OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 2

1 mark · Medium difficulty · Multiple Choice

Determine the oxidation numbers of iodine and antimony in the ionic compound [ICl2]+[SbCl6]- given that the oxidation number of chlorine is -1.

Practise this question

Question

Multiple choice question asking for the oxidation numbers of I and Sb in the compound [ICl2]+[SbCl6]- with chlorine having an oxidation number of -1. A table provides four options A, B, C, and D with varying oxidation number pairs for I and Sb, alongside an answer box.
Question text

2 In the compound [ICl ]+ [SbCl ]–, the oxidation number of chlorine is –1.

What are the oxidation numbers of I and Sb in the compound?

I Sb

A +1 +5

B +1 +7

C +3 +5

D +3 +7

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 2 is option C.

2 C 1 AO2.2

How to answer it

Determining Oxidation Numbers in Ionic Compounds

What this question tests

This question assesses your ability to apply rules for assigning oxidation numbers to elements within complex ionic species containing polyatomic ions. Specifically, it tests whether you can split an ionic lattice compound into its constituent cation and anion, and then use algebraic equations where the sum of the oxidation numbers equals the overall charge of the respective ion.

Question 2 Multiple Choice

Full Worked Solution & Breakdown

✅ Correct Answer: C

Iodine has an oxidation number of +3 and Antimony has an oxidation number of +5 .

💡 Key Knowledge

  • Chlorine bonded to other non-metals almost always has an oxidation number of -1 (as stated in the stem).
  • The sum of oxidation numbers in a neutral compound is 0, but in a polyatomic ion, it equals the overall charge of that ion.
  • Split the formula into [ICl₂]⁺ and [SbCl₆]⁻ and solve them independently.

🧠 Exam Technique

Don't let unfamiliar formulas intimidate you! Treat unfamiliar species like Sb (Antimony) algebraically just like any transition metal or main group element.

❌ Common Errors

Students often forget to account for the overall ionic charge ( +1 or -1 ) and mistakenly set the sum of oxidation numbers for the ions to 0 , leading to incorrect options like A or B.

📐 Step-by-Step Calculation

  1. Split the compound into its ions: Cation is [ICl₂]⁺ and anion is [SbCl₆]⁻ .
  2. Find the oxidation number of I in [ICl₂]⁺ : Let oxidation number of I = x .
    x + 2(-1) = +1
    x - 2 = +1 → x = +3 .
  3. Find the oxidation number of Sb in [SbCl₆]⁻ : Let oxidation number of Sb = y .
    y + 6(-1) = -1
    y - 6 = -1 → y = +5 .
  4. Match with options: I = +3 , Sb = +5 points directly to option C.
Examiner Note: This was an AO2.2 question testing application of rules in a novel context. 78% of higher-tier candidates successfully separated the ions, while errors were primarily driven by sign confusion with the ionic charges.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.