OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 16

10 marks · Medium difficulty · Structured Questions

Complete a Born-Haber cycle for barium iodide, calculate its lattice enthalpy using thermodynamic data, and explain trends in the first and second ionisation energies of Group 2 elements.

Practise this question

Question

A multipart chemistry exam question about energy changes. Part (a) provides a table of energy terms and values for barium iodide, an incomplete Born-Haber cycle diagram to complete with species and state symbols, and a calculation for the lattice enthalpy. Part (b) provides a table of first and second ionisation energies for Mg and Sr, followed by two explanation questions comparing ionisation energies.
Question text

16 This question is about energy changes.

(a) Lattice enthalpies can be determined indirectly using Born-Haber cycles.

The table below shows the energy changes that are needed to determine the lattice enthalpy

of barium iodide, BaI2.

Energy term Energy change / kJ mol–1

formation of barium iodide –602

1st electron affinity of iodine –296

1st ionisation energy of barium +503

2nd ionisation energy of barium +965

atomisation of iodine +107

atomisation of barium +180

(i) The diagram below shows an incomplete Born-Haber cycle that can be used to

calculate the lattice enthalpy of barium iodide.

On the dotted lines, add the species present, including state symbols.

Ba2+(g) + 2I(g) + 2e–

Ba(s) + 2I(g)

BaI2(s)

[4]

(ii) Calculate the lattice enthalpy of barium iodide.

lattice enthalpy = … kJ mol–1 [2]

(b) The first and second ionisation energies of magnesium, Mg, and strontium, Sr, in Group 2

are given in the table below.

First ionisation energy Second ionisation energy

Element –1 –1

/ kJ mol / kJ mol

Mg +738 +1451

Sr +550 +1064

• Explain why the first ionisation energy of Mg is greater than the first

ionisation energy of Sr.

• Explain why the second ionisation energy of Sr is greater than the first

ionisation energy of Sr.

… [4]

Mark scheme

Show the mark scheme The mark scheme displays the completed Born-Haber cycle with correct species and state symbols on the dotted lines, the calculation steps leading to -1872 kJ mol-1 for the lattice enthalpy, and marking points explaining the trends in first and second ionisation energies referencing atomic radius, shielding, and nuclear attraction.

AO

Question Answer Marks Guidance

element

16 (a) (i) 4 AO1.2

✓

✓ ✓

✓

AO

Question Answer 10 Marks Guidance

element

(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 2 AO2.2 ALLOW for 1 mark +1872 (wrong sign on answer)

If answer = –1872 award 2 marks 2

----------------------------------------------------------------- Common errors for 1 mark

H lattice = –3056 (–296 x 2 instead of 296 x2)

–2168 (296 x 1 instead of 296 x2)

2(+ 296) – 965 – 503 – 180 + 2(–107)– 602 ✓ –1765 (–107 x 1 instead of –107 x 2)

–1512 (180 instead of –180)

H lattice = –1872 (kJ mol–1) ✓

–1444 (107 x 2 instead of –107 x 2)

– 866 (503 instead of –503)

– 668 (602 instead of –602)

+58 (965 instead of -965)

For other answers, check for a single transcription

error or calculation error which could merit 1 mark if all

values have been used.

DO NOT ALLOW any answer which involves two errors

(b) 4 ORA throughout

Ist IE of Mg and Sr ALLOW going down the group for comparison of Mg/Sr

(Mg) removes electron from shell closer to the nucleus / Assume ‘it’ means Mg

smaller atomic radius ✓ AO1.1 ALLOW (Mg) fewer shells

ALLOW less shielding

AO1.2 ALLOW removal of electron from 3s rather than 5s

Greater nuclear attraction (between atom and outer ALLOW Greater attraction between nucleus (and

electron) ✓ outer electron)

AO1.1

AO1.2

11 AO

element

2nd/1st IE of Sr

nd ALLOW Sr+ ion smaller (than Sr atom)

2 electron removed from cation/positively charged ion

OR

proton:electron ratio (in (1)+ ion) is greater (than in atom)

✓

Greater nuclear attraction / attraction between ion (and ALLOW same number of protons/nuclear charge

outer electron)✓ attracting one fewer electron

IGNORE repulsion between electrons in the s orbital

IGNORE shielding

How to answer it

Energy Changes, Born-Haber Cycles & Ionisation Energy

What this question tests

This question assesses your understanding of enthalpy changes, specifically constructing and using Born-Haber cycles for ionic lattices (like BaI₂), and applying periodic trends in ionisation energies down Group 2 and between successive ionisation levels.

