OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 5

1 mark · Medium difficulty · Multiple Choice

Identify the statement that best explains why nitrogen has a larger first ionisation energy than oxygen.

Practise this question

Question

Multiple choice question 5 asks which statement best explains why nitrogen has a larger first ionisation energy than oxygen, with four options A to D: A states N atoms have less repulsion between p-orbital electrons than O atoms, B states N atoms have a smaller nuclear charge than O atoms, C states N atoms lose an electron from the 2s subshell while O atoms lose an electron from the 2p subshell, and D states N atoms have an odd number of electrons while O atoms have an even number. An answer box and mark allocation of [1] are at the bottom.
Question text

5 Which statement best explains why nitrogen has a larger first ionisation energy than oxygen?

A N atoms have less repulsion between p-orbital electrons than O atoms.

B N atoms have a smaller nuclear charge than O atoms.

C N atoms lose an electron from the 2s subshell, while O atoms lose an electron from the 2p

subshell.

D N atoms have an odd number of electrons, while O atoms have an even number.

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 5 is A, worth 1 mark.

5 A 1 AO1.2

How to answer it

First Ionisation Energy: Nitrogen vs Oxygen

What this question tests

This question assesses your understanding of periodic trends in first ionisation energies across Period 2, specifically the anomalous dip between Group 15 (nitrogen) and Group 16 (oxygen) due to electron-electron repulsion in shared orbitals.

Question 5 (Multiple Choice)

Anomalous Trends in First Ionisation Energy

✅ Correct Answer: A

N atoms have less repulsion between p-orbital electrons than O atoms.

Nitrogen has the electron configuration 1s² 2s² 2p³ with three unpaired electrons in separate 2p orbitals. Oxygen has the configuration 1s² 2s² 2p⁴ , meaning one 2p orbital contains a paired pair of electrons. This pairing creates mutual repulsion, making it easier to remove an electron from oxygen despite its higher nuclear charge.

💡 Key Knowledge

  • Electron Configurations: N is 1s² 2s² 2p³ ; O is 1s² 2s² 2p⁴ .
  • Hund's Rule: Electrons fill degenerate orbitals singly before pairing up.
  • Spin-Pair Repulsion: Two electrons in the same orbital repel each other, lowering the energy required to remove one of them.

🧠 Exam Technique

When dealing with unexpected drops in ionisation energy trends across a period (such as Be to B, or N to O), never rely solely on nuclear charge. Always write out or visualise the subshell electron configuration and check for orbital pairing.

❌ Common Errors

  • Distractor B: Students often assume higher atomic number always means higher ionisation energy, forgetting about subshell shielding and spin-pairing repulsion.
  • Distractor C: Incorrectly claiming that electrons are removed from different principal energy levels or subshells (both N and O lose their outer electron from the 2p subshell).
Mark Allocation: [1] mark available for selecting option A (AO1.2 - Demonstrate knowledge and understanding of scientific ideas).

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · 3.1 The periodic table · 2.2 Electrons, bonding and structure

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.