AQA A-Level Chemistry AS Paper 1, June 2023: Question 2

19 marks · Medium difficulty · State/Explain/Numerical

Answer questions on Group 2 elements including metallic bonding in magnesium, trends in atomic radius, reaction with steam, sulfate solubility, isotopic abundance of strontium, and a back titration of magnesium hydroxide.

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Question

Question 2 consists of six parts: 02.1 asks to describe the structure and bonding in magnesium (2 marks); 02.2 asks for the trend in atomic radius down Group 2 from Mg to Ba and a reason (2 marks); 02.3 asks for an equation with state symbols for magnesium reacting with steam and two observations (3 marks); 02.4 asks for the least soluble Group 2 sulfate and a medical or industrial use for it (2 marks); 02.5 asks to calculate the percentage abundance of isotopes 86Sr and 87Sr given Ar = 87.73 and 83.00% abundance of 88Sr (4 marks); 02.6 describes a back titration to find the percentage by mass of Mg(OH)2 in an antacid tablet using excess HCl neutralized by NaOH (6 marks).
Question text

02 This question is about the elements in Group 2.

02.1 Describe the structure and bonding in magnesium.

[2 marks]

02.2 State the trend in the atomic radius of the elements down Group 2 from Mg to Ba

Give a reason for this trend.

[2 marks]

Trend

Reason

02.3 Give an equation, including state symbols, for the reaction of magnesium with steam.

State two observations for this reaction.

[3 marks]

Equation

Observation 1

Observation 2

02.4 The sulfates of the elements in Group 2 from Mg to Ba have different solubilities.

State the formula of the least soluble of these sulfates.

Give a use for this sulfate.

[2 marks]

Formula

Use

02.5 A sample of strontium is made up of only three isotopes: 86Sr, 87Sr and 88Sr

*03* This sample contains 83.00% by mass of 88Sr

This sample of strontium has Ar = 87.73

Calculate the percentage abundance of each of the other two isotopes in this sample.

[4 marks]

% abundance 87Sr =

% abundance Sr =

02.6 Mg(OH)2 is used as an antacid to treat indigestion.

A student does an experiment to determine the percentage by mass of Mg(OH)2 in an

indigestion tablet.

*04* 3 –3

40.0 cm of 0.200 mol dm HCl (an excess) is added to 0.200 g of a powdered tablet.

The mixture is swirled thoroughly.

All of the Mg(OH)2 reacts with HCl as shown.

Mg(OH)2 + 2HCl → MgCl2 + 2H2O

The amount of HCl remaining after this reaction is determined by titration with

0.100 mol dm–3 NaOH

29.25 cm3 of 0.100 mol dm–3 NaOH are needed.

Calculate the percentage by mass of Mg(OH)2 in the indigestion tablet.

[6 marks]

Percentage by mass

Mark scheme

Show the mark scheme Mark scheme for Question 2 with marking guidance: 02.1 requires giant lattice of Mg2+ cations and electrostatic attractions with delocalised electrons (2 marks); 02.2 states atomic radius increases because number of electron shells increases (2 marks); 02.3 provides Mg(s) + H2O(g) -> MgO(s) + H2(g) and observations of bright white flame and white powder/smoke (3 marks); 02.4 identifies BaSO4 and barium meal/X-ray use (2 marks); 02.5 sets up algebraic equation for Sr isotopic abundances giving 7% for 87Sr and 10% for 86Sr (4 marks); 02.6 shows complete working for moles of initial HCl (0.00800 mol), excess HCl titrated by NaOH (0.002925 mol), reacted HCl (0.005075 mol), moles of Mg(OH)2 (0.0025375 mol), mass of Mg(OH)2 (0.148 g), and final percentage of 74.0% (6 marks).

Question Marking guidance Additional Comments/Guidelines Mark

M1 (giant) lattice of (Mg2+) cations / (giant) lattice of (Mg) atoms Incorrect structure type loses M1

02.1

2+ (2 x AO1)

M2 (Electrostatic) attractions between cations / Mg ions / nuclei and

delocalised electrons

M1 Trend: increases

02.2

M2 Reason: the number of electron energy levels increases Allow: the number of electron shells increases (2 x AO1)

Ignore increase in shielding

Mg(s) + H2O(g) → MgO(s) + H2(g) State symbols essential

Bright/white flame/light 3

02.3 (1 x AO1,

White/grey ash/powder (allow smoke) Do not allow ppt 2 x AO2)

Ignore black solid

Ignore fumes.

