AQA A-Level Chemistry Paper 1, June 2024: Question 9

17 marks · Medium difficulty · State/Explain/Numerical

Answer questions on metals and their compounds, including Group 2 trends and reactions, an ideal gas calculation to identify a metal nitrate, bonding in sodium tetrahydroaluminate, and lithium electrochemical cells.

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Question

Question 9 consisting of 8 parts. Part 09.1 asks to state why the atomic radius of calcium is greater than magnesium (1 mark). Part 09.2 asks for the balanced equation with state symbols for magnesium reacting with steam (1 mark). Part 09.3 asks to explain why barium initially reacts faster with dilute sulfuric acid than magnesium does, but stops before all barium reacts (3 marks). Part 09.4 provides the thermal decomposition equation 2X(NO3)2(s) -> 2XO(s) + 4NO2(g) + O2(g) and gas data (0.832 g produces 348 cm3 gas at 298 K, 100 kPa) to deduce metal X (6 marks). Part 09.5 asks for the equation forming NaAlH4 and why it has a high melting point (3 marks). Part 09.6 asks for the equation between H3PO4 and excess NaOH (1 mark). Parts 09.7 and 09.8 ask for the positive electrode reaction of a lithium cell and why it can be recharged (1 mark each).
Question text

09 This question is about metals and their compounds.

09.1 State why the atomic radius of calcium is greater than the atomic radius of

magnesium.

[1 mark]

09.2 Magnesium reacts with steam.

Give an equation, including state symbols, for this reaction.

[1 mark]

09.3 Similar-sized pieces of barium and magnesium are added to separate

100 cm3 samples of dilute sulfuric acid. In each case the sulfuric acid is in excess.

The barium reacts quickly at first. After a few minutes the reaction stops, even though

there is still some unreacted barium in the flask.

The magnesium reacts more slowly than the barium, but the reaction continues until

all the magnesium has reacted.

Explain why

• the barium initially reacts more quickly than the magnesium

• the barium reaction stops before all the barium has reacted.

[3 marks]

09.4 A metal nitrate X(NO3)2 completely decomposes when heated.

2X(NO3)2(s) → 2XO(s) + 4NO2(g) + O2(g)

A 0.832 g sample of X(NO3)2 decomposes on heating to produce a total of

*28* 348 cm3 of gas at 298 K and 100 kPa

Deduce the identity of metal X.

The ideal gas constant, R = 8.31 J K–1 mol–1

[6 marks]

Identity of metal31 X

09.5 Sodium reacts with aluminium and hydrogen to form solid NaAlH4

Give an equation for this reaction.

Suggest why NaAlH4 has a high melting point.

[3 marks]

Equation

Suggestion

09.6 Give the equation for the reaction between H3PO4 and an excess of NaOH

[1 mark]

Lithium is an important metal used in cells to power mobile phones.

09.7 In a lithium cell, a lithium cobalt oxide electrode and a lithium electrode are used.

Give the equation for the reaction that occurs at the positive electrode.

[1 mark]

09.8 Commercial electrochemical cells can be rechargeable or non-rechargeable.

State why lithium cells can be recharged.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for Question 9. 09.1 gives 1 mark for more shells or energy levels. 09.2 gives 1 mark for Mg(s) + H2O(g) -> MgO(s) + H2(g). 09.3 awards 3 marks for Ba outer electrons being further from the nucleus, insoluble barium sulfate forming, and barium sulfate coating the metal preventing further reaction. 09.4 awards 6 marks for converting P and V, calculating moles of gas (0.01405 mol), moles of nitrate (5.62x10^-3 mol), Mr (148.0), Ar (24.0) and identifying Mg. 09.5 gives 3 marks for Na + Al + 2H2 -> NaAlH4 and describing strong electrostatic attractions between Na+ and AlH4- ions. 09.6 gives 1 mark for 3NaOH + H3PO4 -> Na3PO4 + 3H2O. 09.7 gives 1 mark for Li+ + CoO2 + e- -> Li+(CoO2)-. 09.8 gives 1 mark for electrode reactions being reversible by applying a reverse potential.