Question 16(a)(i)

Completing a Born-Haber Cycle

✅ Correct Answers (Dotted Lines)

  • Bottom left line: Ba(s) + I₂(s)
  • Lower middle line: Ba(g) + 2I(g)
  • Middle-upper left line: Ba⁺(g) + 2I(g) + e⁻
  • Right side intermediate line: Ba²⁺(g) + 2I⁻(g)

💡 Key Knowledge

A Born-Haber cycle applies Hess's Law. You must build the species step-by-step from elements in their standard states up to gaseous ions:

  • Atomisation of barium and iodine.
  • 1st and 2nd ionisation energies of barium (removing electrons one by one).
  • Electron affinity of iodine (adding electrons to form gaseous iodide ions). Keep stoichiometry ( 2I ) in mind!

🧠 Exam Technique

Always include state symbols (s) , (l) , (g) , and (aq) . Ensure electron numbers balance correctly at each ionisation stage (e.g., adding + e⁻ or + 2e⁻ ).

Mark breakdown: 4 marks available (1 mark for each correctly completed horizontal energy level with proper species and state symbols).
Question 16(a)(ii)

Calculating Lattice Enthalpy

📐 Step-by-Step Calculation

  1. Identify the cycle route: Enthalpy of formation equals the sum of atomisation, ionisation, electron affinity, and lattice enthalpy. Rearrange to solve for lattice enthalpy ( ΔH_lattice ).
  2. Substitute values with correct multipliers:
    ΔH_lattice = ΔH_formation - (ΔH_atom[Ba] + 1st IE[Ba] + 2nd IE[Ba] + 2 × ΔH_atom[I] + 2 × 1st EA[I])
  3. Plug in numbers:
    -602 - (+180 + 503 + 965 + 2(+107) + 2(-296))
  4. Evaluate carefully:
    -602 - (180 + 503 + 965 + 214 - 592) = -602 - 1270 = -1872 kJ mol⁻¹

❌ Common Calculation Traps

  • Stoichiometry errors: Forgetting to multiply the enthalpy of atomisation of iodine and 1st electron affinity of iodine by 2, because formula BaI₂ contains two moles of iodine atoms/ions per mole of compound.
  • Sign errors: Missing negative signs on electron affinities or getting confused rearranging Hess's Law loops.
Mark breakdown: 2 marks total (1 mark for correct working/method, 1 mark for final correct value with units: -1872 kJ mol⁻¹ ).
Question 16(b)

Explaining Ionisation Energy Trends

💡 Part 1: 1st IE of Mg vs Sr

Answer: Mg has a smaller atomic radius / fewer electron shells than Sr. Therefore, the outer electron in Mg is closer to the nucleus and experiences greater nuclear attraction.

Examiner note: Watch out for ORA (Onions/Other Reverse Arguments) — make sure your comparison explicitly targets magnesium versus strontium.

💡 Part 2: 2nd IE of Sr vs 1st IE of Sr

Answer: The second ionisation energy involves removing an electron from a positively charged ion ( Sr⁺ ) rather than a neutral atom. The proton-to-electron ratio is greater, meaning the remaining electrons are pulled closer with greater nuclear attraction.

🧠 Top-Level Exam Phrasing

To secure full marks across ionisation energy explanations, always structure your answer using this winning triad:

  1. Distance: Atomic/ionic radius or number of shells.
  2. Shielding: Inner shell repulsion (if comparing different shell levels).
  3. Nuclear charge: Attraction between the nucleus and the outer electron.
Mark breakdown: 4 marks total (2 marks for the Mg vs Sr comparison, 2 marks for the successive Sr ionisation comparison).

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.1 The periodic table · 3.2 Physical chemistry · 5.2 Energy

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.