Edexcel A-Level Chemistry Paper 1, November 2021: Question 9

12 marks · Hard difficulty · Calculations

Analyze lattice energies of barium and magnesium iodide, complete a Born-Haber cycle for copper(II) oxide, calculate its lattice energy, and determine the lattice energy of calcium bromide using an enthalpy cycle.

Practise this question

Question

The question presents a multipart problem about lattice energies. Part (a) includes a table comparing theoretical and experimental lattice energy values for magnesium iodide and barium iodide, followed by questions asking to deduce bonding from close similarity and explain differences due to polarization. Part (b) provides a table of Born-Haber cycle data for copper(II) oxide, requiring the student to complete a Born-Haber diagram with species and state symbols, and calculate the lattice energy. Part (c) provides an energy cycle linking gaseous ions, aqueous ions, and an ionic solid, along with enthalpy of solution and hydration data, to calculate the lattice energy of calcium bromide.
Question text

9 This question is about lattice energies.

(a) The table shows the theoretical and experimental lattice energy values of

two compounds.

Theoretical lattice energy Experimental lattice energy

Compound –1 –1

/ kJ mol / kJ mol

magnesium iodide –1944 –2327

barium iodide –1831 –1877

(i) State what can be deduced by the close similarity of the lattice energy values

for barium iodide.

(1)

(ii) Explain why there is a significant difference in the lattice energy values for

magnesium iodide.

(4)

(b) Data for the Born-Haber cycle for copper(II) oxide are given in the table.

Label Energy change Value / kJ mol–1

A standard enthalpy change of atomisation of copper*P65463A02328* +338

B standard enthalpy change of atomisation of oxygen +249

C sum of first and second ionisation energies of copper +2704

D first electron affinity of oxygen –141

E second electron affinity of oxygen +798

F standard enthalpy change of formation of copper(II) oxide –157

(i) Complete the diagram of the Born-Haber cycle for copper(II) oxide.

Include labels of enthalpy changes with arrows indicating the direction of

change, and the respective species with state symbols.

(4)

Cu(s) + ½O2(g)

F

CuO(s)

(ii) Calculate the lattice energy of copper(II) oxide.

(1)

*P65463A02428*

(c) A different energy cycle can be used to calculate lattice energy.

Ca2+(g) + 2Br–(g) CaBr (s)

Ca2+(aq) + 2Br–(aq)

Enthalpy change Value / kJ mol–1

enthalpy change of solution of CaBr2 –73

enthalpy change of hydration of Ca2+ –1577

enthalpy change of hydration of Br– –336

Calculate the lattice energy of calcium bromide.

(2)

(Total for Question 9 = 12 marks)

Mark scheme

Show the mark scheme The mark scheme provides detailed answers for each subpart of Question 9. For (a)(i), it awards 1 mark for stating barium iodide is almost 100 percent ionic. For (a)(ii), it allocates 4 marks for explaining magnesium ion's small size/high charge, iodide's large radius, polarization of iodide, and resulting covalent character. For (b)(i), 4 marks are given for completing the Born-Haber cycle including species, state symbols, energy values, and arrows. For (b)(ii), 1 mark is for the calculation resulting in -4105 kJ mol-1. For (c), 2 marks are awarded for applying Hess's law and calculating -2176 kJ mol-1.

Question

Answer Additional Guidance Mark

Number

9(a)(i) An answer that includes (1)

• barium iodide has (almost) 100% ionic (bonds) Allow small amount of/zero covalency

Ignore just it is ‘ionic’

Question

Answer Additional Guidance Mark

Number

9(a)(ii) An answer that includes (4)

• the magnesium ion is small and highly charged (1) Allow magnesium ion has a high charge

density

• the iodide ion has a large ionic radius (1) Allow iodide ion has a much larger radius

Ignore reference to atomic radius

• the iodide ion is polarised by the magnesium ion (1) ALLOW description of polarisation such as

distortion of the iodide electron cloud by the

magnesium ion

• (so) the bonding in magnesium iodide has (partial) covalent Do not award magnesium iodide is covalent

character (which is why the lattice energy values are different) (1) Do not award ‘MgI’

Penalise once only reference to

magnesium/iodine/iodide without ‘ion’ in

marking points 1 to 3

Question

Answer Additional Guidance Mark

Number

9(b)(i) An answer that includes Example of Born-Haber cycle and calculation (4)

• species on lines (1) Allow omission of electrons but if included then must be correct

• state symbols (1)

• energy changes / values (1)

• arrows indicating direction (1) A and B can be drawn in either order or A then C followed by B

Exemplar cycle:

Each different species error can be penalised so four different species

errors scores (0)

Question

Answer Additional Guidance Mark

Number

9(b)(ii) • calculation of lattice energy LE= −4105 (kJ mol-1) (1)

Question

Answer Additional Guidance Mark

Number

9(c) Example of calculation (2)

• application of Hess’s law (1) LE = (−1577 + (2 × −336) – (−73)) =

= −2176 kJ mol−1

• evaluation of lattice energy (1)

Final answer without working scores (2)

(+) 2176 kJ mol−1 scores (1) for TE on

incorrect application of Hess’s law

−1840 kJ mol−1 scores (1) for use of

single −336 instead of double

(Total Question 9 = 12 marks)

How to answer it

Lattice Energies & Born-Haber Cycles Study Guide

Edexcel A-Level Chemistry — Total Marks: 12

What this question tests

This comprehensive question assesses your understanding of ionic bonding models versus reality. It tests your ability to interpret discrepancies between theoretical and experimental lattice energy values using polarization concepts, construct and label multi-step Born-Haber cycles with correct species and state symbols, and apply Hess's Law to energy cycles involving solution and hydration enthalpies.

