 Step-by-Step Calculation

1. State the governing formula:
   ΔE = hν  (or ΔE = hc / λ )
2. Rearrange for frequency (ν):
   ν = ΔE / h
3. Substitute given values:
   ν = (3.98 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ J s)
4. Evaluate to appropriate significant figures (3 s.f.):
   ν = 6.00 × 10¹⁴ s⁻¹ (or 6 × 10¹⁴ s⁻¹ )

❌ Calculation Pitfalls

• Bracket errors on calculator: Typing 3.98 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ without brackets on older calculators can lead to power-of-ten errors (e.g. 10⁻⁵³).
• Inverting the division: Dividing Planck's constant by energy ( h / ΔE ).
• Unnecessary conversions: Energy is already in Joules (J), not kJ mol⁻¹, so do NOT multiply or divide by the Avogadro constant.

✅ Mark Scheme Answers (Any 3)

1. Change in oxidation state (of the metal ion).
2. Change of ligand (type or nature of ligand).
3. Change in coordination number (allow number of ligands).

💡 Why Do These Factors Change Colour?

The colour depends directly on the magnitude of the d-orbital splitting energy gap ΔE ( ΔE = hν ):

• Oxidation state: Higher positive charge pulls ligands closer, increasing repulsion and enlarging ΔE.
• Ligand identity: Different ligands produce different electrostatic field strengths (spectrochemical series: e.g. CN⁻ > NH₃ > H₂O > Cl⁻).
• Coordination number & geometry: Tetrahedral splitting is roughly 4/9 of octahedral splitting for the same metal and ligand.

✅ Correct Answer

Tetrahedral

(Accept: tetrahedron)

💡 Examiner Tip: Why Not Square Planar or Octahedral?

• Chloride ligands (Cl⁻) are large and negatively charged. Only 4 Cl⁻ ions can fit sterically around the Cu²⁺ centre.
• Four-coordinate complexes with large ligands adopt the minimally repulsive tetrahedral shape with 109.5° bond angles (unlike Pt²⁺/Pd²⁺ complexes such as cisplatin, which are square planar).

✅ Balanced Ionic Equation

[CuCl₄]²⁻ + 6H₂O → [Cu(H₂O)₆]²⁺ + 4Cl⁻

🧠 Identifying Complex P

Adding excess water shifts the equilibrium completely to displace the Cl⁻ ligands. Water molecules are neutral and uncharged, so the overall complex ion charge changes from 2− to 2+.

✅ Correct Colour & Equation

Colour: Deep blue (allow dark blue; do not accept just "blue")

Equation:

[CuCl₄]²⁻ + 4NH₃ + 2H₂O → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4Cl⁻

❌ Common Errors in 07.6

• Writing just "blue": Unacceptable because hexaaquacopper(II) is blue/pale blue. The ammonia complex is distinctly deep blue / dark blue / royal blue.
• Forgetting the 2 H₂O ligands: Writing [Cu(NH₃)₄]²⁺ instead of the octahedrally coordinated [Cu(NH₃)₄(H₂O)₂]²⁺ . In copper(II), incomplete ligand exchange replaces only 4 equatorial ligands.
• Missing the reactant water: Omitting + 2H₂O on the left side of the equation causes an atom imbalance.

✅ Correct Identity

[Cu(EDTA)]²⁻

(Accept: [CuEDTA]²⁻, ignore absence of square brackets)

💡 The Chelate Effect

EDTA⁴⁻ is a hexadentate ligand with 6 donor atoms (2 nitrogen lone pairs, 4 carboxylate oxygen lone pairs):

[Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O

Because 2 particles produce 7 particles, there is a large increase in entropy (ΔS° is positive), making ΔG° highly negative and driving the substitution forwards irreversibly.