Edexcel A-Level Chemistry AS Paper 1, June 2018: Question 6

✅ Model Answers

(a)(i) Complete configuration for Cl [1 Mark]:
2s² 2p⁶ 3s² 3p⁵
(Full: 1s² 2s² 2p⁶ 3s² 3p⁵)

(a)(ii) Why iodine and chlorine react similarly [2 Marks]:
• Both iodine and chlorine have 7 electrons in their outer shell (or both are ns² np⁵ ). [1 Mark]
• Chemical reactions / chemical properties are determined by the number of outer-shell electrons (electronic configuration). [1 Mark]

🧠 Exam Technique & Examiner Notes

• Check the prompt carefully: 1s² was already printed on the exam paper. Repeating it won't lose marks, but don't accidentally write 1s² 1s²... .
• Dependency condition: In part (a)(ii), Mark 2 is strictly dependent on scoring Mark 1. You cannot simply state "they react similarly because they are in the same group" — you must mention outer electrons.

❌ Common Errors to Avoid

• Vague phrasing like "both elements need 1 electron to be stable" without explicitly mentioning 7 outer electrons.
• Stopping at chlorine's 3p orbital without counting total electrons (Cl has 17 electrons: 2 + 2 + 6 + 2 + 5 = 17).

✅ Model Table

📐 Step-by-Step Calculation

Oxygen is always -2 in these ions.

• S₂O₃²⁻:
  2(S) + 3(-2) = -2  ⇒  2(S) - 6 = -2  ⇒  2(S) = +4  ⇒  S = +2
• SO₄²⁻:
  S + 4(-2) = -2  ⇒  S - 8 = -2  ⇒  S = +6
• S₄O₆²⁻:
  4(S) + 6(-2) = -2  ⇒  4(S) - 12 = -2  ⇒  4(S) = +10  ⇒  S = +2.5

❌ Common Errors

• Omitting the sign: Writing 2 or 6 instead of +2 or +6 . The mark scheme penalises omitting the positive sign.
• Assuming oxidation states must always be whole integers. Tetrathionate features an average oxidation state of +2.5 across its four sulfur atoms.

✅ Model Answers

(b)(ii) Why iodine is an oxidising agent [1 Mark]:
Iodine (I₂) gains electrons (to form I⁻ ions).
(Alternatively: thiosulfate / sulfur loses electrons).

(b)(iii) Why chlorine is a stronger oxidising agent [1 Mark]:
Chlorine oxidises sulfur from +2 to +6, whereas iodine only oxidises sulfur from +2 to +2.5 (chlorine causes a greater increase in sulfur's oxidation number).

💡 Key Knowledge

• OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.
• An oxidising agent oxidises another substance by removing electrons from it; therefore, the oxidising agent itself gains electrons (is reduced).
• Stronger oxidising agents can force elements into higher positive oxidation states. +6 in sulfate is a far higher oxidation state than +2.5 in tetrathionate.

❌ Common Misconceptions

• Do NOT say: "Chlorine is stronger because it gains more electrons." The mark scheme specifically rejects this statement! Explain it in terms of the sulfur: chlorine causes a greater loss of electrons from sulfur or a greater increase in oxidation number.
• Ignoring the requirement in (b)(ii) to answer "in terms of electrons" — explaining in terms of oxidation state will not gain the mark here.

📐 Step-by-Step Construction for (b)(iv)

Step 1: Balance main element (S):
S₂O₃²⁻ → 2SO₄²⁻

Step 2: Balance oxygen atoms using H₂O:
Left has 3 O, Right has 8 O  ⇒  add 5 H₂O to left:
S₂O₃²⁻ + 5H₂O → 2SO₄²⁻

Step 3: Balance hydrogen atoms using H⁺:
Left has 10 H  ⇒  add 10 H⁺ to right:
S₂O₃²⁻ + 5H₂O → 2SO₄²⁻ + 10H⁺

Step 4: Balance charges using electrons (e⁻):
Left charge = -2.
Right charge = 2(-2) + 10(+1) = +6.
To go from +6 to -2, add 8 e⁻ to the right side:
S₂O₃²⁻ + 5H₂O → 2SO₄²⁻ + 10H⁺ + 8e⁻

✅ Model Answers & Mark Breakdown

(b)(iv) Half-equation [2 Marks]:
S₂O₃²⁻ + 5H₂O → 2SO₄²⁻ + 10H⁺ + 8e⁻

(b)(v) Overall Ionic Equation [1 Mark]:
Chlorine half-equation: Cl₂ + 2e⁻ → 2Cl⁻
Multiply by 4 to balance electrons (8 e⁻):
4Cl₂ + 8e⁻ → 8Cl⁻

Combine with sulfur half-equation:

4Cl₂ + S₂O₃²⁻ + 5H₂O → 8Cl⁻ + 2SO₄²⁻ + 10H⁺

🧠 Exam Technique: Combining Half-Equations

• Electrons MUST cancel: Never leave electrons in an overall redox reaction. If your final equation has electrons, you will score 0 marks for part (b)(v).
• State symbols: The question states "State symbols are not required", so do not waste time writing (aq) or (l) unless confident.

❌ Common Errors in (b)(iv) & (b)(v)

• Forgetting to double the sulfate: writing S₂O₃²⁻ → SO₄²⁻ . Sulfur must be balanced first!
• Miscounting electrons: calculating difference in oxidation numbers for one sulfur atom (+2 to +6 = 4 e⁻) and forgetting there are two sulfur atoms involved (2 × 4 = 8 e⁻).
• Incorrect stoichiometry on chlorine when combining: forgetting to multiply Cl₂ by 4.