AQA A-Level Chemistry AS Paper 1, June 2023: Question 13

1 mark · Medium difficulty · Multiple Choice

Identify which molecule among NCl3, CCl4, PF5, and SF6 possesses a permanent dipole.

Practise this question

Question

Question 13 asks: 'Which molecule has a permanent dipole?' followed by four multiple-choice options with checkboxes: A NCl3, B CCl4, C PF5, and D SF6. The question is worth 1 mark.
Question text

13 Which molecule has a permanent dipole?

[1 mark]

A NCl3

B CCl4

C PF5

D SF6

Mark scheme

Show the mark scheme Mark scheme for question 13 indicates that the correct answer is option A (NCl3), awarded 1 mark for AO1.

13 A 1 (AO1) NCl3

How to answer it

Determining Molecular Polarity and Permanent Dipoles

📌 What this question tests

This question assesses your ability to apply VSEPR theory and electronegativity differences to determine whether a molecule possesses an overall permanent dipole moment.

  • Deducing electron pair geometry (bonding pairs vs lone pairs).
  • Distinguishing between polar bonds and polar molecules.
  • Recognising symmetrical molecular geometries that lead to bond dipoles cancelling out.

Question 13

Which molecule has a permanent dipole?

✅ Correct Answer

A: NCl₃

Award 1 mark for option A.

🧠 Exam Technique: 2-Step Polarity Test

  1. Check for polar bonds: Does the molecule contain atoms with different electronegativities?
  2. Check for symmetry: Do the individual bond dipoles cancel out due to symmetrical arrangement?
    • Symmetrical shape (no lone pairs on central atom) → Dipoles cancel → Non-polar
    • Asymmetrical shape (lone pair present) → Dipoles do not cancel → Polar (permanent dipole)

📐 Molecular Breakdown of All Options

Molecule Central Atom Valence e⁻ Bonding Pairs (BP) : Lone Pairs (LP) Molecular Shape Symmetry & Dipole Status
NCl₃ 5 (Group 15) 3 BP : 1 LP Trigonal pyramidal Asymmetrical: Dipoles do not cancel out → Permanent dipole
CCl₄ 4 (Group 14) 4 BP : 0 LP Tetrahedral Symmetrical: 4 equal C–Cl dipoles cancel → Non-polar
PF₅ 5 (Group 15) 5 BP : 0 LP Trigonal bipyramidal Symmetrical: 5 equal P–F dipoles cancel → Non-polar
SF₆ 6 (Group 16) 6 BP : 0 LP Octahedral Symmetrical: 6 equal S–F dipoles cancel → Non-polar

💡 Key Knowledge: Why NCl₃ is Polar

  • Nitrogen has 5 valence electrons and forms 3 single covalent bonds with chlorine atoms, leaving one non-bonding lone pair.
  • According to VSEPR theory, lone pair–bonding pair repulsion is greater than bonding pair–bonding pair repulsion, pushing the bonds down into a trigonal pyramidal shape with a bond angle of ~107°.
  • Because the shape is not completely symmetrical, the bond dipoles (along with the effect of the lone pair) do not sum to zero, leaving a net permanent dipole moment.

❌ Common Misconceptions & Traps

  • Confusing polar bonds with polar molecules: C–Cl, P–F, and S–F bonds are all highly polar due to substantial electronegativity differences. However, high symmetry in CCl₄, PF₅, and SF₆ causes the vector sum of these dipoles to equal zero.
  • Looking only at electronegativity: Nitrogen and chlorine have very similar electronegativity values (both ~3.0 on the Pauling scale), leading some students to incorrectly assume NCl₃ cannot be polar. However, chlorine is more polarisable and the lone pair creates an uneven distribution of charge, resulting in an overall permanent molecular dipole.
  • Overlooking the lone pair: Forgetting that N in NCl₃ has a lone pair leads to mistaking it for a flat, trigonal planar molecule.

Topics

Physical Chemistry · 3.1.3 Bonding

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