M1 BaSO4

02.4

(2 x AO1)

M2 X-rays (of internal organs) / barium meal

M1 Abundance of 87 Sr = X Allow M1 for

and Abundance of 86Sr = 1 – 0.83 – X Abundance of 87Sr = X and Abundance of 86Sr = Y

= 0.17 – X if also states that X + Y = 17

M2 87.73 = (88 × 0.83) + (87 × X) + (86 × (0.17 – X)) 87.73 = (88 × 0.83) + (87 × X) + (86 × Y)

02.5

(4 x AO2)

87.73 = 73.04 + 87X + 14.62 – 86X

87.73 = 87.66 + X

M3 87Sr = 0.07 = 7 %

M4 Abundance of 86Sr = 1 – 0.83 - 0.07 = 0.1 = 10 % M4 = 17 – M3

M1 Amount of HCl added = 0.200 × 0.040 = 0.00800 mol

M2 Amount of NaOH = 0.100 × 0.02925 = 0.002925 mol

(Amount of HCl = 0.002925 mol)

M3 Amount of HCl reacted with Mg(OH)2 = 0.00800 – 0.002925 = M3 = M1 – M2

0.005075 mol 6

02.6

(6 x AO2)

M4 Amount of Mg(OH)2 = 0.005075 ÷ 2 = 0.0025375 mol M4 = M3 ÷ 2

M5 Mass of Mg(OH)2 = 58.3 × 0.0025375 = 0.148 g M5 = M4 × 58.3

0.148 M5

M6 % by mass = × 100 = 74.0 % M6 = × 100

0.200 0.200

Do not allow M6 if >100%

How to answer it

Group 2 Trends, Reactions & Quantitative Titrations

📌 WHAT THIS QUESTION TESTS

This question brings together foundational physical and inorganic chemistry concepts across Group 2:

  • Bonding: Describing metallic bonding accurately using standard AQA nomenclature.
  • Periodic Trends: Explaining atomic radius changes down a group.
  • Redox & Reactivity: Writing balanced equations with state symbols for magnesium reacting with steam, alongside visual observations.
  • Solubility & Medical Uses: Recalling sulfate solubility trends and applications (barium meal).
  • Isotopic Abundances: Formulating and solving simultaneous equations using relative atomic mass (Aᵣ).
  • Back Titration Calculations: Multi-step quantitative analysis determining reactant moles, stoichiometry, mass, and percentage purity.
Part 02.1 • 2 Marks

Structure and Bonding in Magnesium

Describing metallic structure and forces

✅ Correct Answer

  • M1: (Giant) lattice of Mg²⁺ cations / (giant) lattice of magnesium atoms.
  • M2: Electrostatic attraction between cations / Mg²⁺ ions / nuclei and delocalised electrons.

🧠 Exam Technique

Always state both the type of particle and the arrangement when describing structure:

  • Structure = giant metallic lattice.
  • Bonding = electrostatic force between positive metal ions and delocalised electrons.

❌ Common Errors & Misconceptions

  • Stating "covalent bonds" or "intermolecular forces" instantly forfeits M1.
  • Forgetting the word delocalised when referencing electrons loses M2.
  • Saying "attraction between magnesium atoms and electrons" (atoms are neutral; the attraction is to cations/nuclei).

💡 Key Knowledge

Magnesium has metallic bonding: Mg atoms each donate 2 valence electrons into a delocalised "sea" of electrons, forming an array of closely packed Mg²⁺ ions held tightly by electrostatic forces.

Mark scheme: 1 mark for lattice of cations/atoms; 1 mark for attraction between cations/nuclei and delocalised electrons.
Part 02.2 • 2 Marks

Trend in Atomic Radius Down Group 2

Trend from Mg to Ba and underlying explanation

✅ Correct Answer

  • Trend (M1): Increases.
  • Reason (M2): The number of electron shells (or electron energy levels) increases.

❌ Common Errors

  • Writing "shielding increases" as the primary reason. While shielding does increase, AQA explicitly requires mentioning the increase in shells / main energy levels for the mark.
  • Confusing group trends with period trends (e.g. arguing that nuclear charge increases so radius decreases).
Mark scheme: M1 for "increases"; M2 for "number of electron shells / energy levels increases". Note: "Ignore increase in shielding".
Part 02.3 • 3 Marks

Reaction of Magnesium with Steam

Balanced equation with state symbols and two observations

✅ Correct Answer

Equation:

Mg(s) + H₂O(g) → MgO(s) + H₂(g)

Observations (any two):

  • Bright white flame / white light
  • White / grey solid / ash / powder (smoke allowed)

💡 Key Knowledge: Steam vs Cold Water

  • With steam: Mg burns vigorously to form magnesium oxide (MgO) and hydrogen (H₂). State symbol for steam is (g) !
  • With cold water: Reaction is very slow; forms magnesium hydroxide: Mg(s) + 2H₂O(l) → Mg(OH)₂(aq/s) + H₂(g) .

❌ Common Errors

  • Writing H₂O(l) instead of H₂O(g) — state symbols are explicitly stated as essential.
  • Producing Mg(OH)₂ instead of MgO .
  • Calling the solid product a "precipitate" — precipitates only form in aqueous solution, not when a gas reacts with a solid!
  • Vague observations like "fumes", "effervescence", or "black solid".

🧠 Exam Technique

Always inspect the question prompt: when it specifies "including state symbols", full equation marks depend strictly on correct states: (s) , (g) .

Mark scheme: 1 mark for correct balanced equation with state symbols; 2 marks for two distinct visual observations.
Part 02.4 • 2 Marks

Sulfate Solubility & Medical Application

Identification of the least soluble Group 2 sulfate

✅ Correct Answer

  • Formula: BaSO₄
  • Use: Barium meal / X-ray imaging (of digestive system / gut / internal organs).

