AQA A-Level Chemistry Paper 1, June 2024: Question 3

14 marks · Hard difficulty · Long Answer

Describe the structure and chelate effect for an iron(III) ethanedioate complex, explain the catalysis of the peroxodisulfate-iodide reaction by iron(II), and give reactions of iron(II) and iron(III) with ammonia and sodium carbonate.

Practise this question

Question

Question text

03 This question is about aqueous ions of the metal iron.

When an aqueous [Fe(H O) ]3+ ion reacts with ethanedioate ions,

an iron(III) complex ion X is formed.

The only ligands in X are ethanedioate ions.

03.1 Draw the structure of X.

Include the charge.

[2 marks]

03.2 The formation of X is an example of the chelate effect.

Explain the meaning of the chelate effect.

[2 marks]

03.3 Outline how Fe2+ ions catalyse the reaction between S O 2– ions and I– ions in

aqueous solution.

In your answer you should include

• a sketch graph to show how the concentration of S O 2– ions changes over time

• an explanation of how Fe2+ ions catalyse the reaction, including equations

• an overall equation for the reaction.

[6 marks]

03.4 A student adds dilute ammonia solution to a solution containing [Fe(H O) ]2+ ions.

*12* Give the formula of the precipitate that forms.

[1 mark]

03.5 The student adds sodium carbonate solution to a solution containing

[Fe(H O) ]2+ ions.

State one observation the student would make.

Give an equation for the reaction.

[2 marks]

Observation

Equation

03.6 A solution containing [Fe(H O) ]2+ ions changes to a yellow-brown colour after several

hours in contact with air.

The student adds sodium carbonate to the yellow-brown solution.

Give an equation for the reaction with sodium carbonate.

[1 mark]

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

M1: 1 mark for structure

Allow skeletal

M2: 1 mark for charge of 3–

Ignore charges inside bracket

03.1

(2 x AO2)

M1 When bidentate/multidentate ligands replace monodentate

ligands (to form a more stable complex)

M2 Because there is an increase in entropy/positive entropy M2 Allow S increases or ∆S is positive.

change/disorder or more particles formed (so ∆G is negative and

∆H is approximately 0) Do not accept ∆S increases or S is positive (2 x AO1)

03.2

This question is marked using levels of response. Refer to Indicative Chemistry Content

the Mark Scheme Instructions for Examiners for guidance on

how to mark this question. stage 1

Level 3

5–6 marks

All stages are covered and the description of each stage is

generally correct and virtually complete.

Answer is communicated coherently and shows a logical

progression from stage 1 to stage 2 and stage 3.

Level 2

3–4 marks

All stages are covered but the description of each stage may

be incomplete or may contain inaccuracies OR two stages are 6

covered and the explanations are generally correct and

03.3 virtually complete. (2 × AO1,

2 x AO2, 2

Answer is mainly coherent and shows progression from stage x AO3)

1a labelled axes

1 to stage 2 and/or stage 3. 2–

and concentration (of S2O8 ions) decreasing with

Level 1 time (ignore units)

1–2 marks 1b downwards curve of reducing steepness

Two stages are covered but the description of each stage

may be incomplete or may contain inaccuracies, OR only one

stage is covered but the explanation is generally correct and

virtually complete.

Answer includes isolated statements and these are presented

in a logical order.

Level 0

0 marks

Insufficient correct chemistry to gain a mark.

stage 2 explanation 19

2a (reaction slow) because S O 2– and I– repel/high

(reaction slow) because two negative ions repel/high

2b Fe2+ attracts the S O 2– so lower E

28 a

Fe2+ and S O 2– oppositely charged so lower Ea

2c Iron/Fe has a variable oxidation state

Fe2+ oxidised to Fe3+

Fe2+ → Fe3+ + e–

stage 3 equations

3a 2 Fe2+ + S O 2– → 2 SO 2– + 2 Fe3+

28 4

3b 2 Fe3+ + 2 I– → 2 Fe2+ + I

3c S O 2– + 2 I– → 2 SO 2– + I

28 4 2

allow equations with hexaaqua ions

[Fe(H2O)4(OH)2] 1

03.4

(1 x AO2)

green precipitate

2+ + 2

[Fe(H2O)6] + Na2CO3 → FeCO3 + 6H2O + 2Na

03.5 ignore state symbols (1 × AO1,

Or 1 × AO2)

[Fe(H O) ]2+ + CO 2– → FeCO + 6 H O

26 3 3 2

2[Fe(H O) ]3+ + 3 Na CO → 2 [Fe(H O) (OH) ] + 3 CO + 3 H O + 6Na+ ignore state symbols

