Edexcel A-Level Chemistry Paper 1, June 2023: Question 9

13 marks · Hard difficulty · Calculations

Complete a Born-Haber cycle for silver(I) oxide, explain electron affinities, calculate standard enthalpy of atomisation, state assumptions of theoretical lattice energy, and explain differences between experimental and theoretical lattice energies for silver chloride.

Practise this question

Question

Exam question about silver compounds, featuring a Born-Haber cycle diagram for silver(I) oxide with missing boxes to complete, followed by questions about electron affinities of oxygen, enthalpy of atomisation calculation, assumptions of theoretical lattice energy for silver chloride, and an explanation of the difference between experimental and theoretical lattice energies.
Question text

9 This question is about silver compounds.

(a) The diagram shows a Born–Haber cycle for the formation of silver(I) oxide, Ag2O.

All quantities are measured in kJ mol–1.

+ 2– –2969

2Ag (g) + O (g) Ag2O(s)

+798

+731×2

–141.1

–31

O(g)

Δ H d(Ag(s)) × 2 +249.2

at

2Ag(s) + ½O2(g)

(i) Complete the diagram by adding appropriate species and state symbols to

the empty boxes.

(2)

(ii) Explain why the value for the first electron affinity of oxygen is negative and

the value for the second electron affinity is positive.

(3)

(iii)Calculate a value for the standard enthalpy change of atomisation of silver,*P71912A02028*

Δ H d, using the Born–Haber cycle.

at

(3)

(b) Another silver compound is silver chloride, AgCl. Values for its lattice energy can

be found by experiment or by theoretical calculation.

Experimental lattice energy Theoretical lattice energy

Compound –1 –1

/ kJ mol / kJ mol

Silver chloride –905 –833

(i) Give two assumptions used in the model to calculate the

theoretical lattice energy.

(2)

(ii) Explain the difference in the two values for the lattice energy of silver chloride

by considering the possible bonding models.

(3)

(Total for Question 9 = 13 marks) 21

Mark scheme

Show the mark scheme Mark scheme showing the completed Born-Haber cycle with species and state symbols, marking points for explaining oxygen electron affinities, step-by-step calculation for the enthalpy of atomisation of silver, assumptions for theoretical lattice energy models, and explanations of covalent character and ion polarisation causing the difference between experimental and theoretical lattice energy values.

Question

Answer Additional Guidance Mark

Number

9(a)(i) (2)

Allow O1−(g)

Ignore + ½O2(g) in left hand box

Penalise missing state symbols once only

Penalise incorrect / missing charges in each

species

Penalise addition of electrons once only

Question

Answer Additional Guidance Mark

Number

9(a)(ii) An explanation that makes reference to the following points: Reference to removing electron(s) scores (3)

(0) overall

• (first electron affinity is exothermic / negative) due to attraction between

(positive) nucleus (in neutral atom) and (incoming) electron (1)

• (second electron affinity is endothermic / positive) due to repulsion (1)

• between negative ion and (incoming) electron (1) Ignore just ‘negative species’

Question

Answer Additional Guidance Mark

Number

9(a)(iii) (3)

• expression based on Hess cycle (1) 2Δ Ho + (1462 +249.2 – 141.1 + 798 – 2969) = −31

at

This can be subsumed in M2

2Δ Ho = −31 − (1462 +249.2 – 141.1 + 798 – 2969) = 569.9

• calculation of missing value from cycle (1) at

No TE on incorrect expression in M1

569.9 ÷ 2 = (+) 284.95 (kJ mol−1)

• calculation of Δ Ho (1)

at

TE on M2

Correct answer, with or without working, scores (3)

Ignore SF except 1 SF

Ignore units

Question

Answer Additional Guidance Mark

Number

9(b)(i) An answer that makes reference to two of the following points: Penalise use of atom / molecule / compound (2)

instead of ion once only

Ignore reference to standard conditions

• model assumes bonding is 100% / completely / purely ionic (1) Allow there is no covalent bonding / character

Allow no polarisation of ions / electron cloud

Allow electrons are fully transferred

Ignore almost 100% ionic

• the ions are (perfect) spheres (1) Allow no distortion of ions / electron cloud

Ignore circular

• the charge is distributed evenly across the ions / the ions are

Allow charge dispersed equally

point charges (1) Ignore ions with fixed charges

• ions are in contact with one another (1)

Question

Answer Additional guidance Mark

Number

9(b)(ii) Penalise use of incorrect ions in M1 only (3)

An answer that makes reference to the following points Ignore reference to electronegativity

Ignore silver chloride is a polar molecule

• as the silver ion polarises / distorts the chloride ion / electron cloud Do not award if incorrect charges on ions

(1) Do not award if silver ion is larger than chloride ion

• so the bonding (is ionic and) has some covalent character (1) Ignore just ‘bonding is not purely ionic’

Do not award covalent with ionic character

• which is stronger (than ionic alone so experimental value is more Allow ‘releases more energy when bond forms’ (than

exothermic / more negative) (1) ionic alone so experimental value is more exothermic /

more negative)

How to answer it

Born-Haber Cycles and Ionic Lattice Models

What this question tests

This question assesses your understanding of energetic cycles (Born-Haber cycles), definitions of electron affinity, enthalpy calculations using Hess's Law, and limitations of the perfect ionic model (including ion polarisation and covalent character).

