Edexcel A-Level Chemistry Paper 1, June 2024: Question 8

12 marks · Medium difficulty · Synoptic Questions

Draw the dot-and-cross diagram of magnesium hydroxide, define enthalpy of solution, complete a Born–Haber cycle to calculate enthalpy of formation, and explain differences in experimental and theoretical lattice energies.

Practise this question

Question

Question 8 consists of parts (a) to (c). Part (a) asks for dot-and-cross diagrams for the ions in magnesium hydroxide. Part (b) is a multiple-choice question asking for the definition of the enthalpy change of solution. Part (c) provides energetic data for magnesium fluoride and asks students to complete a Born–Haber cycle diagram, calculate the standard enthalpy change of formation of MgF2, and explain the differences between experimental and theoretical lattice energies for MgF2 and MgI2 using given data in a table.
Question text

8 This question is about ionic compounds.

(a) Draw dot-and-cross diagrams of the ions in magnesium hydroxide, showing the

outer shell electrons only.

Use × for magnesium electrons, ● for oxygen electrons and for each

hydrogen electron.

(2)

(b) Which definition correctly describes the enthalpy change of solution, ΔsolH?

(1)

Enthalpy change of solution, ΔsolH

The enthalpy change when 1mol of gaseous ions dissolves in

A

sufficient water to give an infinitely dilute solution.

The enthalpy change when 1mol of an ionic substance dissolves in

B

water to give an infinitely dilute solution.

The enthalpy change when 1mol of gaseous ions dissolves in

C −3

sufficient water to give a solution of concentration 1moldm .

The enthalpy change when 1mol of an ionic substance dissolves in

D −3

water to give a solution of concentration 1moldm .

(c) The table shows the information needed to calculate the

standard enthalpy change of formation of magnesium fluoride.

Label Description Value / kJ mol−1

A enthalpy change of formation of magnesium fluoride

B lattice energy of magnesium fluoride −2957

C enthalpy change of atomisation of magnesium +148

D 1st ionisation energy of magnesium +738

E 2nd ionisation energy of magnesium +1451

F enthalpy change of atomisation of fluorine +79

G *P76895A01928*1st electron affinity of fluorine −328

(i) Complete the Born–Haber cycle for magnesium fluoride with formulae,

state symbols, electrons and correctly labelled arrows.

The cycle is not drawn to scale.

(4)

Mg(g) + F2(g)

C

A

Mg(s) + F2(g) MgF2(s)

(ii) Calculate the value of ΔfH [MgF2(s)].

20 (1)

(iii) The experimental and theoretical values of the lattice energy for MgF*P76895A02028*2and

MgI2 are given in the table.

Explain the differences in these values.

(4)

Experimental lattice energy Theoretical lattice energy

Compound −1 −1

/kJmol /kJmol

MgF2 −2957 −2913

MgI2 −2327 −1944

… 21

… *P76895A02128*

(Total for Question 8 = 12 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 8 detailing: (a) 2 marks for correct dot-and-cross representations of Mg2+ and two OH- ions; (b) 1 mark for answer B; (c)(i) 4 marks for completing the Born–Haber cycle with correct species, state symbols, and enthalpy labels; (c)(ii) 1 mark for the calculated enthalpy of formation value of -1118 kJ/mol; (c)(iii) 4 marks for explaining that MgF2 is almost purely ionic due to close theoretical and experimental lattice energy values, while MgI2 has significant covalent character because the larger iodide ion is more polarisable, and noting that MgF2 has a more exothermic lattice energy due to the smaller size of fluoride ions.

Question

Acceptable Answer Additional Guidance Mark

Number

8(a) An explanation that makes reference to the following points: (2)

• magnesium ion with correct structure and charge (1)

• structure of two hydroxide ions with correct charge (1) Allow Mg2+ with no electrons in the outer shell

Ignore inner shell electrons

Allow circle(s) of electron shells to be shown

Allow square brackets missing

Penalise incorrect symbols once

Question

Answer Mark

Number

8(b) The only correct answer is B (The enthalpy change when 1mol of an ionic substance dissolves in water to give an (1)

infinitely dilute solution.)

