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

11 marks · Medium difficulty · Extended Writing

Identify Group 3 elements from successive ionisation energies, write the equation for the second ionisation of beryllium, explain trends in ionisation energies for Group 2 elements, and relate these to reactivity with chlorine.

Practise this question

Question

Question text

9 This question is about ionisation energy.

(a) The first four ionisation energies, in kJ mol–1, of four elements are shown.

Which element is in Group 3?

(1)

A 738 1451 7753 10541

B 578 1817 2745 11578

C 789 1577 3232 4356

D 1012 1903 2912 4957

(b) The second ionisation energy of beryllium is more endothermic than the

first ionisation energy.

(i) Write an equation for the second ionisation of beryllium.

Include state symbols.

(2)

*(ii) Explain how the nuclear structure and the electronic structure of the Group 2

elements affect ionisation energies.

Include:

• an explanation of the trend of the first ionisation energies down the group

• a comparison of the first two ionisation energies for an element.

(6)

… *P71926A02228*

… *P71926A02328*

(iii)The value of the sum of the first and second ionisation energies changes on*P71926A02428*

descending Group 2.

Explain how this changing value might contribute to the relative reactivity of

the elements with chlorine.

(2)

(Total for Question 9 = 11 marks)

Mark scheme

Show the mark scheme

Question

Answer Mark

Number

9(a) (1)

The only correct answer is B (578 1817 2745 11578)

A is not correct because there is a big jump between the second and third ionisation energies so it is in Group 2

C is not correct because there is no big jump between the third and fourth ionisation energies

D is not correct because there is no big jump between the third and fourth ionisation energies

Question Answer Additional Guidance Mark

Number

9(b)(i) An answer that makes reference to the following Example of equation (2)

points:

• equation (1) Be+(g) → Be2+(g) + e‒

/ Be+(g) - e‒ → Be2+(g)

• state symbols (1) Dependent on M1 or near miss e.g.

Be(g) → Be2+(g) + 2e‒

Be(g) → Be+(g) + e‒

Question

Acceptable Answer Additional Guidance Mark

Number

*9(b)(ii) This question assesses a student’s ability to show a Guidance on how the mark scheme should be 6

coherent and logically structured answer with linkages and fully- applied:

sustained reasoning.

The mark for indicative content should be

Marks are awarded for indicative content and for how the answer added to the mark for lines of reasoning.

is structured and shows lines of reasoning.

For example, an answer with five indicative

The following table shows how the marks should be marking points, which is partially structured

awarded for indicative content. with some linkages and lines of reasoning,

scores 4 marks (3 marks for indicative

Number of indicative marking Number of marks awarded for content and 1 mark for partial structure and

points seen in answer indicative marking points some linkages and lines of reasoning).

5–4 3 If there are no linkages between points, the

3–2 2 same five indicative marking points would

11 yield an overall score of 3 marks (3 marks for

00 indicative content and no marks for linkages).

The following table shows how the marks should be

awarded for structure and lines of reasoning.

Number of marks awarded

for structure of answer and In general it would be expected

sustained line of reasoning that 5 or 6 indicative points would get 2 reasoning marks, and 3

Answer shows a coherent and logical 2 or 4 indicative points would get 1 mark

structure with linkages and fully for reasoning, and 0, 1 or 2

sustained lines of reasoning indicative points would score zero marks for reasoning.

demonstrated throughout.

Answer is partially structured with 1

some linkages and lines of reasoning. Reasoning marks may be reduced for extra incorrect chemistry

Answer has no linkages between 0

points and is unstructured.

Question

Acceptable Answer Additional Guidance Mark

Number

*9(b)(ii)

contd Indicative content:

• IP1 trend

ionisation energy becomes less endothermic on descending the group Ignore more exothermic

Allow decreases

• IP2 distance

electrons are removed from a shell that is further from the nucleus Allow increasing atomic radius

Ignore atoms get larger/bigger

• IP3 shielding

electrons more shielded / experience greater repulsion from inner Ignore there are more shells

electrons

• IP4 nuclear charge

(despite the) nuclear charge / number of protons being greater (down the

group)

• IP5 position of electrons

(the second ionisation energy is more endothermic because) the electron is Allow energy level

removed from the same shell/subshell

• IP6 repulsion

(for the second ionisation energy) there is less repulsion by electrons / Allow repulsion from electrons in the

shielding (than in the first) same orbital

Allow the electron is being removed from

a positive ion

Allow the same number of protons attracts

one fewer electrons

Number

9(b)(iii) An explanation that makes reference to the (2)

following points:

• (the sum of the first two ionisation Allow smaller for decrease

energies) decreases down the group so the Allow less energy is required for electrons to be lost so reactions

reactivity increases (with chlorine) (1) might be faster down the group

• because Group 2 elements form ionic

bonds / react by losing their (outer 2)

electrons (1)

How to answer it

Ionisation Energies & Group 2 Periodic Trends

📋 What this question tests

This question assesses your understanding of successive ionisation energies to identify group numbers, writing exact equations with state symbols for specific ionisation steps, constructing a coherent 6-mark explanation for atomic trends down Group 2 versus within an atom, and linking ionisation enthalpy values to chemical reactivity.

