AQA A-Level Chemistry Paper 1, June 2025: Question 7

15 marks · Medium difficulty · State/Explain/Numerical

Analyze various properties and reactions of vanadium and its compounds, including its position in the periodic table, catalytic activity, electrode potentials, acidity of its aqueous ions, mass spectrometry, and isotope abundance calculations.

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Question text

07 This question is about vanadium and some of its compounds.

07.1 State the block in the Periodic Table that contains vanadium.

[1 mark]

07.2 Suggest why vanadium can be pulled into thin wires.

[1 mark]

07.3 Vanadium(V) oxide is a catalyst in the Contact process.

2SO2(g) + O2(g) ⇌ 2SO3(g)

Give two equations to show how vanadium(V) oxide acts as a catalyst in this process.

[2 marks]

Equation 1

Equation 2

07.4 Table 6 shows some electrode half-equations and their standard electrode potentials.

Table 6

Electrode half-equation Eo / V

VO +(aq) + 2 H+(aq) + e− → VO2+(aq) + H O(l) +1.00

VO2+(aq) + 2 H+(aq) + e− → V3+(aq) + H O(l) +0.34

Sn2+(aq) + 2 e− → Sn(s) –0.14

V3+(aq) + e– → V2+(aq) –0.26

Sn reduces VO + in acidic solution.

*23* 2

Use the data in Table 6 to deduce an overall equation for the reaction that occurs.

[2 marks]

Vanadium(III) sulfate is a pale yellow solid at room temperature.

07.5 Explain why vanadium(III) sulfate has a high melting point.

[2 marks]

07.6 When vanadium(III) sulfate dissolves in water, the vanadium-containing species

formed is acidic.

Give an equation to show how vanadium(III) sulfate dissolves in water to form the

acidic vanadium-containing species.

Explain why the vanadium-containing species is acidic.

[3 marks]

Equation

Explanation

07.7 Compound Q is found in crude oil and has the molecular formula C33H34N4VO4

A sample of Q is ionised by electrospray ionisation in a

time of flight (TOF) mass spectrometer. The spectrum produced shows that the

vanadium in Q is 51V

Give an equation to show how Q is ionised using electrospray ionisation.

Determine the m/z value of the molecular ion formed in this ionisation process.

[2 marks]

Equation

m/z

07.8 A sample of vanadium contains only 49V and 51V

The relative atomic mass of vanadium in this sample is 50.97

Calculate the percentage abundance of 51V in this sample.

[2 marks]

Percentage abundance of 51V

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

07.1 d (block) Do not accept 3d

(1 x AO1)

allow layers of (positive) ions can slide over one 1

07.2 layers of atoms can slide over one another

another (1 x AO1)

M1 V2O5 + SO2 → V2O4 + SO3 allow multiples

07.3

𝟏

M2 V2O4 + 2 O2 → V2O5 allow 1 mark for equations in wrong order (2 x AO1)

M1 identification of V3+ as only vanadium product

07.4 + + 3+ 2+ Ignore state symbols

M2 VO2 + 4H + Sn → V + Sn + 2H2O(l) (2 x AO3)

– A-LEVEL CHEMISTRY – –

M1 ions are V3+ and SO 2–

OR 2

07.5

contains oppositely charged ions (2 x AO1)

M2 strong attraction between them

M1 V2(SO4)3 + 12H2O → 2 [V(H2O)6] 3+ + 3SO42– Allow as product [V(H2O)6]2(SO4)3

M2 V3+ has a high charge and small size

07.6 OR V3+ has a high charge density (1 x AO2,

2 x AO3)

M3 V3+ weakens the O–H bond (in water ligands and donates H+ to

M3 V3+ attracts electrons from the O-H bond (in

water or forms H O+ ions)

3 ligand and releases H+ or H O+)

