AQA A-Level Chemistry Paper 3, June 2022: Question 16

✅ Correct Answer: A

Equation: Al(g) → Al⁺(g) + e⁻

The first ionisation energy of aluminium (578 kJ mol⁻¹) is lower than that of magnesium (738 kJ mol⁻¹). Removing an electron from aluminium requires less energy than the target process.

💡 Key Knowledge

• Magnesium: [Ne] 3s². The electron is removed from a full 3s orbital.
• Aluminium: [Ne] 3s² 3p¹. The electron is removed from the higher-energy 3p subshell.
• The 3p electron in Al is:
  • At a slightly higher energy level than the 3s electrons.
  • Shielded from the nuclear charge by the inner core and the 3s² subshell.
• These factors outweigh the increased nuclear charge (+13 vs +12), making the 3p electron easier to remove.

🧠 Exam Technique: Option Elimination

• B — Ar(g) → Ar⁺(g) + e⁻: Argon is at the far right of Period 3. It has a much higher nuclear charge (+18) with similar shielding, so its IE₁ is much greater (1521 kJ mol⁻¹).
• C — Be(g) → Be⁺(g) + e⁻: Beryllium is directly above magnesium in Group 2. Its valence electrons are in the 2s subshell (closer to the nucleus, fewer inner shielding shells), so its IE₁ is greater (900 kJ mol⁻¹).
• D — Mg⁺(g) → Mg²⁺(g) + e⁻: This is the second ionisation energy of magnesium. Removing an electron from a positive ion against an increased effective nuclear attraction always requires significantly more energy (1451 kJ mol⁻¹).

❌ Common Errors & Pitfalls

• Over-generalising the periodic trend: Assuming ionisation energy always increases monotonically across Period 3, leading students to wrongly conclude that Al must be higher than Mg.
• Confusing the drops: Mixing up the Group 2 → 3 drop (subshell: 3s vs 3p) with the Group 5 → 6 drop (orbital repulsion: paired electrons in 3p⁴, e.g. P vs S).
• Misinterpreting option D: Forgetting that successive ionisation energies always increase ( IE₂ > IE₁ ).