Question 9 (a)(i) - Completing the Born-Haber Cycle

Filling in missing species and state symbols for Ag₂O

✅ Correct Answers

  • Lower empty box: 2Ag(g) + O(g)
  • Upper empty box: 2Ag⁺(g) + O⁻(g)

❌ Common Errors

  • Leaving out state symbols or writing incomplete states like (s) instead of (g) .
  • Failing to balance moles correctly (forgetting the coefficient 2 for silver species).
  • Adding stray electrons into the intermediate species boxes.
Mark Allocation: 2 marks total. 1 mark for the lower box species/states; 1 mark for the upper box species/states. Penalise missing state symbols once only across the whole diagram.

Question 9 (a)(ii) - Electron Affinities of Oxygen

Explaining why 1st EA is negative and 2nd EA is positive

💡 Key Knowledge

  • 1st Electron Affinity: Exothermic (negative) because energy is released when an incoming electron is attracted by the positive nucleus of the neutral oxygen atom.
  • 2nd Electron Affinity: Endothermic (positive) because energy must be put in to overcome the strong electrostatic repulsion between the negative ion ( O⁻(g) ) and the incoming electron.

🧠 Exam Technique

Use precise language. Always mention the interaction between the nucleus and the electron for the first step, and the repulsion between the negative ion/electron cloud and the incoming electron for the second step. Avoid vague terms like "negative species".

Mark Allocation: 3 marks. Point 1: 1st EA is exothermic / negative due to nucleus-electron attraction. Point 2: 2nd EA is endothermic / positive due to repulsion. Point 3: Repulsion is specifically between the negative ion and the incoming electron. *(Note: mentioning removal of electrons scores 0 overall).*

Question 9 (a)(iii) - Born-Haber Calculation

Calculating the standard enthalpy change of atomisation of silver

📐 Calculation Steps (Hess's Law)

  1. Set up the cycle equation:
    2ΔₐₜH°(Ag) + 1462 + 249.2 - 141.1 + 798 - 2969 = -31
  2. Sum the known energy terms:
    1462 + 249.2 - 141.1 + 798 - 2969 = -569.9 kJ mol⁻¹
  3. Rearrange to solve for 2ΔₐₜH°(Ag):
    2ΔₐₜH°(Ag) = -31 - (-569.9) = 538.9 *(Note: using diagram values: -31 - (-600.9) = 569.9 depending on route)*
  4. Divide by 2 to find standard atomisation of 1 mole of Ag:
    ΔₐₜH°(Ag) = 569.9 ÷ 2 = +284.95 kJ mol⁻¹ (accept 285).

❌ Common Calculation Traps

  • Forgetting to divide the final answer by 2 (since the equation contains 2Ag(s) ).
  • Sign errors when transposing values across the equals sign in Hess cycles.
Mark Allocation: 3 marks. 1 mark for correct Hess expression; 1 mark for calculating the missing cycle value; 1 mark for correct final division to get +284.95 kJ mol⁻¹ .

Question 9 (b)(i) - Theoretical Lattice Energy Assumptions

State two assumptions used in the theoretical model

💡 Key Knowledge

  • The model assumes the bonding is 100% / purely ionic with no covalent character.
  • The ions are treated as perfect, hard spheres.
  • Charge is distributed evenly across the ions (point charges).
  • Ions are assumed to be in direct contact with one another.

🧠 Exam Technique

Make sure you refer specifically to ions rather than atoms, molecules, or general compounds. Examiners heavily penalise references to "atoms" in this context.

Mark Allocation: 2 marks (1 mark per valid assumption, max 2).

Question 9 (b)(ii) - Lattice Energy Discrepancy

Explaining the difference between experimental (-905) and theoretical (-833) values

💡 Key Knowledge

  • The experimental lattice energy is more exothermic (more negative) than the theoretical value.
  • The small, polarizing silver ion ( Ag⁺ ) distorts the electron cloud of the chloride ion ( Cl⁻ ).
  • This introduces significant covalent character alongside ionic bonding.
  • Additional energy is released due to this covalent contribution, making the experimental value larger (more negative).

❌ Common Errors

  • Stating that silver chloride is a covalent molecule (it has covalent character, not pure covalency).
  • Getting ions mixed up or claiming the chloride ion polarises the silver ion.
Mark Allocation: 3 marks. Point 1: Silver ion polarises / distorts the chloride ion's electron cloud. Point 2: Bonding gains covalent character. Point 3: Experimental value is more exothermic / negative because extra energy is released when covalent bonding character is present.

Topics

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

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