OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 8

1 mark · Easy difficulty · Multiple Choice

Identify which of the given chloride compounds possesses molecules with a trigonal planar shape.

Practise this question

Question

Question 8 asks: 'Which compound has molecules with a trigonal planar shape?' followed by four multiple choice options: A BeCl2, B BCl3, C NCl3, and D PCl3. A response box is provided next to 'Your answer' with a total mark of 1 indicated at the bottom right.
Question text

8 Which compound has molecules with a trigonal planar shape?

A BeCl2

B BCl3

C NCl3

D PCl3

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme row for question 8 showing the correct answer as 'B' with an allocation of 1 mark.

8 B 1

How to answer it

Determining Molecular Shapes Using VSEPR Theory

📋 What this question tests

This question assesses your ability to apply Valence Shell Electron Pair Repulsion (VSEPR) theory to deduce the 3D shapes of simple covalent molecules. Specifically, it tests counting outer shell bonding pairs and lone pairs around central atoms (Groups 2, 13/3, and 15/5) and linking those arrangements to standard geometric names and bond angles.

Question 8 Breakdown

Multiple Choice: Identification of a Trigonal Planar Molecule

✅ Correct Answer

B: BCl₃

Awarded: [1 mark] for selecting option B.

  • Boron (B) is in Group 13 (Group 3) and has 3 outer electrons.
  • It shares each electron to form 3 single covalent bonds with chlorine atoms.
  • 3 bonding pairs + 0 lone pairs around the central B atom.
  • The 3 electron pairs repel each other equally to positions of maximum separation, giving a trigonal planar shape with bond angles of 120°.

📐 Breakdown of All Options

Molecule Central Atom Group Bond / Lone Pairs Shape & Bond Angle
A: BeCl₂ Group 2 (2 e⁻) 2 BP, 0 LP Linear (180°)
B: BCl₃ Group 13 (3 e⁻) 3 BP, 0 LP Trigonal planar (120°)
C: NCl₃ Group 15 (5 e⁻) 3 BP, 1 LP Pyramidal (107°)
D: PCl₃ Group 15 (5 e⁻) 3 BP, 1 LP Pyramidal (107°)

💡 Key Knowledge: VSEPR Principles

  • Electron repulsion: Electron pairs surround the central atom and repel each other to get as far apart as possible to minimise repulsion.
  • Lone pair vs Bonding pair repulsion:
    LP–LP > LP–BP > BP–BP .
  • Each lone pair reduces ideal tetrahedral angles (109.5°) by approximately 2.5°. Hence, both NCl₃ and PCl₃ have a pyramidal shape with bond angles of around 107°.
  • Incomplete octet: Boron in BCl₃ only has 6 electrons in its outer shell (electron-deficient) and has no lone pairs left over.

🧠 Exam Technique & Diagrammatic Details

  • 3-Step Method for Shapes:
    1. Find the group number of the central atom (= outer shell electrons).
    2. Add 1 electron for each bonded monovalent atom (e.g. Cl, H, F).
    3. Divide by 2 to get total electron pairs, then subtract bonded atoms to find lone pairs.
  • Drawing BCl₃: If asked to draw in an extended response, draw the central B atom with 3 coplanar bonds spaced symmetrically at 120° angles in the plane of the page (no wedges or dashes required because all atoms lie in the same flat plane).

❌ Common Misconceptions & Traps

  • Confusing "Trigonal Planar" with "Pyramidal (Trigonal Pyramidal)": Both have three peripheral atoms attached to a central atom, but molecules like NCl₃ and PCl₃ have a lone pair which pushes the bonding pairs down out of the plane.
  • Assuming all three-coordinate species are planar: Only molecules with 3 bonding pairs and zero lone pairs are trigonal planar.
  • Octet Rule Trap: Students often mistakenly force an extra lone pair onto Boron, assuming every stable non-metal must follow the strict octet rule. Boron commonly forms stable compounds with only a sextet (6 outer electrons).

Topics

Module 2: Foundations in chemistry · 2.2 Electrons, bonding and structure

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.