Question Answers Additional comments/Guidelines Mark

more shells Allow Ca has 4 shells and Mg has 3 shells

Or Do not accept more outer shells 1

09.1

more energy levels Ignore shielding (1 x AO3)

Ignore subshells/orbitals/more electrons

Mg(s) + H2O(g) ⟶ MgO(s) + H2(g) State symbols required 1

09.2

Allow multiples (1 x AO3)

M1 (Ba is more reactive) because outer/valence electrons further

from nucleus/less attracted to the nucleus/lost more easily

M2 Insoluble barium sulfate (is formed)

Or

Ba + H2SO4 → BaSO4(s) + H2 3

09.3 (2 x AO1,

M3 Barium sulfate prevents further reaction (with sulfuric acid) 1 x AO3)

Or

Barium gets coated with barium sulfate (so no more barium

reacts)

M1 P = 100 000 Pa and V = 348 × 10–6 m3

PV 100 000 𝑥𝑥 348 𝑥𝑥 10−6

M2 n = or

RT 8.31 𝑥𝑥 298

M3 n = 0.01405 mol

M4 n metal nitrate = 0.01405 × = 5.62 × 10–3 mol M4 = M3 x

55 6

09.4 0.832 (1 x AO1,

M5 Mr metal nitrate = = 148(.0) M5 = 0.832 ÷ M4

5.62×10–3 5 x AO2)

M6 Ar of metal = 148.0 – (2 × 14 +2 × 48) = 24(.0) = Mg M6 = M5 – 124 and identity of a metal with 2+

oxidation state

M1 Na + Al + 2H2 → NaAlH4

09.5 M2 contains oppositely charged ions/ Na+ and AlH – ions

4 (2 x AO2,

1 x AO3)

M3 strong attraction between (oppositely charged) ions

3 NaOH + H3PO4 → Na3PO4 + 3H2O Allow multiples and ignore state symbols 1

09.6

(1 x AO3)

Li++ CoO + e– → Li+(CoO )– allow Li(CoO ) as product 1

22 2

09.7

(1 x AO1)

The electrode reactions can be reversed (by applying a reverse Allow reaction is reversible (by applying a

09.8 potential) reverse potential)

(1 x AO1)

How to answer it

Metals, Group 2 Trends, and Electrochemical Cells

What this question tests

This multi-topic question assesses core Physical and Inorganic Chemistry concepts: Group 2 periodic trends (atomic radius, reactivity down the group, and sulfate solubility); chemical equations with state symbols (magnesium with steam); the ideal gas equation ( pV = nRT ) with stoichiometric ratios to identify an unknown metal; bonding and structure in metal hydrides; neutralisation stoichiometry of polyprotic acids; and commercial lithium-ion cell chemistry (electrode half-equations and recharging principles).

Question 09.1

Atomic Radius Trend: Calcium vs Magnesium (1 Mark)

✅ Correct Answer

Calcium has more shells (or more energy levels).

Also accepted: Ca has 4 shells and Mg has 3 shells.

❌ Common Errors & Trapdoors

  • Writing "more outer shells" — this contradicts chemical principles and scores zero.
  • Focusing purely on shielding or nuclear charge without explicitly stating that calcium has more occupied electron shells.

Question 09.2

Reaction of Magnesium with Steam (1 Mark)

✅ Correct Answer

Mg(s) + H₂O(g) → MgO(s) + H₂(g)

State symbols are compulsory for this mark.

🧠 Exam Technique: Steam vs Cold Water

  • With cold water: Mg reacts very slowly to form magnesium hydroxide:
    Mg(s) + 2H₂O(l) → Mg(OH)₂(aq/s) + H₂(g)
  • With steam: Mg burns with a bright white flame to form magnesium oxide (MgO) and hydrogen gas. Remember H₂O(g) !

Question 09.3

Reactivity and Sulfate Solubility Trends in Group 2 (3 Marks)

✅ Mark Scheme Breakdown

  • Mark 1 (Initial rate): Barium's outer electrons are further from the nucleus (more shielded) so are less attracted to the nucleus / lost more easily.
  • Mark 2 (Product identity): Insoluble barium sulfate (BaSO₄) is formed (or write: Ba + H₂SO₄ → BaSO₄(s) + H₂ ).
  • Mark 3 (Passivation effect): The insoluble BaSO₄ forms a layer that coats the barium, preventing further contact/reaction with the sulfuric acid.

💡 Group 2 Sulfate Solubility Trend

Sulfate solubility decreases down Group 2:

  • MgSO₄: Soluble → Magnesium continues reacting until fully consumed.
  • BaSO₄: Highly insoluble white precipitate → Coats metal surface, creating an impermeable barrier.