26 2 3 2 3 3 2 2

Or 1

03.6

2 [Fe(H O) ]3+ + 3 CO 2– → 2 [Fe(H O) (OH) ] + 3 CO + 3 H O (1 × AO3)

26 3 2 3 3 2 2

How to answer it

Reactions & Catalysis of Aqueous Iron Ions

📌 What This Question Tests

This question examines core concepts in Year 2 Transition Metal Chemistry (AQA 3.2.5):

Ligand substitution & complex geometry: Drawing octahedral complexes with bidentate ethanedioate ligands and deducing the overall ionic charge.
Thermodynamics of coordination: Explaining the chelate effect in terms of entropy change (ΔS) and number of particles in solution.
Homogeneous catalysis: Mechanism, activation energy considerations, variable oxidation states, rate curve sketch, and stoichiometric half/full redox equations for the Fe²⁺-catalysed reaction between S₂O₈²⁻ and I⁻.
Aqueous transition metal ions & acidity: Hydrolysis and precipitation reactions of [Fe(H₂O)₆]²⁺ vs [Fe(H₂O)₆]³⁺ with NH₃ and Na₂CO₃, including the effect of aerial oxidation.

Question 03.1 • 2 Marks

Drawing the Iron(III) Ethanedioate Complex X

Octahedral coordination with bidentate ligands

✅ Structure & Charge

Formula: [Fe(C₂O₄)₃]³⁻

How to draw the structure:

Place Fe at the centre.
Show an octahedral arrangement of 6 coordination bonds (co-ordinate/dative covalent bonds) from oxygen atoms to the central Fe.
Draw three bidentate ethanedioate ligands. Each ligand has two carbonyl groups linked together: -O-C(=O)-C(=O)-O- . Both terminal single-bonded oxygen atoms coordinate to the Fe centre forming a 5-membered chelate ring.
Enclose the entire complex in large square brackets: [ ... ]³⁻.

❌ Common Errors

Incorrect overall charge: Writing no charge, 3+, or 2-. Fe is in oxidation state +3; three ethanedioate ligands contribute 3 × (-2) = -6. Overall charge = +3 - 6 = 3-.
Wrong coordinating atom: Bonding through the carbon atoms or double-bonded oxygens instead of the negatively charged oxygen atoms.
Non-octahedral geometry: Showing planar or tetrahedral shapes rather than 6-coordinate octahedral.

Mark Scheme Breakdown:
• M1: Correct octahedral structure with three bidentate ethanedioate ligands bonded via oxygens (skeletal or displayed is acceptable).
• M2: Overall charge of 3- outside brackets (charges on individual atoms inside the bracket are ignored).

Question 03.2 • 2 Marks

The Chelate Effect

Thermodynamic driving force behind multidentate ligand substitution

✅ Correct Answer

Definition & Driving Force:

M1: Substitution of monodentate ligands by bidentate (or multidentate) ligands to form a more stable complex.
M2: There is an increase in entropy / a positive entropy change (ΔS > 0) because more moles of particles are generated in solution (e.g. 1 complex + 3 ethanedioate → 1 complex + 6 H₂O; 4 particles become 7 particles), making ΔG negative (as ΔH ≈ 0).

🧠 Exam Technique: Exact Phrasing

The mark scheme is exceptionally strict regarding wording:

Say "entropy increases" or "ΔS is positive".
NEVER say "ΔS increases" — this is penalized because ΔS is the change itself, not the state function.
Mention particle numbers: substituting 6 monodentate ligands with 3 bidentate ligands increases total species from 4 to 7.

Mark Scheme Breakdown:
• M1: Bidentate/multidentate ligands replace monodentate ligands.
• M2: Positive entropy change / entropy increases / more particles formed (Note: "ΔS increases" is rejected).

Question 03.3 • 6 Marks (Level of Response)

Catalysis of Peroxodisulfate and Iodide by Fe²⁺

Homogeneous redox catalysis: rate curve, mechanism, and equations

💡 Stage 1: Concentration-Time Graph

Graph characteristics:

y-axis: [S₂O₈²⁻] (concentration of peroxodisulfate).
x-axis: Time.
Curve: Starts high on the y-axis and exhibits a downward curve of decreasing steepness (gradient decreases as [S₂O₈²⁻] drops and reaction rate slows). Curve levels off towards the x-axis.

💡 Stage 2: Why the Catalyst Works

Uncatalysed reaction is slow: Both S₂O₈²⁻ and I⁻ are negatively charged ions. They repel each other, resulting in a very high activation energy (E_a).
Role of Fe²⁺: Oppositely charged ions attract. Positive Fe²⁺ attracts negative S₂O₈²⁻, lowering E_a.
Variable oxidation states: Transition metals make effective catalysts because they can readily change oxidation states (Fe²⁺ ⇄ Fe³⁺).

