Edexcel A-Level Chemistry Paper 1, June 2024: Question 2

7 marks · Medium difficulty · Open Response

Explain the group trend in first electron affinities of halogens and identify elements from isoelectronic ions and successive ionisation energies data.

Practise this question

Question

Question text

2 This question is about the formation of ions.

(a) Explain the trend in the values of the first electron affinities of the

elements shown.

(4)

Element First electron affinity / kJ mol−1

chlorine −349

bromine −325

iodine −295

(b) Which of these isoelectronic ions has the smallest ionic radius?

(1)

A S2−

B Cl−

C K+

D Ca2+

The letters do not refer to the symbols of the elements.

Element Successive ionisation energies / kJ mol−1

A 578 1817 2745 11578 14831

B *P76895A0328*653159229874740 6686

C 738 1451 7733 10541 13629

D 1086 2353 4621 6223 37832

(i) Which element in the table could be in Group 4?

(1)

(ii) Which element in the table could be described as an s-block element?

(1)

(Total for Question 2 = 7 marks)

Mark scheme

Show the mark scheme

Question

Acceptable Answer Additional Guidance Mark

Number

2(a) An explanation that makes reference to the following points: Allow reverse arguments up the group / (4)

comparison of specific elements.

Penalise ‘losing an electron’ or incorrect

reference to oxidation once.

• electron affinity becomes less negative / less exothermic / (1) Allow the electron affinity becomes more

less energy is released going from chlorine to iodine / down positive / more endothermic going from

the group chlorine to iodine

Ignore electron affinity increases /

decreases

Do not award requires / produces energy

• atomic radius increases / number of shells increases /

increased distance between the nucleus and (1) Do not award any reference to ions, ionic

the outer / valence electron(s) radius, charge or charge density for M2

and unless point is clearly made that ions are

there is less attraction between the nucleus and the being formed from atoms

incoming / added electron / valence electron(s)

• (there is) increased shielding (from inner electron shells) (1) Allow there is an increase in repulsion

between the (inner) electron shells and the

incoming electron(s)

• (increased) shielding outweighs the effect of increasing (1)

nuclear charge

(increased) repulsion between the (inner) electron shells and

the incoming electron outweighs the effect of increasing

nuclear charge

(increased) distance of the outer shell / energy level

outweighs the effect of increasing nuclear charge

Question

Answer Mark

Number

2(b) The only correct answer is D (Ca2+) (1)

A is not correct because S2– has fewer protons than Ca2+ and has the largest ionic radius of those listed

B is not correct because Cl− has fewer protons than Ca2+ and has the second largest ionic radius

C is not correct because K+ has fewer protons than Ca2+ and has the second smallest ionic radius

Question

Answer Mark

Number

2(c)(i) The only correct answer is D (1086, 2353, 4621, 6223, 37832) (1)

A is not correct because these are the successive ionisation energies of a Group 3 element

B is not correct because these are the successive ionisation energies of a transition element

C is not correct because these are the successive ionisation energies of a Group 2 element

Question

Answer Mark

Number

2(c)(ii) The only correct answer is C (738, 1451, 7733, 10541, 13629) (1)

A is not correct because these are the successive ionisation energies of a p-block element

B is not correct because these are the successive ionisation energies of a d-block element

D is not correct because these are the successive ionisation energies of a p-block element

(Total for Question 2 = 7 marks)

How to answer it

Ion Formation, Electron Affinities & Ionisation Energies

📋 What this question tests

This question evaluates your foundational knowledge of atomic structure and periodic trends:

Explaining trends in first electron affinity down Group 7 using atomic radius, nuclear charge, and shielding.
Comparing the ionic radii of isoelectronic ions by examining effective nuclear charge (proton number).
Deducing periodic groups and blocks from successive ionisation energy (IE) data by identifying large jumps between electron shells.

Part (a) — 4 Marks

Trend in First Electron Affinities Down Group 7

Explaining why EA becomes less negative from Cl to I

✅ Model Answer (Mark Scheme Breakdown)

Trend: First electron affinity becomes less negative / less exothermic (less energy is released) from chlorine to iodine / down the group. [1 mark]
Atomic Radius: The atomic radius increases / number of electron shells increases, so the distance between the nucleus and the incoming electron increases, resulting in weaker attraction between the nucleus and the added electron. [1 mark]
Shielding: There is increased shielding of the incoming electron by inner electron shells. [1 mark]
Net Effect: The increased distance and increased shielding outweigh the effect of the increasing nuclear charge. [1 mark]

💡 Key Knowledge

First Electron Affinity definition:

X(g) + e⁻ → X⁻(g)

Because energy is released when a neutral gaseous atom attracts an electron, values are exothermic (negative). As you descend the group:

Number of inner shells increases → greater shielding.
Outer shell is further from the nucleus → weaker electrostatic pull on the incoming electron.
Although nuclear charge increases (more protons), shielding and distance dominate.

