Edexcel A-Level Chemistry AS Paper 1, June 2024: Question 5

18 marks · Medium difficulty · Synoptic Questions

Calculate reacting masses, gas volumes, intermolecular forces, dot-and-cross diagrams, bond angles, and oxidation numbers involving chlorine dioxide and chlorate(III) ions.

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Question

Question text

5 This question is about chlorine dioxide, ClO , and the chlorate(III) ion, ClO2– .

Chlorine dioxide can be used to sterilise drinking water.

Chlorine dioxide is a gas at room temperature and pressure (r.t.p.).

Chlorine dioxide can be prepared by reacting sodium chlorate(III) with

hydrochloric acid.

The equation for this reaction is shown.

5NaClO2 + 4HCl → 5NaCl + 4ClO2 + 2H2O

(a) Chlorine dioxide is very toxic by inhalation and skin absorption.

State two precautions that must be taken when preparing chlorine dioxide in

a laboratory.

You may assume that a lab coat and eye protection are worn.

(2)

(b) Calculate the mass of sodium chlorate(III) needed to make

5.40g of chlorine dioxide.

[Ar values: H = 1.00 O = 16.0 Na = 23.0 Cl = 35.5]

(4)

The equation for this decomposition is shown.

2ClO2(g) → Cl2(g) + 2O2(g)

Calculate the increase in volume, in cm3 , when 0.125 mol of

chlorine dioxide gas completely decomposes.

[Molar gas volume = 24.0 dm3 mol–1]

(2)

*P76893A0624*

(ii) A swimming pool contains 400 m3 water. Chlorine dioxide has been suggested

as a disinfectant for use in swimming pools.

Calculate the mass of chlorine dioxide needed to produce a concentration of

chlorine of 7.82 × 10–8 mol dm–3 in this pool.

Give your answer to an appropriate number of significant figures.

(3)

(d) The strongest of the attractions between molecules in liquid chlorine dioxide is

(1)

A covalent bonding

B hydrogen bonding

C ionic bonding

D permanent dipoles

(e) (i) Complete a dot‑and‑cross diagram for the chlorate(III) ion, ClO2– .

Use crosses (×) for chlorine electrons, dots (••) for oxygen electrons and a

triangle ( ) for the extra electron.

(2)

*P76893A0724*

O O

(ii) Predict the bond angle in this ion. Justify your answer.

(3)

(f ) What is the oxidation number of oxygen in the chlorate(III) ion, ClO– ?

(1)

8 A −1

B +1 *P76893A0824*

C −2

D +2

(Total for Question 5 = 18 marks)

Mark scheme

Show the mark scheme

Question

Acceptable Answer Additional Guidance Mark

Number

5(a) An answer that makes reference to the following points: (2)

• use a fume cupboard (1) Allow an answer that recognises the problem of a toxic

gas

Allow fume hood/box

Ignore use of mask, respirator, breathing equipment (or

anything that uses all/part of the available air).

• gloves (1)

Allow an answer that recognises the problem of skin

absorption

Ignore type of glove (nitrile, plastic, gauntlet etc.)

Question

Acceptable Answer Additional Guidance Mark

Number

5(b) Example of calculation (4)

• molar mass of chlorine dioxide (1) 67.5 (g mol−1)

• moles of chlorine dioxide (1) 5.40 ÷ 67.5 = 0.08 / 0.080 (mol)

• moles of NaClO2 required (1) 5 ÷ 4 × 0.08(0) = 0.1 / 0.10 (mol)

• molar mass of NaClO2 90.5

and and

calculation of mass of NaClO2 (1) 90.5 × 0.1 = 9.05 / 9.1 (g)

Ignore SF except 1 SF in final answer only

TE at each stage

Question

Acceptable Answer Additional Guidance Mark

Number

5(c)(i) Example of calculation (2)

EITHER

• calculation of increase in moles of gas (1) 0.125 × 0.5 = 0.0625 (mol)

3 (1) 0.0625 × 24 000 = 1500 (cm3)

• convert increase in moles of gas to volume (cm )

• calculation of product volume (1) 0.125 × 1.5 = 0.1875

0.1875 × 24 000 = 4500 (cm3)

(1) 0.125 × 24 000 = 3000 (cm3)

• calculation of reactant volume and increase 3

4500 – 3000 = 1500 (cm )