Question 09.4

Deducing Metal X using the Ideal Gas Equation (6 Marks)

📐 Step-by-Step Calculation

Equation: 2 X(NO₃)₂(s) → 2 XO(s) + 4 NO₂(g) + O₂(g)

  1. Convert given values to SI units (Mark 1):
    Pressure: P = 100 kPa = 100 000 Pa = 1.00 × 10⁵ Pa
    Volume: V = 348 cm³ = 348 × 10⁻⁶ m³ = 3.48 × 10⁻⁴ m³
    Temperature: T = 298 K
  2. Calculate total moles of gas produced (Marks 2 & 3):
    n(gas) = pV / RT = (100 000 × 348 × 10⁻⁶) / (8.31 × 298)
    n(gas) = 34.8 / 2476.38 = 0.01405 mol
  3. Apply molar ratio to find moles of X(NO₃)₂ (Mark 4):
    Total gas produced per 2 moles of nitrate = 4 NO₂ + 1 O₂ = 5 moles of gas.
    Mole ratio is 2 moles of X(NO₃)₂ : 5 moles of gas:
    n(metal nitrate) = 0.01405 × (2 / 5) = 5.62 × 10⁻³ mol
  4. Calculate Mᵣ of the metal nitrate (Mark 5):
    Mᵣ = mass / moles = 0.832 g / (5.62 × 10⁻³ mol) = 148.0 g mol⁻¹
  5. Find Aᵣ and identify Metal X (Mark 6):
    Mass of two nitrate ions, 2 × NO₃⁻ = 2 × [14.0 + (3 × 16.0)] = 2 × 62.0 = 124.0
    Aᵣ(X) = 148.0 - 124.0 = 24.0
    Identity: Magnesium (Mg) (Group 2 metal with Aᵣ = 24.3 / 24.0)

❌ Common Errors in this Question

  • Unit conversions: Forgetting that cm³ → m³ requires multiplying by 10⁻⁶ (dividing by 1 000 000).
  • Gas stoichiometry trap: Counting only NO₂ (4 moles) or forgetting that O₂ is also a gas! Total gas produced = 4 + 1 = 5 mol .
  • Formula mass mistake: Forgetting that X(NO₃)₂ contains two nitrate groups ( 2 × 62 = 124 ).

Question 09.5

Formation and High Melting Point of NaAlH₄ (3 Marks)

✅ Mark Scheme Breakdown

Equation (1 mark):

Na + Al + 2 H₂ → NaAlH₄

High Melting Point Suggestion (2 marks):

  • It contains oppositely charged ions (specifically Na⁺ and AlH₄⁻ ). (1 mark)
  • There are strong electrostatic attractions between the oppositely charged ions requiring large amounts of energy to overcome. (1 mark)

🧠 Exam Technique: Explaining High Melting Points

Always structure ionic melting point explanations around two essential points:

  1. Identify the nature of the particles: ions (and state their charges where possible: Na⁺ and AlH₄⁻).
  2. Identify the type and strength of bonding: strong electrostatic attraction between oppositely charged ions.

Question 09.6

Neutralisation of Phosphoric(V) Acid (1 Mark)

✅ Correct Answer

3 NaOH + H₃PO₄ → Na₃PO₄ + 3 H₂O

Multiples accepted. State symbols are not required.

💡 Triprotic Acid Insight

H₃PO₄ is a triprotic (tribasic) acid because it has 3 ionisable protons. When reacted with excess sodium hydroxide, all three protons are neutralised to produce the phosphate salt Na₃PO₄ .

Questions 09.7 & 09.8

Rechargeable Lithium-Ion Cells (2 Marks)

✅ 09.7: Positive Electrode Half-Equation (1 Mark)

Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻

Also acceptable: Li⁺ + CoO₂ + e⁻ → LiCoO₂

Reduction takes place at the positive electrode during cell discharge.

✅ 09.8: Why Lithium Cells are Rechargeable (1 Mark)

The electrode reactions are reversible (or can be reversed by applying an external reverse electrical potential/voltage).

🧠 Electrochemical Cell Classification

Cell Type Reversibility Key Characteristic
Non-rechargeable (Primary) Irreversible Reactions cannot be reversed; discarded once reactants are spent (e.g. standard alkaline zinc-MnO₂).
Rechargeable (Secondary) Reversible Applying external potential forces electrons in opposite direction, regenerating original chemicals.

Topics

Inorganic Chemistry · Physical Chemistry · 3.1.2 Amount of Substance · 3.1.3 Bonding · 3.1.11 Electrode Potentials · 3.2.2 Group 2, The Alkaline Earth Metals · 3.2.4 Properties of Period 3 Elements and Their Oxides

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.