Edexcel A-Level Chemistry Paper 1, June 2023: Question 3

9 marks · Medium difficulty · Short Open Response

Answer questions on Group 7 elements covering reactions of potassium bromide with concentrated sulfuric acid, testing for bromide ions using silver nitrate, identifying the shape of the ICl4- ion, and predicting Kc for the dissociation of molten I2Cl6.

Practise this question

Question

A three-part structured exam question about Group 7 compounds. Part (a) is a multiple-choice question on the reaction of potassium bromide with concentrated sulfuric acid. Part (b) involves testing for bromide ions with silver nitrate, explaining why the reagent is acidified, and identifying the correct acid. Part (c) explores the dissociation of molten I2Cl6 into ICl2+ and ICl4-, asking for an experimental setup to confirm dissociation, the shape of the ICl4- ion, and the likely numerical value of Kc for the equilibrium.
Question text

3 This question is about compounds containing elements from Group 7.

(a) Which change occurs when concentrated sulfuric acid is added to

potassium bromide?

(1)

A bromide ions oxidise sulfuric acid forming sulfur

B bromide ions oxidise sulfuric acid forming sulfur dioxide

C bromide ions reduce sulfuric acid forming sulfur

D bromide ions reduce sulfuric acid forming sulfur dioxide

(b) Chemists can test for the presence of bromide ions in solution by adding a small

amount of acidified silver nitrate solution.

The solubility of the precipitate in aqueous ammonia is then tested.

(i) Which statement is correct for bromide ions?

(1)

A a white precipitate forms that dissolves in concentrated

ammonia only

B a white precipitate forms that dissolves in both dilute and

concentrated ammonia

C a cream precipitate forms that dissolves in concentrated

ammonia only

D a cream precipitate forms that dissolves in both dilute and

concentrated ammonia

(ii) Give a reason why the silver nitrate must be acidified.

(1)

(iii) Explain which acid needs to be used to acidify the silver nitrate solution and

why other acids are unsuitable.

(2)

(c) Iodine trichloride forms a dimer,*P71912A0428*I2Cl6, in the solid state.

When molten, it is suggested that it breaks down as shown.

I Cl ICl+ + ICl–

26 2 4

(i) Draw a labelled diagram of a simple experiment to confirm this dissociation

has occurred, stating the positive result.

(2)

Result …

(ii) What is the shape of the ICl– ion?

(1)

A octahedral

B square planar

C tetrahedral

D trigonal bipyramidal

(iii) The equilibrium position for the dissociation of molten I2Cl6 lies to the left.

I Cl ICl+ + ICl–

26 2 4

What is the most likely numerical value of Kc for this equilibrium?

(1)

A 1.0 × 106

B 5.0 × 103

C 1.0

D 5.0 × 10–3

(Total for Question 3 = 9 marks)

Mark scheme

Show the mark scheme The mark scheme providing correct answers and guidance for all parts of Question 3. It lists the correct multiple-choice options, acceptable reagents and reasons for acidification in the silver nitrate test, a diagram and observation criteria for testing the conductivity of molten I2Cl6, the shape of ICl4-, and the correct numerical value for Kc.

Question

Answer Mark

Number

3(a) The only correct answer is D (bromide ions reduce sulfuric acid forming sulfur dioxide) (1)

A is not correct because bromide ions reduce sulfuric acid

B is not correct because bromide ions reduce sulfuric acid

C is not correct because bromide ions are not strong enough reducing agents to form sulfur

Question

Answer Mark

Number

3(b)(i) The only correct answer is C (a cream precipitate forms that dissolves in concentrated ammonia only) (1)

A is not correct because bromide ions do not form a white precipitate

B is not correct because bromide ions do not form a white precipitate

D is not correct because the cream precipitate does not dissolve in dilute ammonia

Question

Answer Additional Guidance Mark

Number

3(b)(ii) An answer that makes reference to the following point: Allow name or formula of ion but if both are (1)

2− − given, both must be correct

• to react with / remove carbonate / CO3 / hydroxide / OH (ions, which

could lead to a false positive) Allow other specified anions that would form

a precipitate with Ag+/ AgNO e.g.

hydrogencarbonate / sulfite

Ignore just ‘ to prevent a false positive’

Ignore just ‘ it reacts with impurities’

Ignore just ‘ so it only reacts with Br− ‘

Question

Answer Additional Guidance Mark

Number

3(b)(iii) An explanation that makes reference to the following points: Mark independently (2)

• nitric acid / HNO3 (1) Do not award incorrect formula

for nitric acid / use of NH3

• so that the anion (from the acid) does not form a precipitate (with silver Allow forms a precipitate with HCl / H2SO4 /

ions) (1) other acids (giving a false positive)

Allow so it doesn’t form precipitates with HCl

/ H2SO4 / other acids (giving a false positive)

Allow solid / ppt / ppte for precipitate

Ignore colour of precipitate

Question

Answer Additional Guidance Mark

Number

3(c)(i) An answer that makes reference to the following points: (2)

molten

compound

heat

• a diagram of a simple electrical circuit, containing electrodes, Allow alternatives to bulb e.g. buzzer / ammeter

power supply and a bulb in series (1) Allow low voltage supply instead of cell

Allow a cell and electrodes dipping into the liquid

but with no bulb / ammeter etc.

Allow cell with connecting wires to filter paper on

microscope slide

Ignore missing heat / labels for molten compound /

/electrodes, wires, etc.

