Edexcel A-Level Chemistry AS Paper 1, June 2025: Question 7

✅ Correct Answer & Mark Scheme

• Mark 1: The outer electrons are in the same quantum shell / have the same number of inner shielding shells (similar shielding).
• Mark 2: The number of protons (nuclear charge) increases, pulling the outer electrons closer to the nucleus.

💡 Key Knowledge

• Across Period 3, each successive element adds one proton to the nucleus and one electron to the 3rd principal energy level (n = 3).
• Inner electron shielding remains roughly constant (10 core electrons: 1s² 2s² 2p⁶).
• Greater positive pull from the nucleus on electrons with similar shielding pulls outer electrons inwards.

🧠 Exam Technique

Always structure period-trend explanations using the 3 Core Factors:

1. Nuclear Charge: State clearly that proton number increases.
2. Shielding: State that shielding is constant / electrons are in the same shell.
3. Net Attraction: State that electrostatic attraction to outer electrons increases, pulling them closer.

❌ Common Errors

• Saying "chlorine has more electrons so it's bigger" — forgetting nuclear charge dominates across a period.
• Failing to mention shielding or stating that shielding increases across the period.
• Saying "attraction increases" without specifying it is between the nucleus and the outer electrons.

✅ Correct Option

Option C: ½N₂(g) − e⁻ → N⁺(g)

💡 Key Knowledge

First Ionisation Energy: The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.

X(g) → X⁺(g) + e⁻ (or X(g) − e⁻ → X⁺(g))

Options A, B, and D all use isolated gaseous atoms: B(g), C(g), and Ne(g). Only C begins with molecular nitrogen.

🧠 Exam Technique

• Notice the minus sign format: X(g) − e⁻ → X⁺(g) is mathematically equivalent to X(g) → X⁺(g) + e⁻ . Don't let this distract you!
• Always check the state symbols and chemical formula first: it must be a single atom, never a diatomic molecule like N₂, O₂, or Cl₂.

✅ Correct Option

Option B: Mg²⁺

📐 Proton vs Electron Breakdown

• O²⁻: 8 protons pulling 10 electrons (weakest pull → largest radius)
• F⁻: 9 protons pulling 10 electrons
• Na⁺: 11 protons pulling 10 electrons
• Mg²⁺: 12 protons pulling 10 electrons (strongest pull → smallest radius)

❌ Common Errors

• Assuming negative ions are smaller because they are non-metals. (Anions are actually larger than their parent atoms due to electron repulsion; cations are smaller).
• Confusing atomic radius with ionic radius.

✅ The 6 Indicative Content Points (IP1 – IP6)

1. IP1 (General Trend): First ionisation energy increases overall across Period 3 because the number of protons / nuclear charge increases.
2. IP2 (Shielding across period): The outer electrons are in the same shell (n = 3) and experience similar shielding / same distance from nucleus.
3. IP3 (Al anomaly observed): Aluminium has a lower first ionisation energy than magnesium.
4. IP4 (Al explanation): The electron removed from Al is in a 3p orbital/subshell, which is higher in energy and shielded by the 3s² electrons.
5. IP5 (S anomaly observed): Sulfur has a lower first ionisation energy than phosphorus.
6. IP6 (S explanation): In sulfur, the electron is removed from a paired 3p orbital. Mutual electron spin-pair repulsion makes this electron easier to remove.

💡 Electron Configurations to Quote

• Mg: [Ne] 3s² vs Al: [Ne] 3s² 3p¹
  The 3p electron in Al is in a higher subshell and shielded by the 3s electrons.
• P: [Ne] 3s² 3p_x¹ 3p_y¹ 3p_z¹ (all half-filled, no pairing)
  S: [Ne] 3s² 3p_x² 3p_y¹ 3p_z¹ (paired in one 3p orbital)
  Spin-pair repulsion between two electrons in the same 3p orbital destabilises the electron in S.

❌ Costly Examiner Pitfalls

• Vague Al explanation: Saying "Al is in a new shell" — it is in a new subshell (3p), NOT a new shell! Both are in shell n = 3.
• Saying Al electron is simply "further away": Examiners do not award credit for saying the Al electron is further from the nucleus than the Mg electron. Focus on shielding by the 3s subshell.
• Vague S explanation: Writing "phosphorus is more stable because it's half full" without explaining the repulsion between paired electrons in sulfur.
• Lack of structure: Jumping between Al and S without first establishing the general background trend (increasing nuclear charge with similar shielding).

🎯 Model Full-Mark Structure

Organise your written answer into three distinct paragraphs to guarantee the 2 reasoning marks:

1. Paragraph 1 (General Trend): State that IE increases overall due to increasing proton number/nuclear charge with similar shielding across the 3rd shell.
2. Paragraph 2 (Drop at Al): Identify Mg → Al drop. State electron in Al is lost from 3p, shielded by 3s subshell, requiring less energy.
3. Paragraph 3 (Drop at S): Identify P → S drop. State that in S, the electron is removed from an orbital with paired electrons; spin-pair repulsion eases removal.