A is not correct as the mention of gaseous ions is incorrect

C is not correct as the mention of gaseous ions is incorrect and the concentration is incorrect

D is not correct as the concentration is incorrect

Question

Answer and Additional Guidance Mark

Number

8(c)(i) (4)

• correct letters B, D, E, F, G or correct values corresponding to their cycle, G and B must be on the

correct boxes (1)

• multiplying F and G by 2 (must be seen on the cycle for this mark) (1)

• correct formulae in every (1)

• state symbols and electrons (charge does not need to be shown on electrons) (1)

Accept order of alternative order of boxes 2 – 3 – 1 with accompanying letters etc

Question

Acceptable Answer Additional Guidance Mark

Number

8(c)(ii) Example of calculation (1)

• correct answer (148 + (2 × 79) + 738 + 1451 + (−328 × 2) + (−2957))

= −1118 (kJ mol−1 )

Comment if −1118 seen then award the mark no matter what

their cycle indicates

No TE on incorrect values mis-transcribed on their cycle

Allow TE if matches cycle

−790 scores 1 (electron affinity not doubled)

−869 scores 1 (both values not doubled)

−1197 scores 1 (atomisation not doubled)

Question

Acceptable Answer Additional Guidance Mark

Number

8(c)(iii) An explanation that makes reference to the following points: Allow reverse argument throughout (4)

• (theoretical and experimental values of MgF2 are close / similar as) (1)

MgF2 is (nearly) 100% ionic Allow MgF2 is more ionic than

MgI2

• (theoretical and experimental values of MgI2 are different as) MgI2 (1)

has some covalent character / more covalent character than MgF2

• I− / iodide ions are more polarisable (than F− (ions)) / highly (1)

− Allow iodine ions / fluorine ions

polarisable / because the I ions are larger / have a larger radius Allow I− can be distorted more

Allow bond between Mg and I more

polarised as long as ions mentioned

in response.

Do not award species other than

ions (e.g. atom, molecule)

Do not award atomic radius is larger

• values for MgF2 are more negative / more exothermic than for MgI2

as the (ionic) bonding is stronger in MgF (because the F− ion is (1)

2 Allow ‘higher’ for ‘more negative’

smaller than the I− ion)

(Total for Question 8 = 12 marks)

How to answer it

Question 8: Ionic Bonding, Energetics & Born–Haber Cycles

Edexcel A-Level Chemistry | Total: 12 Marks

📋 What this question tests

This multi-topic question assesses core physical and inorganic chemistry concepts across bonding and thermodynamics:

  • Dot-and-cross diagrams: Representing compound ions (hydroxide) and metal cations with specific electron symbols and square brackets.
  • Enthalpy definitions: Precisely distinguishing standard enthalpy change of solution from hydration.
  • Born–Haber cycles: Correct sequencing of atomisation, ionisation, electron affinity, and lattice energy steps; stoichiometry doubling (F₂ → 2F).
  • Hess's Law calculations: Applying indirect thermodynamic cycles to calculate standard enthalpy of formation.
  • Lattice energy & polarization: Comparing theoretical (purely ionic) vs. experimental (Born–Haber) lattice energies to explain polarization, covalent character, and ionic radii trends.
Part (a) — 2 Marks

Dot-and-Cross Diagram of Magnesium Hydroxide

Outer shell electrons of Mg²⁺ and OH⁻ ions using specific symbols

✅ Correct Answer & Diagram Structure

The formula of magnesium hydroxide is Mg(OH)₂, requiring one magnesium ion and two hydroxide ions:

  • Magnesium ion: [Mg]²⁺ containing an octet of 8 crosses (×) from its full shell (level 2), OR an empty outer shell with charge 2+.
  • Two Hydroxide ions (drawn twice or written with a "2 ×"):
    Square brackets around each ion with a 1− charge: [OH]⁻ .
    Inside the bracket:
    • Oxygen atom sharing 1 pair of electrons with hydrogen (1 dot • from O, 1 triangle △ from H).
    • Oxygen retains 3 lone pairs of its own electrons (5 dots •).
    • 1 gained electron from magnesium shown as a cross (×).
    • Total electrons around O = 8 (6 dots •, 1 triangle △, 1 cross ×).