🧠 Exam Technique: Securing all 4 marks

Always state the direction of the trend first: Use terms like "less negative" or "less exothermic". Avoid vague terms like "decreases" because -295 is numerically greater than -349.
Reference the correct particle: Electron affinity is about adding an electron to an atom. Discuss the attraction to the incoming electron, not electrons being lost.
Include the "outweighs" statement: Whenever comparing opposing factors (increasing protons vs. increasing shielding/distance), always state which one wins!

❌ Common Errors to Avoid

Describing ionisation energy instead: Writing about "removing" or "losing" an electron scores zero for that explanation point.
Mentioning ionic radius: Do not talk about the radius of the halide ion (X⁻); electron affinity is an action performed on a neutral atom.
Saying "EA increases" or "decreases" without clarity: In chemistry, -349 to -295 can be confusing. State clearly: "becomes less exothermic" or "less energy released".

Total for (a): 4 marks — 1 mark for stating the trend; 1 mark for radius/attraction; 1 mark for shielding; 1 mark for shielding/distance outweighing nuclear charge.

Part (b) — 1 Mark

Isoelectronic Ions and Ionic Radius

Comparing S²⁻, Cl⁻, K⁺, and Ca²⁺

✅ Correct Answer

D : Ca²⁺

Ca²⁺ has the smallest ionic radius.

💡 Why Ca²⁺ is Smallest

Ion	Electrons	Protons (Z)	Relative Size
S²⁻	18	16	Largest (least pull)
Cl⁻	18	17	Large
K⁺	18	19	Small
Ca²⁺	18	20	Smallest (greatest pull)

All four ions are isoelectronic (each has 18 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ ). Ca²⁺ has the greatest number of protons (20), exerting the strongest electrostatic attraction on the electrons and pulling them closest to the nucleus.

Total for (b): 1 mark (Multiple Choice)

Part (c) — 2 Marks

Deducing Element Identity from Successive Ionisation Energies

Analysing Jumps in Successive Ionisation Energies

Element	1st IE (kJ mol⁻¹)	2nd IE (kJ mol⁻¹)	3rd IE (kJ mol⁻¹)	4th IE (kJ mol⁻¹)	5th IE (kJ mol⁻¹)	Group / Block
A	578	1817	2745	11 578	14 831	Group 3 (jump between 3rd & 4th)
B	653	1592	2987	4740	6686	Transition metal / d-block (steady increase)
C	738	1451	7733	10 541	13 629	Group 2 / s-block (jump between 2nd & 3rd)
D	1086	2353	4621	6223	37 832	Group 4 / p-block (jump between 4th & 5th)

(i) Element in Group 4

Correct Answer: D

Reasoning:

IE₁ to IE₄ rise steadily: 1086 → 2353 → 4621 → 6223 kJ mol⁻¹.
Huge jump between 4th and 5th IE (6223 → 37 832 kJ mol⁻¹), which is roughly a 6× increase!
This indicates that the 5th electron is removed from a shell closer to the nucleus. Therefore, there are 4 valence electrons, placing it in Group 4 (Group 14).

(ii) Element in the s-block

Correct Answer: C

Reasoning:

Large jump between 2nd and 3rd IE (1451 → 7733 kJ mol⁻¹).
This means there are 2 valence electrons in the outermost shell ( ns² configuration).
Having 2 valence electrons means it belongs to Group 2, which is part of the s-block (e.g., Magnesium).

🧠 Exam Technique: Spotting the Shell Boundary

Always calculate the ratio or difference between consecutive values:

For A: 11 578 / 2745 ≈ 4.2× (Jump after 3 → Group 3, p-block).
For B: Gradual increments; no massive shell jump in the first 5 values → d-block transition element.
For C: 7733 / 1451 ≈ 5.3× (Jump after 2 → Group 2, s-block).
For D: 37 832 / 6223 ≈ 6.1× (Jump after 4 → Group 4, p-block).

❌ Common Errors in Successive IE Questions

Confusing the jump number with the group number: the jump happens after all valence electrons are removed. If the jump is between 4 and 5, the element has 4 valence electrons (not 5).
Confusing Group 2 with p-block: remember s-block consists of Group 1 ( s¹ ) and Group 2 ( s² ).

Total for (c): 2 marks — (i) 1 mark for D; (ii) 1 mark for C.

Topics

Physical Chemistry · Topic 1: Atomic Structure and the Periodic Table · 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.