V3+ polarises the O–H bond/water molecule

– A-LEVEL CHEMISTRY – –

Allow C H N VO + H+ → C H N VO +

+ + 33 34 4 4 33 35 4 4

M1 Q + H → QH

Allow C H N VO + H+ → C H N VO H+ 2

33 34 4 4 33 34 4 4

07.7 Ignore state symbols (1 x AO1,

1 x AO3)

M2 602

Alternatives for M1

51x + 49(100 − x)

M1 100 = 50.97 51x + 49(1-x) = 50.97, giving x = 0.985

49x + 51(100 − x) 49 2

07.8 = 50.97, giving x = 1.5% for V

100 (2 x AO2)

M2 51V = x = 98.5%

98.5% scores M1 and M2

How to answer it

Vanadium Chemistry, Redox, and Mass Spectrometry

What this question tests

This comprehensive question assesses core transition metal chemistry, physical properties, and fundamental calculations. Key areas include:

Classification of elements in the Periodic Table (d-block) and metallic bonding.
Catalytic mechanisms of transition metals, specifically the heterogeneous Contact Process.
Deducing complex redox equations using standard electrode potentials (E°).
Structure, bonding, and the acidity of hexaaqua metal ions (charge density and polarisation).
Electrospray ionisation equations and calculating m/z values in mass spectrometry.
Isotopic abundance calculations using relative atomic mass.

Parts 7.1 & 7.2: Classification & Metallic Bonding

Basic Properties of Vanadium

Correct Answers

7.1: d (or d block )

7.2: layers of atoms (or positive ions) can slide over one another.

Common Pitfalls

7.1: Do NOT write 3d . The question asks for the block, which is simply the d-block.
7.2: Simply saying "atoms can slide" is not enough. You must explicitly mention layers of atoms/ions sliding.

Mark Scheme Breakdown:
• 7.1: 1 Mark for identifying the d-block.
• 7.2: 1 Mark for describing the sliding layers of metallic structure.

Part 7.3: The Contact Process Catalyst

Heterogeneous Catalysis by Vanadium(V) Oxide

Key Knowledge

Transition metals make excellent catalysts because they can change oxidation states. In the Contact Process, V₂O₅ acts as a heterogeneous catalyst by being reduced to V₂O₄, and then re-oxidised back to V₂O₅ by oxygen.

Correct Equations

Equation 1:
V₂O₅ + SO₂ → V₂O₄ + SO₃

Equation 2:
V₂O₄ + ½O₂ → V₂O₅ (or 2V₂O₄ + O₂ → 2V₂O₅ )

Exam Technique & Order

Always write the equations in the logical sequence of the reaction: first, the catalyst reacts with the reactant (SO₂); second, the intermediate catalyst species (V₂O₄) is regenerated by oxygen. Note: The mark scheme allows 1 mark if you write the correct equations but in the wrong order.

Part 7.4: Electrode Potentials & Redox Equations

Deducing the Reduction of VO₂⁺ by Tin (Sn)

Step-by-Step Deduction

Identify the Reactants & Feasibility: Sn is a reducing agent. Its oxidation half-equation is:
Sn(s) → Sn²⁺(aq) + 2e⁻ (E° = -0.14 V, so oxidation potential is +0.14 V).
Determine the Final Vanadium Product: Sn can reduce any vanadium species whose reduction half-equation has an E° more positive than -0.14 V.
- VO₂⁺ to VO²⁺ (E° = +1.00 V) — Feasible
- VO²⁺ to V³⁺ (E° = +0.34 V) — Feasible
- V³⁺ to V²⁺ (E° = -0.26 V) — Not Feasible (since -0.26 V is more negative than -0.14 V).
Therefore, the vanadium is reduced all the way to V³⁺, but no further.
Combine the Half-Equations:
- Overall Reduction: VO₂⁺ + 4H⁺ + 2e⁻ → V³⁺ + 2H₂O
- Overall Oxidation: Sn → Sn²⁺ + 2e⁻
Since the electron transfer is balanced (2e⁻ each), simply add them together.

