AQA A-Level Chemistry AS Paper 1, June 2022: Question 11
1 mark · Easy difficulty · Multiple Choice
Identify an element given its first six ionisation energies.
Practise this questionQuestion
Question text
11 The first six ionisation energies, in kJ mol–1, of an element are:
1090, 2350, 4610, 6220, 37 800, 47 000
What is the element?
[1 mark]
A Boron
B Carbon
C Nitrogen
D Oxygen
Mark scheme
Show the mark scheme
11 B (AO3) 1 Carbon
How to answer it
Identifying an Element from Successive Ionisation Energies
This question assesses your ability to analyse successive ionisation energy data to deduce electronic configurations and group numbers in the Periodic Table (AO3 - analyse and interpret data).
- Recognising that a massive jump between successive ionisation energies indicates the removal of an electron from an inner, less shielded quantum shell closer to the nucleus.
- Counting the number of outer-shell (valence) electrons before the huge jump occurs.
- Matching the number of valence electrons to the correct Group and element.
Deducing the Element from the Data
AQA Chemistry A-Level • Section 3.1.1 Atomic Structure
📐 Step-by-Step Data Analysis
Look at the progression of the first 6 ionisation energies (in kJ mol⁻¹):
| 1ˢᵗ | 2ⁿᵈ | 3ʳᵈ | 4ᵗʰ | 5ᵗʰ | 6ᵗʰ |
|---|---|---|---|---|---|
| 1090 | 2350 | 4610 | 6220 | 37 800 | 47 000 |
- Calculate differences:
IE₁ to IE₂: +1260 kJ mol⁻¹ (steady increase)
IE₂ to IE₃: +2260 kJ mol⁻¹ (steady increase)
IE₃ to IE₄: +1610 kJ mol⁻¹ (steady increase)
IE₄ to IE₅: +31 580 kJ mol⁻¹ (a massive jump ×6 larger than IE₄!)
IE₅ to IE₆: +9200 kJ mol⁻¹ (steady increase) - Identify the valence shell: The first 4 electrons are removed with relatively small increases. The 5ᵗʰ electron requires drastically more energy because it is removed from an inner quantum shell (closer to the nucleus, experiences significantly less shielding).
- Deduce the group number: Since 4 electrons can be removed before the jump into the inner shell, the element has 4 valence electrons and belongs to Group 14 (Group 4).
- Match to the correct option:
Boron: Group 13 (3 outer electrons)
Carbon: Group 14 (4 outer electrons) → Correct
Nitrogen: Group 15 (5 outer electrons)
Oxygen: Group 16 (6 outer electrons)
✅ Correct Answer
B (Carbon)
• 1 mark awarded for selecting B.
• AO3 (Analysis & Evaluation): 1 mark total.
💡 Key Knowledge
- Successive ionisation energies always increase because electrons are removed from increasingly positive ions (greater effective nuclear attraction per remaining electron).
- A sharp, dramatic increase (often a 5–10 fold jump) pinpoints the boundary between two principle energy levels (quantum shells, n).
- Carbon electronic configuration: 1s² 2s² 2p² . The first 4 electrons removed come from the second shell ( n = 2 ); the 5ᵗʰ comes from the inner 1s orbital ( n = 1 ).
🧠 Exam Technique
- Quick Rule: The number of electrons before the huge jump equals the Group Number (in the old Group 1–8 naming) and the number of outer-shell electrons.
- Count the values: 1ˢᵗ, 2ⁿᵈ, 3ʳᵈ, 4ᵗʰ are all under 7000. The 5ᵗʰ jumps to ~38 000. Four values before the jump → 4 outer electrons → Group 4 (Carbon).
- Takes less than 20 seconds to answer if you look directly for the order-of-magnitude jump.
❌ Common Errors
- Off-by-one counting error: Selecting Nitrogen (Option C) by misinterpreting the jump at the 5ᵗʰ electron as meaning there are 5 valence electrons. Remember: the jump happens after the valence electrons have already been removed!
- Confusing Boron and Carbon: Forgetting the group number of Boron (Group 3/13) versus Carbon (Group 4/14). Always verify with the provided Periodic Table in the Data Booklet.
Topics
Physical Chemistry · 3.1.1 Atomic Structure
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.