Edexcel A-Level Chemistry AS Paper 1, November 2021: Question 4

✅ Correct Answer

C: S(g) -> S⁺(g) + e⁻

💡 Key Knowledge

Ionisation energy equations must always represent the removal of one mole of electrons from one mole of gaseous atoms. State symbols are crucial ( (g) is mandatory).

❌ Common Errors

Students often fail by using S₈(s) or forgetting that sulfur exists as isolated atoms in the gaseous state for ionisation energy definitions, making options A, B, and D incorrect.

💡 Key Knowledge

First ionisation energy decreases down a group (e.g., in the group containing sulfur: O, S, Se, Te).

• Outer electrons occupy a principal quantum shell further from the nucleus.
• Increased shielding from inner electron shells.
• These factors outweigh the increasing nuclear charge.

🧠 Exam Technique

Make sure you explicitly mention shielding, distance/shell number, and nuclear charge to capture all 3 marking points. Never just say "shielding increases" without context.

✅ Correct Answer

Sulfur has a lower first ionisation energy than chlorine because sulfur has a smaller nuclear charge (fewer protons, 16 vs 17), while the electron is removed from the same sub-shell (3p) with similar shielding.

❌ Common Errors

Do not state that sulfur has a larger atomic radius or increased shielding compared to chlorine—both are in Period 3 and experience similar shielding effects.

✅ Correct Answer

Sulfur's first ionisation energy is lower than phosphorus because sulfur experiences spin-pairing in one of its 3p orbitals (the first pairing in the sub-shell). The resulting electron-electron repulsion makes the paired electron easier to remove.

🧠 Exam Technique

Keywords required: pairing, repulsion, and specifying the orbital/sub-shell. Avoid referencing shielding or distance here, as phosphorus has a higher nuclear charge yet a lower ionisation energy due to this sub-shell anomaly.

📐 Step-by-Step Calculation

Step 1: Rearrange the ideal gas equation

n = pV / RT

Step 2: Convert units into SI units

Pressure p = 105,000 Pa (from 105 kPa)
Volume V = 0.000132 m³ (from 132 cm³ ÷ 10⁶)
Temperature T = 420 K
Gas constant R = 8.31 J mol⁻¹ K⁻¹

Step 3: Calculate moles (n)

n = (105000 × 0.000132) / (8.31 × 420) = 0.0039711 mol

Step 4: Calculate Molar Mass (Mᵣ)

Mᵣ = mass / moles = 0.318 / 0.0039711 = 80.078 g mol⁻¹ (or 80.1)

Step 5: Deduce Formula

Sulfur (32.1) + Three Oxygens (3 × 16.0 = 48.0) = 80.1. Formula: SO₃

❌ Common Calculation Traps

• Forgetting to convert cm³ to m³ (multiplying by 10⁻⁶ instead of dividing).
• Failing to convert kPa to Pa by multiplying by 10³.
• Rounding intermediate values too early, which skews the final Mᵣ.

💡 Key Knowledge & Graph Skills

Plotting (c)(i): Ensure correct scale selection so plots cover at least half of the grid space, and plot points accurately to within 1 small grid square. Do not force a line of best fit through unrelated points unless asked.

Estimation (c)(ii): Based on the trend and table values, the melting temperature of H₂S is estimated to be between 170 K and 200 K (actual value is ~188 K).

🧠 Exam Technique

Always use a sharp pencil and a ruler for plotting points cleanly with crosses (×) or encircled dots (⊙). Double-check axis labels and units.