📐 Stage 3: Step-by-Step Equations

Step 1: Oxidation of catalyst by peroxodisulfate

2Fe²⁺ + S₂O₈²⁻ → 2SO₄²⁻ + 2Fe³⁺

Step 2: Regeneration of catalyst by iodide

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

Overall Equation:

S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂

❌ Common Misconceptions on this 6-Marker

Drawing an autocatalysis curve: Do not draw a sigmoidal ("S-shaped") curve! This is homogeneous catalysis, not autocatalysis (unlike Mn²⁺ in ethanedioate titrations). Rate does not start slow and speed up; it is fastest at t = 0.
Forgetting ion-ion repulsion: Saying simply "bonds are strong" misses the crucial point that two anions repel each other.
Unbalanced equations: Make sure you have 2 Fe²⁺ reacting with 1 S₂O₈²⁻ to balance electrons transferred (2e⁻).

Levels of Response Marking (5–6 marks for Level 3):
• All 3 stages (Graph, Explanation of repulsion/variable oxidation states, and 3 Balanced Equations) covered correctly and logically presented.

Question 03.4 • 1 Mark

Precipitation with Dilute Ammonia

Deprotonation of hexaaquairon(II)

✅ Correct Formula

[Fe(H₂O)₄(OH)₂]

(or written simply as Fe(H₂O)₄(OH)₂ )

💡 Key Chemistry Principle

Dilute aqueous ammonia acts as a Bronsted-Lowry base (not a ligand here). It removes two protons (H⁺) from coordinated water molecules in [Fe(H₂O)₆]²⁺ to form the uncharged, insoluble green precipitate:

[Fe(H₂O)₆]²⁺ + 2NH₃ → [Fe(H₂O)₄(OH)₂] + 2NH₄⁺

Mark Scheme Breakdown:
• 1 Mark for [Fe(H₂O)₄(OH)₂] . Brackets around the complex are preferred.

Question 03.5 • 2 Marks

Reaction of [Fe(H₂O)₆]²⁺ with Sodium Carbonate

Precipitation vs Hydrolysis

✅ Correct Answer

Observation: Green precipitate

Equation:

[Fe(H₂O)₆]²⁺ + CO₃²⁻ → FeCO₃ + 6H₂O

Full formula equation also accepted:
[Fe(H₂O)₆]²⁺ + Na₂CO₃ → FeCO₃ + 6H₂O + 2Na⁺

❌ Major Trap: Effervescence / CO₂ Gas

DO NOT state: "Effervescence", "bubbles", or "fizzing".
Why? The 2+ iron ion has a relatively low charge density compared to 3+ ions. It is not acidic enough to protonate carbonate to form CO₂ gas. Instead, simple precipitation of iron(II) carbonate (FeCO₃) occurs.

Mark Scheme Breakdown:
• M1: Observation = green precipitate.
• M2: Balanced equation forming FeCO₃ (state symbols not required).

Question 03.6 • 1 Mark

Reaction of Oxidised Solution with Sodium Carbonate

Hydrolysis of the acidic hexaaquairon(III) ion

✅ Correct Equation

2[Fe(H₂O)₆]³⁺ + 3CO₃²⁻ → 2[Fe(H₂O)₃(OH)₃] + 3CO₂ + 3H₂O

Or using sodium carbonate:
2[Fe(H₂O)₆]³⁺ + 3Na₂CO₃ → 2[Fe(H₂O)₃(OH)₃] + 3CO₂ + 3H₂O + 6Na⁺

💡 Why does this reaction differ from 03.5?

After standing in air, Fe²⁺ is oxidised by atmospheric oxygen to yellow-brown [Fe(H₂O)₆]³⁺.
Fe³⁺ has a higher charge-to-size ratio (greater charge density) than Fe²⁺.
Fe³⁺ strongly polarises the coordinated O-H bonds in water, releasing H⁺ ions into solution (it is sufficiently acidic).
Therefore, carbonate acts as a base, accepting protons to release CO₂ gas and precipitating brown [Fe(H₂O)₃(OH)₃] rather than forming Fe₂(CO₃)₃.

Mark Scheme Breakdown:
• 1 Mark for the correct balanced equation with stoichiometry 2 : 3 → 2 : 3 : 3 (state symbols ignored).

Topics

Inorganic Chemistry · Physical Chemistry · 3.2.5 Transition Metals · 3.2.6 Reactions of Ions in Aqueous Solution · 3.1.5 Kinetics

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