🧠 Exam Technique & Notation

  • Symbol Checklist: The prompt explicitly specifies:
    • × = magnesium electron
    • • = oxygen electron
    • △ = hydrogen electron
  • Check the stoichiometry: magnesium forms Mg²⁺, so each magnesium transfers two electrons—one to each of the two hydroxide ions.
  • Always put charges outside square brackets.

❌ Common Errors to Avoid

  • Symbol confusion: Drawing generic circles or mixing up which electron came from which atom (marks are deducted for incorrect symbols).
  • Missing the second OH⁻: Forgetting that Mg²⁺ needs two OH⁻ ions to achieve charge neutrality.
  • Incorrect charge: Writing Mg⁺ or drawing OH²⁻.
Mark Scheme Breakdown:
• Mark 1: Magnesium ion with correct structure and 2+ charge.
• Mark 2: Structure of two hydroxide ions with correct 1− charge and correct electron distribution.
Part (b) — 1 Mark

Definition: Enthalpy Change of Solution (ΔsolH)

Multiple Choice Question

✅ Correct Answer: Option B

B: "The enthalpy change when 1 mol of an ionic substance dissolves in water to give an infinitely dilute solution."

💡 Key Knowledge: Enthalpy of Solution vs. Hydration

  • ΔsolH: Starts with 1 mol of solid ionic compound dissolving in water to infinite dilution.
    Equation: NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)
  • ΔhydH: Starts with 1 mol of gaseous ions dissolving in water.
    Equation: Na⁺(g) + aq → Na⁺(aq)
  • Infinite dilution: Means adding enough water so that further addition of solvent produces no further heat change (ions are far enough apart to not interact). It does not mean 1 mol dm⁻³.

❌ Why Other Options Are Wrong

  • A is incorrect: Refers to gaseous ions, which is the definition for enthalpy of hydration.
  • C is incorrect: Refers to gaseous ions and incorrect concentration.
  • D is incorrect: Standard state/definition requires infinitely dilute solution, not a 1 mol dm⁻³ solution.
Mark Scheme Breakdown: 1 Mark for selecting B.
Part (c)(i) — 4 Marks

Completing the Born–Haber Cycle for MgF₂

Filling in intermediate states, electron stoichiometry, and step labels

📐 Cycle Architecture & Required Contents

Step 4 (Top): [ Mg²⁺(g) + 2F(g) + 2e⁻ ] | \ Step 3: | \ Arrow downwards: 2 × G [or 2 × (−328) = −656] Label: E [(+)1451] | \ ^ v Step 2: [ Mg⁺(g) + 2F(g) + e⁻ ] [ Mg²⁺(g) + 2F⁻(g) ] (Step 5) ^ | Label: D [(+)738] | | Arrow downwards: B [or −2957] | v Step 1: [ Mg(g) + 2F(g) ] [ MgF₂(s) ] ^ ^ Label: 2 × F [(+)158]| | | | Arrow rightwards: A (ΔfH) Given Base Level: [ Mg(g) + F₂(g) ] | ^ | Label: C [(+)148] | | | | Starting Level: [ Mg(s) + F₂(g) ] ----------------+

🧠 Essential Details for Full Marks

  • Box 1 (Atomisation of F₂): Must be Mg(g) + 2F(g) . The arrow from Mg(g) + F₂(g) represents atomisation of 1 mole of F₂ to form 2 moles of F(g), so label must be 2 × F or +158 kJ mol⁻¹.
  • Box 2 (1st IE of Mg): Mg⁺(g) + 2F(g) + e⁻ . Arrow label: D or +738.
  • Box 3 (2nd IE of Mg - topmost box): Mg²⁺(g) + 2F(g) + 2e⁻ . Arrow label: E or +1451.
  • Box 4 (1st EA of Fluorine): Downward arrow to Mg²⁺(g) + 2F⁻(g) . Must show 2 × G or 2 × (−328) = −656 because 2 moles of F atoms gain electrons!
  • Downward arrow to MgF₂(s): Labelled B or −2957 (Lattice energy of formation).