Part (a)(i): Barium Iodide Lattice Energies

State what can be deduced by the close similarity of the lattice energy values for barium iodide.

✅ Correct Answer

  • Barium iodide has (almost) 100% ionic bonding.

💡 Key Knowledge

When theoretical lattice energy (calculated assuming a perfect point-charge spherical ion model) matches experimental lattice energy closely, it proves there is negligible covalent character.

Mark allocation: 1 mark for stating the bonding is nearly 100% ionic.

Part (a)(ii): Magnesium Iodide Discrepancy

Explain why there is a significant difference in the lattice energy values for magnesium iodide.

✅ Correct Answer

  • The magnesium ion ( Mg²⁺ ) is small and has a high charge density.
  • The iodide ion ( I⁻ ) has a large ionic radius.
  • The large iodide ion is heavily polarized (distorted electron cloud) by the small, highly charged magnesium ion.
  • This introduces significant partial covalent character, making the experimental lattice energy more exothermic than the theoretical value.

❌ Common Errors

  • Saying "magnesium iodide is covalent" (it is predominantly ionic with covalent character, not fully covalent).
  • Forgetting to use the word ion when describing the species ( Mg instead of Mg²⁺ ).
Mark allocation: 4 marks total (1 mark per detailed bullet point covering ion size/charge, polarization, and covalent character).

Part (b)(i): Born-Haber Cycle Construction

Complete the diagram of the Born-Haber cycle for copper(II) oxide.

🧠 Exam Technique & Steps

  • Species on lines: Ensure you build up step-by-step from elements to gaseous atoms, ions, and back to the lattice.
  • State symbols: Must be included on every species (e.g., (s) , (g) ).
  • Energy changes & values: Clearly label arrows with letters/values provided in the table (A through E).
  • Direction of arrows: Point upwards for endothermic changes (atomization, ionization, electron affinity steps that absorb energy) and downwards for exothermic changes.

❌ Common Errors

  • Missing electrons ( 2e⁻ ) when adding ionisation energies or electron affinities.
  • Failing to include state symbols, which immediately caps your marks.
Mark allocation: 4 marks (1 for species, 1 for state symbols, 1 for energy changes/values, 1 for correct arrow directions).

Part (b)(ii): Copper(II) Oxide Lattice Energy Calculation

Calculate the lattice energy of copper(II) oxide.

📐 Calculation Breakdown

Using the data table values:

  • A (atomisation of Cu) = +338
  • B (atomisation of O) = +249
  • C (1st + 2nd IE of Cu) = +2704
  • D (1st EA of O) = -141
  • E (2nd EA of O) = +798
  • F (formation of CuO) = -157

Rearranging the Born-Haber cycle for Lattice Energy (LE):

LE = F - (A + B + C + D + E)

LE = -157 - (338 + 249 + 2704 - 141 + 798)

LE = -157 - 3948 = -4105 kJ mol⁻¹

Mark allocation: 1 mark for the correct final numerical answer with units.

Part (c): Calcium Bromide Solution/Hydration Cycle

Calculate the lattice energy of calcium bromide.

📐 Step-by-Step Hess's Law Calculation

Step 1: Set up the Hess's Law loop vector equation

From the provided cycle:

ΔH_solution = ΔH_lattice + ΔH_hydration(Ca²⁺) + 2 × ΔH_hydration(Br⁻)

Step 2: Rearrange to solve for Lattice Energy

ΔH_lattice = ΔH_solution - [ΔH_hydration(Ca²⁺) + 2 × ΔH_hydration(Br⁻)]

Step 3: Substitute values carefully

ΔH_lattice = (-73) - [(-1577) + 2 × (-336)]

ΔH_lattice = -73 - [-1577 - 672]

ΔH_lattice = -73 - (-2249)

ΔH_lattice = -73 + 2249 = -2176 kJ mol⁻¹

❌ Common Calculation Traps

  • Stoichiometry multiplier: Forgetting to multiply the bromide hydration enthalpy by 2 ( CaBr₂ releases two bromide ions per mole). This yields the common error value of -1840 kJ mol⁻¹ (scoring only 1 mark).
  • Sign errors: Incorrectly combining negative signs during subtraction steps.
Mark allocation: 2 marks (1 mark for correct application of Hess's Law / working, 1 mark for correct evaluation). ECF applies if stoichiometry was missed.

Topics

Physical Chemistry · Topic 13: Energetics II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.