Ignore SF except 1 SF

Question

Acceptable Answer Additional Guidance Mark

Number

5(c)(ii) Example of calculation (3)

• calculation of moles of Cl (1) (7.82 × 10−8 × 400) × 1000 = 0.03128

• calculation of moles of ClO2 (1) 0.03128 × 2 = 0.06256

• calculation of mass of ClO2 to 2 or 3 SF (1) 0.06256 × 67.5 = 4.2228

= 4.2 / 4.22 (g)

Allow alternative method for M1, M2 and M3:

M1 concentration of ClO (= 7.82 x 10-8 mol dm-3 x 2)

moles of ClO in 1 dm3 ( = 1.564 x 10-7 mol)

M2 mass of ClO in 1 dm3 ( = 1.564 x 10-7 x 67.5

= 1.0557 x 10-5 g

M3 mass in 400 m3 ( = 1.0557 x 10-5 x 400000) g

= 4.22/4.2 g

TE at each stage except for a final answer/M3 of a mass

greater than 4220g

Question

Answer Mark

Number

5(d) The only correct answer is D (permanent dipoles) (1)

A is not correct because there are no covalent bonds between molecules

B is not correct because this molecule does not contain hydrogen so there are no hydrogen bonds between molecules

C is not correct because there are no ionic bonds between molecules

Question

Acceptable Answer Additional Guidance Mark

Number

5(e)(i) An explanation that makes reference to the (2)

following points: × × ×

• 8 electrons around both oxygen atoms (1) Cl

including 6 dots and 2 other electron symbols •• × × •

× •

• 8, 10 or 12 electrons around the chlorine atom (1) O

including 7 crosses and 1, 3 or 5 other electron •

symbols respectively • ••

• • ••

10 electrons around the chlorine result from 1 Cl=O

12 electrons around the chlorine result from 2 Cl=O

Do not allow the triangle electron to be placed as a bonded

electron between the chlorine and oxygen

Question

Acceptable Answer Additional Guidance Mark

Number

5(e)(ii) An explanation that makes reference to the following points: (3)

• predicted bond angle = 104.5 (o) (1) Ignore shape even if incorrect

• 4 pairs of electrons around the chlorine suggests a Allow answers that mention 4 pairs of electrons

tetrahedral shape / bond angle 109.5 (o) (1) arranged to minimise repulsion

Do not award repulsion of atoms

• however lone pair repulsion greater (than bond pair

repulsion so angle reduced) (1) Ignore just ‘lone pairs reduce bond angle’

Allow reference to molecular shape rather than ion

Question

Answer Mark

Number

5(f) The only correct answer is C (−2) (1)

A is not correct because −1 is the overall charge on the chlorate (III) ion

B is not correct because +1 is not a possible oxidation state for oxygen in this substance

D is not correct because +2 is present in OF2. And O is more electronegative than Cl, so O is assigned a negative

oxidation number

(Total for Question 5 = 18 marks)

How to answer it

Chemistry Study Guide: Chlorine Dioxide & Chlorate(III)

What this question tests

This exam question assesses core AS chemistry competencies including practical safety precautions, stoichiometry and reacting mole ratios, gas volume calculations, solution concentration conversions, intermolecular forces in covalent molecules, dot-and-cross bonding diagrams, VSEPR theory for bond angles and shapes, and oxidation numbers.

Question 5(a)

Laboratory Safety Precautions

✅ Correct Answer

Any two of the following:

Use a fume cupboard (recognising the hazard of a toxic gas).
Wear gloves (recognising the hazard of skin absorption).

❌ Common Errors

Students often lose marks by writing vague instructions like "be careful" or "wear a mask". Respirators or breathing equipment that filter room air are not acceptable because toxic gases require sealed fume extraction.

Marks: 2 | 1 mark per valid precaution.

Question 5(b)

Stoichiometry & Mass Calculation

📐 Step-by-Step Calculation

Molar mass of ClO₂: 35.5 + (2 × 16.0) = 67.5 g mol⁻¹
Moles of ClO₂: 5.40 ÷ 67.5 = 0.08 mol
Moles of NaClO₂ required: Using the 5:4 ratio from 5NaClO₂ → 4ClO₂ , scale moles: 0.08 × (5 ÷ 4) = 0.10 mol
Mass of NaClO₂: Molar mass of NaClO₂ = 23.0 + 35.5 + (2 × 16.0) = 90.5 g mol⁻¹ . Mass = 0.10 × 90.5 = 9.05 g (or 9.1 g ).

