AQA A-Level Chemistry AS Paper 1, June 2022: Question 12

1 mark · Medium difficulty · Multiple Choice

Identify which pair has an atom Y with a greater first ionisation energy than atom X based on their electron configurations.

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Question

Multiple choice question asking: 'In which pair is the first ionisation energy of atom Y greater than that of atom X?'. A table shows four options with electron configurations: Option A: X = 1s² 2s², Y = 1s² 2s² 2p¹; Option B: X = 1s² 2s² 2p³, Y = 1s² 2s² 2p⁴; Option C: X = 1s² 2s² 2p⁵, Y = 1s² 2s² 2p⁶; Option D: X = 1s² 2s² 2p⁶, Y = 1s² 2s² 2p⁶ 3s¹.
Question text

12 In which pair is the first ionisation energy of atom Y greater than that of atom X?

[1 mark]

Electron Electron

configuration configuration

of atom X of atom Y

A 1s22s2 1s22s22p1

B 1s22s22p3 1s22s22p4

C 1s22s22p5 1s22s22p6

D 1s22s22p6 1s22s22p63s1

Mark scheme

Show the mark scheme Mark scheme for question 12 showing the correct answer is C (AO1), worth 1 mark, with the electron configurations X = 1s² 2s² 2p⁵ and Y = 1s² 2s² 2p⁶.

12 C (AO1) 1 2 2 5 2 2 6

C 1s 2s 2p 1s 2s 2p

How to answer it

First Ionisation Energy Across Period 2

📌 What this question tests

This question evaluates your understanding of Period 2 periodic trends in first ionisation energy (IE₁), specifically requiring you to:

  • Identify elements from their full electron configurations.
  • Recall the general increase in IE₁ across a period due to increasing nuclear charge and similar shielding.
  • Recognise the two classic exceptions in Period 2 (Group 2 to Group 3, and Group 5 to Group 6).
  • Account for the sharp decrease in IE₁ when beginning a new quantum shell.

Question 12 Breakdown

Identifying the pair where IE₁(Y) > IE₁(X)

✅ Correct Answer: C

Option C:

Atom X: 1s²2s²2p⁵ (Fluorine, F)

Atom Y: 1s²2s²2p⁶ (Neon, Ne)

Neon has a higher first ionisation energy than Fluorine because it follows the standard general trend across Period 2: Neon has an extra proton (+10 vs +9), identical shielding (both have outer electrons in the second shell), and a smaller atomic radius. Therefore, the outermost electron is held more tightly.

🧠 Exam Technique: Decode the Elements First

Count the electrons or look at the valence subshell to immediately write down the identity of each element:

  • A: X = Be (Z=4), Y = B (Z=5)
  • B: X = N (Z=7), Y = O (Z=8)
  • C: X = F (Z=9), Y = Ne (Z=10)
  • D: X = Ne (Z=10), Y = Na (Z=11)

Once named, you can spot the anomalies instantly without getting overwhelmed by configurations.

📐 Option-by-Option Analysis

Pair Identity of X & Y Comparison of First IE Why?
A X = Beryllium (Be)
Y = Boron (B)
IE₁(Y) < IE₁(X)
(IE drops)
Boron's outer electron is in a higher-energy 2p subshell, which is further from the nucleus and partially shielded by the 2s² pair. Less energy is needed to remove it.
B X = Nitrogen (N)
Y = Oxygen (O)
IE₁(Y) < IE₁(X)
(IE drops)
Oxygen has a paired electron in one of its 2p orbitals ( 2p⁴ ). Inter-electron spin-pair repulsion makes this electron easier to remove than Nitrogen's singly occupied orbitals ( 2p³ ).
C X = Fluorine (F)
Y = Neon (Ne)
IE₁(Y) > IE₁(X)
(IE rises)
Normal periodic trend. Both have electrons removed from 2p. Neon has greater nuclear charge (+10 vs +9) with similar shielding, so attraction is stronger. Correct!
D X = Neon (Ne)
Y = Sodium (Na)
IE₁(Y) << IE₁(X)
(Huge drop)
Sodium’s outer electron occupies the 3s orbital in a new shell. It experiences significantly increased shielding from inner shells and is much further from the nucleus.

💡 Key Knowledge: The Two Period 2 Drops

Across Period 2 (Li to Ne), first ionisation energy generally increases, but there are two specific dips you must memorise:

  • Be → B (Group 2 → 3): Drop due to subshell change (from 2s to higher-energy 2p).
  • N → O (Group 5 → 6): Drop due to mutual repulsion between paired electrons in the same 2p orbital.
  • Ne → Na (Period 2 → 3): Massive drop due to entering a new quantum shell (3s), with vastly increased shielding and radius.

❌ Common Student Errors

  • Assuming IE always increases: Students forget the anomalies at Boron (Option A) and Oxygen (Option B), incorrectly assuming higher proton number always means higher IE.
  • Confusing the question requirement: Forgetting the inequality asked for IE(Y) > IE(X) and accidentally choosing an anomaly where IE decreases.
  • Assuming noble gases always have lower IE: Believing noble gases are unreactive and therefore don't follow trends, rather than having the highest IE in their period.
Examiner Insight (AQA Specification): This is a standard AO1 multiple-choice recall and application question. Questions contrasting subshell shielding (Be/B) and spin-pair repulsion (N/O) against the general periodic trend appear frequently on Paper 1.

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.