❌ Common Examiner Traps

  • Forgetting to double F and G: Atomisation of fluorine is given as +79 kJ mol⁻¹ (per mole of F atoms formed). Since MgF₂ has 2 fluorines, you need 2 × 79 = 158 . Similarly, 1st electron affinity must be doubled: 2 × (−328) .
  • Missing state symbols: Every species in every box must have (g) or (s) .
  • Missing electrons: Losing electrons when ionizing ( + e⁻ , + 2e⁻ ) will cost marks.
Mark Scheme Breakdown (4 Marks):
• Mark 1: Correct letters B, D, E, F, G or corresponding numerical values.
• Mark 2: Multiplying F and G by 2 clearly shown on cycle.
• Mark 3: Correct chemical formulae in every box.
• Mark 4: Correct state symbols throughout and electrons shown.
Part (c)(ii) — 1 Mark

Calculation: Standard Enthalpy of Formation of MgF₂

Applying Hess's Law to the Born–Haber Cycle

📐 Step-by-Step Calculation

  1. State the Hess's Law expression:
    Formation route = sum of all indirect atomisation, ionisation, electron affinity, and lattice energy steps:
    ΔfH = ΔatH(Mg) + 2[ΔatH(F)] + 1st IE(Mg) + 2nd IE(Mg) + 2[1st EA(F)] + Lattice Energy
  2. Substitute the values:
    ΔfH = (+148) + 2(+79) + (+738) + (+1451) + 2(−328) + (−2957)
    ΔfH = 148 + 158 + 738 + 1451 − 656 − 2957
  3. Evaluate the result:
    ΔfH = 2495 − 3613 = −1118 kJ mol⁻¹
Mark Scheme Note: Award 1 mark directly if −1118 is seen. Transfer of error (TE) allowed only if matching incorrect doubling on the student's diagram (e.g. −790 if EA not doubled; −1197 if atomisation not doubled; −869 if neither doubled).
Part (c)(iii) — 4 Marks

Explaining Experimental vs. Theoretical Lattice Energies

MgF₂ vs. MgI₂: Polarization, Covalent Character, and Ionic Radius

✅ Full-Mark Response Structure (4 Points)

  1. MgF₂ comparison: The experimental (−2957) and theoretical (−2913) lattice energies for MgF₂ are very close / similar, which means MgF₂ is nearly 100% ionic (almost no covalent character).
  2. MgI₂ comparison: The experimental (−2327) and theoretical (−1944) lattice energies for MgI₂ show a significant discrepancy; this indicates MgI₂ has substantial covalent character.
  3. Polarisability reason: The iodide ion (I⁻) has a larger ionic radius than the fluoride ion (F⁻), so the outer electron cloud of I⁻ is much more polarisable (easily distorted by the high charge density of Mg²⁺).
  4. Magnitude comparison: Both values for MgF₂ are significantly more negative (more exothermic) than for MgI₂ because F⁻ is smaller than I⁻, resulting in a smaller inter-ionic distance and stronger ionic bonding in MgF₂.

💡 The Theoretical (Ionic) Model

  • Theoretical Model assumes: Perfectly spherical, non-polarisable point charges with 100% electrostatic attraction.
  • Experimental (Born–Haber) value: Reflects real-world bonding. If experimental is more negative than theoretical, there is orbital overlap (covalent character), which adds extra bonding strength!
  • Fajan's Rules: Polarization is favoured by small, highly charged cations (like Mg²⁺) and large, easily distorted anions (like I⁻).

❌ Critical Examiner Pitfalls to Avoid

  • Confusing atoms with ions: Writing "iodine atom is larger" instead of "iodide ion (I⁻) is larger / has larger radius" loses the mark immediately! Always specify ions.
  • Saying Mg²⁺ is polarised: Mg²⁺ does the polarising (it has high polarising power); the anion (I⁻) gets polarised (it is polarisable).
  • Vague comparisons: Never just state "values are different". You must specify which compound has covalent character and explain why the forces in MgF₂ are stronger overall.
Mark Scheme Breakdown (4 Marks):
• Point 1: Experimental and theoretical values for MgF₂ are close → (nearly) 100% ionic.
• Point 2: Values for MgI₂ are significantly different → has covalent character.
• Point 3: I⁻ is more polarisable than F⁻ because I⁻ has a larger ionic radius.
• Point 4: MgF₂ lattice energy is more negative / exothermic because F⁻ is smaller than I⁻, leading to stronger ionic bonds.

Topics

Physical Chemistry · Topic 2: Bonding and Structure · Topic 13: Energetics II

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