🧠 Exam Technique

Always state your intermediate molar masses clearly and track your stoichiometric ratio straight from the balanced equation (5 moles of chlorate(III) form 4 moles of chlorine dioxide).

Marks: 4 | 1 mark for each correct stage (molar mass ClO₂, moles ClO₂, moles NaClO₂, final mass NaClO₂).

Question 5(c)(i)

Gas Decomposition Volume Calculation

📐 Step-by-Step Calculation

Find increase in moles of gas: From 2ClO₂(g) → Cl₂(g) + 2O₂(g) , 2 moles of gas yield 3 moles of gas (net increase of 1 mole of gas for every 2 moles of ClO₂ decomposed). Alternatively, calculate products ( 0.125 × 1.5 = 0.1875 mol total products) minus reactants ( 0.125 mol ). Increase = 0.0625 mol .
Convert moles to volume: 0.0625 mol × 24.0 dm³ mol⁻¹ = 1.50 dm³ = 1500 cm³ .

❌ Common Errors

Many students calculate the total volume of products instead of the increase in volume. Always check whether the question asks for total volume or change in volume!

Marks: 2 | 1 mark for calculating increase in moles, 1 mark for converting to cm³.

Question 5(c)(ii)

Solution Concentration & Mass Scaling

📐 Step-by-Step Calculation

Moles of Cl₂ needed: Concentration × Volume in dm³ = 7.82 × 10⁻⁸ × 400,000 = 0.03128 mol (since 400 m³ = 400,000 dm³ ).
Moles of ClO₂ needed: Stoichiometry shows 2ClO₂ generates 1Cl₂ , so 0.03128 × 2 = 0.06256 mol of ClO₂.
Mass of ClO₂: 0.06256 mol × 67.5 g mol⁻¹ = 4.22 g (to 3 significant figures).

💡 Key Knowledge

Remember your volume conversion: 1 m³ = 1,000 dm³ . Failing to scale cubic metres to cubic decimetres is a frequent trap.

Marks: 3 | 1 mark for moles of Cl₂, 1 mark for moles of ClO₂, 1 mark for final mass to 2 or 3 significant figures.

Question 5(d)

Intermolecular Forces

✅ Correct Answer

D: permanent dipoles

Why? Liquid chlorine dioxide consists of polar covalent molecules held together by permanent dipole-dipole interactions. There are no hydrogen atoms bonded to N, O, or F (ruling out hydrogen bonding), no ions (ruling out ionic bonding), and the forces between separate molecules are intermolecular, not covalent bonds.

Marks: 1

Question 5(e)(i)

Dot-and-Cross Diagram

💡 Key Knowledge

For the chlorate(III) ion ( ClO₂⁻ ):

Oxygen atoms: 8 electrons around each oxygen (6 dots + 2 shared bonding electrons).
Chlorine atom: Surrounded by 10 or 12 electrons depending on single/double bonding representation (typically 7 crosses from Cl + 1 triangle for the extra electron + shared electrons).
Crucial rule: Do not place the extra electron triangle inside the covalent bond overlap area.

Marks: 2 | 1 mark for oxygen shells, 1 mark for chlorine shell and extra electron placement.

Question 5(e)(ii)

Bond Angle & VSEPR Theory

✅ Correct Answer

Predicted bond angle: 104.5° (or 104°–105°)

🧠 Justification Guide

State electron pairs: There are 4 pairs of electrons around the central chlorine atom (2 bonding pairs, 2 lone pairs), which repels into a tetrahedral arrangement (base angle 109.5°).
Apply VSEPR repulsion rules: Lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion.
Conclusion: The two lone pairs compress the bond angle down from 109.5° to 104.5° (a reduction of 2.5° per lone pair).

Marks: 3 | 1 mark for angle, 1 mark for 4 pairs/tetrahedral basis, 1 mark for lone pair repulsion explanation.

Question 5(f)

Oxidation Numbers

✅ Correct Answer

C: -2

Why? Oxygen almost always exhibits an oxidation state of -2 in compounds (except in peroxides or fluorides). In ClO₂⁻ , each of the two oxygen atoms carries a -2 oxidation number.

Marks: 1

Topics

Physical Chemistry · Inorganic Chemistry · Organic Chemistry · Topic 2: Bonding and Structure · Topic 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.