💡 Key Knowledge: Group 2 Solubility Trends

  • Hydroxides [Mg(OH)₂ → Ba(OH)₂]: Solubility increases down the group. (Mg(OH)₂ is sparingly soluble; Ba(OH)₂ is soluble).
  • Sulfates [MgSO₄ → BaSO₄]: Solubility decreases down the group. (MgSO₄ is soluble; BaSO₄ is insoluble).
  • Even though Ba²⁺ ions are toxic, BaSO₄ can be ingested because its extreme insolubility prevents it from entering the bloodstream.
Mark scheme: 1 mark for formula BaSO₄; 1 mark for medical use: barium meal / X-rays of internal organs.
Part 02.5 • 4 Marks

Isotope Abundance Calculation

Determining unknown percentages from relative atomic mass

📐 Step-by-Step Calculation

Step 1: Set up percentage variables
Total abundance = 100%
Abundance of ⁸⁸Sr = 83.00%
Remaining abundance for (⁸⁶Sr + ⁸⁷Sr) = 100 − 83.00 = 17.00%
Let % abundance of ⁸⁷Sr = x, then % abundance of ⁸⁶Sr = 17.00 − x
Step 2: Substitute into relative atomic mass expression
Aᵣ = [ (88 × 83.00) + (87 × x) + (86 × (17.00 − x)) ] / 100 = 87.73
Step 3: Solve linear equation for x
8773 = 7304 + 87x + 1462 − 86x
8773 = 8766 + x
x = 8773 − 8766 = 7.00%
Step 4: Find abundance of the second isotope
% ⁸⁶Sr = 17.00 − 7.00 = 10.00%

✅ Final Answers

% abundance ⁸⁷Sr = 7% (or 7.00%)
% abundance ⁸⁶Sr = 10% (or 10.00%)

❌ Common Errors

  • Algebraic slip when expanding 86(17 − x) .
  • Forgetting to divide by 100 or mixing fractional abundances with percentages.
  • Swapping the final two answers around at the end.
Mark scheme: M1 for setting up variables (x and 17 − x); M2 for setting up expression with 87.73; M3 for ⁸⁷Sr = 7%; M4 for ⁸⁶Sr = 10%.
Part 02.6 • 6 Marks

Back Titration: Percentage Mass of Mg(OH)₂

Determining purity using excess acid and standard base titration

📐 Step-by-Step Back Titration Method

Step 1: Calculate initial moles of HCl added
n(HCl) initial = conc × vol = 0.200 mol dm⁻³ × (40.0 / 1000) dm³ = 0.00800 mol
Step 2: Calculate moles of unreacted (excess) HCl from titration
Reaction: HCl + NaOH → NaCl + H₂O (1 : 1 ratio)
n(NaOH) = 0.100 mol dm⁻³ × (29.25 / 1000) dm³ = 0.002925 mol
Therefore, n(HCl) unreacted = 0.002925 mol
Step 3: Calculate moles of HCl that reacted with Mg(OH)₂
n(HCl) reacted = 0.00800 − 0.002925 = 0.005075 mol
Step 4: Use stoichiometry to find moles of Mg(OH)₂
From equation: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O (1 : 2 ratio)
n(Mg(OH)₂) = 0.005075 / 2 = 0.0025375 mol
Step 5: Calculate mass of Mg(OH)₂
Mᵣ(Mg(OH)₂) = 24.3 + (2 × 16.0) + (2 × 1.0) = 58.3 g mol⁻¹
mass = 0.0025375 mol × 58.3 g mol⁻¹ = 0.147936 g (0.148 g)
Step 6: Calculate percentage by mass in 0.200 g tablet
% by mass = (0.147936 g / 0.200 g) × 100 = 73.97% ≈ 74.0%

❌ Common Traps in Back Titrations

  • Missing the 1 : 2 ratio: Forgetting to divide moles of HCl by 2 when converting to Mg(OH)₂ (M4).
  • Wrong subtraction: Adding the titration moles instead of subtracting them from initial moles to find the reacted quantity.
  • Mᵣ Errors: Calculating Mᵣ of Mg(OH)₂ without doubling both oxygen and hydrogen atoms (must be 58.3).
  • Sanity Check: Any final percentage over 100% automatically scores 0 for M6.

🧠 Top Tip: The Back Titration Map

Always draw a quick mental timeline:

[Total Acid Added] − [Excess Acid Reacted with NaOH] = [Acid Reacted with Tablet]

Once you have moles of acid consumed by the tablet, apply the chemical equation's stoichiometric ratio to get tablet active ingredient moles.

Mark scheme breakdown: M1 for n(initial HCl); M2 for n(excess HCl); M3 for n(reacted HCl); M4 for dividing by 2; M5 for mass of Mg(OH)₂ (using 58.3); M6 for 74.0%.

Topics

Inorganic Chemistry · Physical Chemistry · 3.2.2 Group 2, The Alkaline Earth Metals · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance · 3.1.3 Bonding

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