Part (a) · 1 Mark

Deducing Group Number from Successive Ionisation Energies

Identifying an element in Group 3 using successive values in kJ mol⁻¹

✅ Correct Answer

B (578, 1817, 2745, 11578)

1 Mark: Correct letter selected.

💡 Key Knowledge

A huge jump in ionisation energy occurs when an electron is removed from an inner principal quantum shell that is closer to the nucleus and experiences significantly less shielding.

For B, the jump occurs between the 3rd and 4th ionisation energies (2745 → 11578 kJ mol⁻¹), proving there are 3 electrons in the outermost shell.

❌ Common Errors & Distractor Analysis

A (738, 1451, 7753, 10541): Big jump is between 2nd and 3rd → Group 2 element (Magnesium).
C & D: Show no large jump between 3rd and 4th ionisations (both represent elements further right in the Periodic Table, e.g., Group 4 or 5).

Part (b)(i) · 2 Marks

Second Ionisation Energy Equation

Writing the exact equation for the second ionisation of beryllium

✅ Correct Answer

Be⁺(g) → Be²⁺(g) + e⁻

Also accepted: Be⁺(g) - e⁻ → Be²⁺(g)

Mark 1: Correct species and balancing ( Be⁺ → Be²⁺ + e⁻ )
Mark 2: Correct state symbols: (g) on both beryllium species.

❌ Common Errors

Starting from neutral Be(g): Writing Be(g) → Be²⁺(g) + 2e⁻ . This represents the sum of the first two ionisation energies, not the second ionisation energy.
Missing or incorrect state symbols: Omitting (g) loses Mark 2 entirely. State symbols are strictly assessed here.
Wrong charges: Writing Be → Be⁺ + e⁻ (this is the 1st IE).

Part (b)(ii) · 6 Marks (Extended Response)

Explaining Ionisation Energy Trends

Nuclear and electronic structure effects: down Group 2 vs. successive ionisations

🧠 Exam Technique: Structuring a 6-Mark Question

This is a level-of-response question marked out of 6 (4 marks for indicative scientific content + 2 marks for coherent reasoning and linkages). You must structure your answer into two clear halves as instructed by the prompt:

💡 Theme 1: 1st IE Trend Down Group 2 (3 Points)

Trend: First ionisation energy decreases (becomes less endothermic) descending Group 2.
Atomic radius: Outer electrons occupy shells further from the nucleus (atomic radius increases).
Shielding: There are more inner shells, so outer electrons experience greater shielding / repulsion from inner electrons.
Nuclear charge comparison: Although nuclear charge increases (more protons), the increase in distance and shielding outweighs this effect.

💡 Theme 2: 1st vs. 2nd IE in an Element (2 Points)

Electron position: The second electron is removed from the same principal shell / subshell (same energy level).
Electron repulsion / effective attraction: In the positive ion ( Be⁺ ), there is less electron–electron repulsion (or fewer electrons being attracted by the same number of protons). The electron is held more tightly, making the 2nd IE more endothermic.

❌ Common Misconceptions & Examiner Traps

Vague sizing: Saying "atoms get bigger" rather than explicitly stating that the atomic radius increases or outer electrons are further from the nucleus.
Shell counting without explanation: Writing simply "more shells" instead of identifying increased electron shielding / repulsion from inner shells.
Ignoring nuclear charge: High-scoring answers explicitly note that nuclear charge increases down the group, but explain that increased shielding and distance outweigh it.

✅ Mark Scheme Breakdown (Indicative Content)

Indicative Points Met	Content Marks
5 – 6 points	4 marks
3 – 4 points	3 marks
1 – 2 points	2 marks

Add +2 marks for a fully sustained, logically structured line of reasoning with clear links; +1 mark for partially structured.

Part (b)(iii) · 2 Marks

Linking Ionisation Energy to Group 2 Reactivity

Explaining how the sum of 1st and 2nd IEs affects reactivity with chlorine

✅ Correct Answer

Point 1: The sum of the first two ionisation energies decreases down the group, so reactivity increases down the group (less energy required to lose electrons / reaction occurs more readily).
Point 2: Group 2 elements react by losing their two outer electrons to form 2+ ions (forming ionic bonds with chlorine).

Mark 1: Sum decreases → reactivity increases / less energy needed to remove 2 electrons.
Mark 2: Explaining that Group 2 elements react by losing 2 valence electrons / forming ionic bonds.

🧠 Exam Technique: Making the Chemical Connection

Always connect the physical property (ionisation energy) to the chemical mechanism of the reaction:

M(s) + Cl₂(g) → M²⁺(Cl⁻)₂(s)

Because the reaction involves forming M²⁺ ions, it requires both the 1st and 2nd electrons to be removed. When this combined energy barrier is lower, the activation energy is generally lower and the reaction is more energetically favourable.

Topics

Physical Chemistry · Inorganic Chemistry · Topic 1: Atomic Structure and the Periodic Table · Topic 4: Inorganic Chemistry and the Periodic Table

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