Do not award voltmeter instead of cell unless

electrodes made from two different materials are

specified

Allow correct observation from alternatives

• bulb lights (1)

buzzer – rings

ammeter - shows a current

Allow observation for formation of iodine or

chlorine e.g. brown colour / purple vapour / bubbles

/ green gas

Question

Answer Mark

Number

3(c)(ii) The only correct answer is B (square planar) (1)

A is not correct because the central iodine atom only has four atoms attached and has two lone pairs of electrons

C is not correct because the central iodine atom has four bond pairs and two lone pairs of electrons

D is not correct because the central iodine atom only has four atoms attached and has two lone pairs of electrons

Question

Answer Mark

Number

3(c)(iii) The only correct answer is D (5.0 × 10−3) (1)

A is not correct because the value is greater than 1, which implies equilibrium lies to product side

B is not correct because the value is greater than 1, which implies equilibrium lies to product side

C is not correct because the value of 1 implies equilibrium concentrations of reactants and products are similar in magnitude

(Total for Question 3 = 9 marks)

How to answer it

Compounds Containing Elements from Group 7

📌 What this question tests

This question assesses your knowledge of Group 7 (halogens) chemistry, specifically the reducing ability of halide ions, halide identification tests using acidified silver nitrate, VSEPR theory for shape determination, and the interpretation of equilibrium constants (Kc).

Part (a): Reactions of Halides with Concentrated Sulfuric Acid

Redox Reactions of Potassium Bromide

✅ Correct Answer

D — bromide ions reduce sulfuric acid forming sulfur dioxide.

💡 Key Knowledge

Bromide ions (Br⁻) act as reducing agents. They reduce sulfuric acid (H₂SO₄ where sulfur is +6) to sulfur dioxide (SO₂ where sulfur is +4), while being oxidized themselves to bromine (Br₂).

❌ Common Errors

Students often confuse the reduction products of concentrated H₂SO₄ down the group: chlorides give misty fumes of HCl (no redox), bromides reduce H₂SO₄ to SO₂, and iodides reduce it further to SO₂, S, and H₂S.

Mark: 1 mark

Part (b)(i): Testing for Bromide Ions

Precipitation with Silver Nitrate and Ammonia Solubility

✅ Correct Answer

C — a cream precipitate forms that dissolves in concentrated ammonia only.

💡 Key Knowledge

Silver bromide (AgBr) is a cream precipitate. Silver chloride is white and dissolves in dilute ammonia; silver iodide is yellow and insoluble in both dilute and concentrated ammonia.

Mark: 1 mark

Part (b)(ii): Acidifying Silver Nitrate

Purpose of Acidification

✅ Correct Answer

To react with / remove carbonate (CO₃²⁻) or hydroxide (OH⁻) ions, which could lead to a false positive.

🧠 Exam Technique

Be specific. Avoid vague phrases like "to prevent a false positive" without stating what causes the false positive (carbonate/hydroxide forming unwanted precipitates with Ag⁺).

Mark: 1 mark

Part (b)(iii): Choosing the Correct Acid

Selecting an Acidification Reagent

✅ Correct Answer

1. Use nitric acid (HNO₃).
2. Reason: So that the anion from the acid does not form a precipitate with silver ions.

❌ Common Errors

Never use hydrochloric acid (HCl) because chloride ions would precipitate with silver ions (AgCl), giving a false positive. Never use sulfuric acid as it forms silver sulfate precipitates.

Mark: 2 marks

Part (c)(i): Confirming Dissociation

Proving Ion Formation in Molten State

✅ Correct Answer

Circuit Diagram: A simple electrical circuit containing a power supply, electrodes dipping into the molten I₂Cl₆, and a bulb/ammeter/buzzer connected in series.
Result: The bulb lights up (or ammeter shows a current / buzzer sounds).

🧠 Exam Technique

When asked to prove ions are present in a molten substance, setting up an electrical conductivity apparatus is standard. Ensure your circuit is a complete loop (in series) with a power source and a detector (like a lamp or LED).

Mark: 2 marks

Part (c)(ii): Shape of the ICl₄⁻ Ion

VSEPR Theory Application

✅ Correct Answer

B — square planar

💡 Key Knowledge

Iodine in ICl₄⁻ has 7 valence electrons, plus 4 from single bonds, plus 1 from the negative charge = 12 electrons in the outer shell (6 electron pairs: 4 bond pairs and 2 lone pairs). To minimise repulsion, the lone pairs take axial positions, resulting in a square planar geometry.

Mark: 1 mark

Part (c)(iii): Interpreting Kc Value

Equilibrium Position and Numerical Kc

✅ Correct Answer

D — 5.0 × 10⁻³

💡 Key Knowledge

The question states the equilibrium position lies to the left (towards reactants: I₂Cl₆). Therefore, the concentration of reactants is much greater than products at equilibrium. Since Kc = [Products] / [Reactants] , a numerator much smaller than the denominator yields a small Kc value significantly less than 1.

❌ Common Errors

Selecting 1.0 × 10⁶ (Option A) or 5.0 × 10³ (Option B). Values greater than 1 indicate that the equilibrium lies far to the right (products favoured), which contradicts the stem of the question.

Mark: 1 mark

Topics

Inorganic Chemistry · Physical Chemistry · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 2: Bonding and Structure · Topic 11: Equilibrium II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.