OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 6

1 mark · Easy difficulty · Multiple Choice

Determine the oxidation number of sulfur in sodium thiosulfate, Na₂S₂O₃.

Practise this question

Question

Question 6 asks: 'What is the oxidation number of S in Na2S2O3?' followed by four multiple-choice options: A is -4, B is -2, C is +2, and D is +4. Below the options is an answer box labeled 'Your answer' worth 1 mark.
Question text

6 What is the oxidation number of S in Na2S2O3?

A –4

B –2

C +2

D +4

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme for question 6 indicates the correct answer is C, with 1 mark awarded, and an additional note to 'ALLOW +2'.

6 C 1 ALLOW +2

How to answer it

Oxidation Number of Sulfur in Na₂S₂O₃

📌 What this question tests

This multiple-choice question assesses your foundational understanding of redox chemistry, specifically:

  • Assigning standard oxidation numbers to Group 1 metals ( Na = +1 ) and combined oxygen ( O = -2 ).
  • Applying the rule that the sum of oxidation numbers in a neutral compound equals zero.
  • Solving for an unknown element when multiple atoms of that element are present in the chemical formula.

Question 6 Analysis

Multiple Choice — 1 Mark

✅ Correct Answer

C: +2

Mark Scheme: Award 1 mark for C (allow writing +2 directly).

📐 Step-by-Step Calculation

  1. Identify fixed oxidation numbers:
    • Sodium (Na) is a Group 1 metal: +1
    • Oxygen (O) in non-peroxides: -2
  2. Set up the charge balance equation:
    Na₂S₂O₃ is a neutral compound, so overall charge = 0.
    2(+1) + 2(S) + 3(-2) = 0
  3. Simplify:
    +2 + 2(S) - 6 = 0
    2(S) - 4 = 0
    2(S) = +4
  4. Solve for a single S atom:
    S = +4 / 2 = +2

💡 Key Rules to Remember

  • Sign convention: Oxidation numbers are written with the sign first (e.g., +2 ), unlike ionic charges which follow the number (e.g., 2+ ).
  • Alternative method (via the thiosulfate ion):
    Separate into ions: 2Na⁺ and S₂O₃²⁻ .
    2(S) + 3(-2) = -2
    2(S) - 6 = -2 ⇒ 2(S) = +4 ⇒ S = +2
  • In Na₂S₂O₃, +2 represents the average oxidation number of sulfur (structurally, one S is +5 and the other is -1). OCR AS Level only requires the average.

❌ Common Student Errors

  • Selecting D (+4): The most common trap. Students correctly work out that the sulfur contributes +4 overall, but forget to divide by 2 for the two sulfur atoms in the formula.
  • Sign mix-up (Selecting B, -2): Incorrect algebra when rearranging 2S - 4 = 0 , mistakenly making the total negative.
  • Treating the polyatomic group incorrectly: Forgetting the sodium atoms entirely and treating S₂O₃ as neutral instead of S₂O₃²⁻ .

🧠 Exam Technique & Examiner Tips

Always circle or underline subscripts in chemical formulas before doing an oxidation state calculation. When you see a formula like Na₂S₂O₃ , immediately flag that any oxidation state you calculate for sulfur will need to be divided by 2 to give the oxidation state per atom.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.