AQA A-Level Chemistry Paper 3, 2024: Question 12

AQA A-Level Chemistry • Bond polarity & molecular shape

Permanent Dipoles: spotting the non-polar molecule

What this question tests

Core idea

Permanent dipole depends on bond polarity + molecular shape/symmetry.
Using shapes from electron pairs (VSEPR) to decide whether dipoles cancel.

Skill

AO2 reasoning: apply known shapes to unfamiliar molecules quickly.
Recognise common symmetric shapes that give no net dipole (e.g. tetrahedral with identical outer atoms).

Examiner focus: This is a 1-mark multiple-choice question. The mark is awarded for selecting the single molecule with no permanent dipole. Students typically lose the mark by thinking “polar bonds ⇒ polar molecule” without checking symmetry.

Part (a) — Which molecule does not have a permanent dipole? 1 mark (AO2)

Options: NH₃, PCl₃, SCl₂, SiCl₄

✅ Correct answer (from mark scheme)

D — SiCl₄

Mark scheme shows: 12 D and specifies SiCl₄ .

💡 Key knowledge: when is there no permanent dipole?

Bond dipoles exist when atoms have different electronegativities (e.g. Si–Cl is polar).
A molecule has no permanent dipole if the vector sum of bond dipoles is zero.
This usually happens in a symmetrical shape with identical surrounding atoms.

🧠 Exam technique (fast method)

Step 1: Identify the shape around the central atom.
Step 2: Ask “Are all outer atoms the same?”
Step 3: If the shape is symmetric and outer atoms identical → dipoles cancel → no permanent dipole.
For a 1-mark MCQ: you don’t need a full dot-and-cross, just shape + symmetry.

❌ Common errors (why marks are lost)

Assuming “polar bonds means polar molecule” (ignoring cancellation).
Forgetting that lone pairs often make shapes asymmetric (so dipoles don’t cancel).
Mixing up shapes: e.g. thinking SiCl₄ is “square planar” (it is tetrahedral).

Shape-by-shape reasoning (how to justify each option)

NH₃: trigonal pyramidal (lone pair on N) → asymmetric → has permanent dipole.
PCl₃: trigonal pyramidal (lone pair on P) → asymmetric → has permanent dipole.
SCl₂: bent / V-shaped (two lone pairs on S) → asymmetric → has permanent dipole.
SiCl₄: tetrahedral, 4 identical Si–Cl bonds arranged symmetrically → dipoles cancel → no permanent dipole.

Examiner-style explanation for SiCl₄ (what distinguishes top responses)

To secure the mark reliably, your thinking should include both parts:

Polar bonds: Si–Cl bonds are polar.
Overall symmetry: tetrahedral arrangement means bond dipoles are equal and point to the corners, so their vectors cancel.

Even though the bonds are polar, the molecule is non-polar due to symmetry.

📐 (No calculations needed)

This question is purely conceptual: shape + symmetry + dipole cancellation.

Mark breakdown (1 mark): Awarded for selecting D (SiCl₄) only.
AO2: applying knowledge of molecular shape and polarity to choose the correct molecule.

If you were asked to draw dipoles (extra exam practice)

💡 How to draw dipole arrows correctly

Use an arrow pointing towards the more electronegative atom (e.g. towards Cl).
Add a small + at the positive end (near the less electronegative atom).
Then consider whether these arrows cancel based on shape.

🧠 What you would draw for SiCl₄

Central Si with four Cl arranged tetrahedrally.
Draw four identical dipole arrows along each Si–Cl bond pointing Si → Cl.
State: “dipoles cancel due to tetrahedral symmetry; net dipole = 0”.

❌ Typical diagram mistakes

Drawing SiCl₄ as flat and square (incorrect geometry).
Forgetting the + end of the dipole arrow or pointing it the wrong way.