Overall Balanced Equation

VO₂⁺ + 4H⁺ + Sn → V³⁺ + Sn²⁺ + 2H₂O

How Marks are Won

M1: Correctly identifying V³⁺ as the only vanadium product.
M2: Writing the fully balanced overall equation (state symbols are not required).

Parts 7.5 & 7.6: Vanadium(III) Sulfate & Acidity

Structure, Dissolution, and Metal-Aqua Ion Acidity

Key Chemistry: Why are 3+ Metal Ions Acidic?

When transition metal ions with a high charge (such as V³⁺) dissolve in water, they form hexaaqua complex ions. The high charge density (high charge and small ionic radius) of the metal ion strongly polarises the O-H bonds in the water ligands. This weakens the O-H bond, allowing a hydrogen ion (H⁺) to be easily donated to a surrounding water molecule, forming an acidic solution.

Correct Answers

7.5 Melting Point:
• Contains oppositely charged ions ( V³⁺ and SO₄²⁻ ).
• There is strong electrostatic attraction between them.

7.6 Dissolution Equation:
V₂(SO₄)₃ + 12H₂O → 2[V(H₂O)₆]³⁺ + 3SO₄²⁻

7.6 Acidity Explanation:
• V³⁺ has a high charge density (or high charge and small size).
• V³⁺ polarises the O-H bond in the water ligands, weakening it and releasing H⁺ ions.

Common Errors to Avoid

7.5: Do not just say "covalent bonds" or "intermolecular forces". Vanadium(III) sulfate is a giant ionic lattice.
7.6 Equation: Ensure you balance the water molecules (12 H₂O are needed to form two [V(H₂O)₆]³⁺ complexes).
7.6 Explanation: You must link the high charge density to the polarisation/weakening of the O-H bond. Simply saying "it releases H⁺" is not enough for full marks.

Part 7.7: Electrospray Ionisation

Mass Spectrometry of Compound Q

Correct Equation & m/z

Equation:
Q + H⁺ → QH⁺
Or: C₃₃H₃₄N₄VO₄ + H⁺ → C₃₃H₃₅N₄VO₄⁺

m/z Value: 602

m/z Calculation Steps

Calculate the Mr of Q (using ⁵¹V):
• C: 33 × 12.0 = 396.0
• H: 34 × 1.0 = 34.0
• N: 4 × 14.0 = 56.0
• V: 1 × 51.0 = 51.0
• O: 4 × 16.0 = 64.0
Total Mr of Q = 601.0
In electrospray ionisation, a proton (H⁺) is added to the molecule: QH⁺ .
Mass of QH⁺ = 601 + 1 = 602. Since the charge is +1, the m/z value is 602.

The Electrospray Trap

Never write the ionisation equation as losing an electron (like electron impact ionisation: Q → Q⁺ + e⁻). Electrospray ionisation always involves protonation (adding H⁺). Consequently, the m/z value is always Mr + 1.

Part 7.8: Isotopic Abundance Calculation

Determining the Percentage of ⁵¹V

Step-by-Step Algebraic Calculation

Let the percentage abundance of ⁵¹V be x % .

Therefore, the percentage abundance of ⁴⁹V must be (100 - x) % .

Set up the Relative Atomic Mass (RAM) equation:
[ 51x + 49(100 - x) ] / 100 = 50.97
Expand and simplify:
51x + 4900 - 49x = 5097
2x + 4900 = 5097
Solve for x:
2x = 197
x = 98.5 %

Abundance of ⁵¹V = 98.5% (and ⁴⁹V = 1.5%)

Tutor Tip: Quick Sanity Check

The relative atomic mass of the sample is 50.97, which is extremely close to 51. This tells you immediately that the abundance of the heavier isotope (⁵¹V) must be very high (close to 100%). If your calculated value is not around 98-99%, you have made an algebraic error!

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.1 Atomic Structure · 3.1.3 Bonding · 3.1.11 Electrode Potentials · 3.2.1 Periodicity · 3.2.5 Transition Metals · 3.2.6 Reactions of Ions in Aqueous Solution

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.