AQA A-Level Chemistry AS Paper 2, June 2025: Question 3

6 marks · Medium difficulty · Long Answer

Explain why CCl4 has a higher boiling point than CHCl3 by considering molecular shape, polarity, and intermolecular forces.

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Question text

03 Molecules of CCl4 and CHCl3 have similar tetrahedral shapes.

The boiling point of CCl4 is 77 °C

The boiling point of CHCl3 is 61 °C

Explain why CCl4 has a higher boiling point than CHCl3

Your answer should

• explain why these molecules have a tetrahedral shape

• explain why the molecules of each compound are polar or non-polar

• compare the intermolecular forces in these compounds.

[6 marks]

Mark scheme

Show the mark scheme

Question Marking guidance Additional Comments/Guidelines Mark

This question is marked using levels of response. Refer to the Mark Scheme Stage 1 - Shape

Instructions for Examiners for guidance on how to mark this question.

1a 4 bonding pairs (& no lone pairs) (allow

Level 3: bonds but not atoms for bonding pairs)

All stages are covered and each stage is generally correct and 1b repel as far as possible / repel equally / to

virtually complete. minimise repulsion

5-6

(6 v 5) Answer is well structured, with no repetition or irrelevant (covered = 1, virtually complete = 1 or 2)

points, and covers all aspects of the question. Accurate and clear

Stage 2 - Polarity

expression of ideas with no errors in use of technical terms.

2a C–Cl bonds polar

Level 2:

2b CCl4 not polar

All stages are covered but stage(s) may be incomplete or may

2c CCl4 dipoles cancel due to symmetry

contain inaccuracies OR

two stages are covered and are generally correct and virtually 2d CHCl3 polar

complete. 3-4 2e CHCl3 dipoles do not cancel out

(4 v 3) Answer has some structure and covers most aspects of the (covered = 2, virtually complete = 3/4/5) (1 x AO1,

03 question. Ideas are expressed with reasonable clarity with,

Stage 3 - Boiling points (no credit for 3 x AO2,

perhaps, some repetition or some irrelevant points. If any, only reference to breaking of covalent bonds) 2 x AO3)

minor errors in use of technical terms.

3a CCl4 has van der Waals’ forces

Level 1: 3b CHCl has dipole-dipole forces and van

Two stages are covered but stage(s) may be incomplete or der Waals’

may contain inaccuracies OR

3c van der Waals’/intermolecular forces in

only one stage is covered but is generally correct and 1-2 CCl stronger than (combined) van der

virtually complete. Waals’ (and dipole-dipole)/ intermolecular

(2 v 1) Answer includes statements which are presented in a forces in CHCl3

logical order and / or linked.

3d as CCl4 has more electrons (or higher Mr

Level 0 or bigger molecule) than CHCl3

Insufficient correct chemistry to gain a mark. Allow dispersion / London forces for van der

Waals’

(covered = 1, virtually complete = 3 or 4)

How to answer it

Explaining Boiling Points: Shape, Polarity & Forces in CCl₄ vs CHCl₃

📋 What this question tests

This 6-mark extended-response question assesses core chemical bonding concepts across three integrated stages:

VSEPR Theory: Explaining molecular geometry using electron pair repulsion principles.
Electronegativity and Dipoles: Determining overall molecular polarity from bond dipoles and molecular symmetry.
Intermolecular Forces (IMFs): Comparing the relative strengths of van der Waals (London dispersion) forces and permanent dipole-dipole forces, and linking them to boiling points.

Question 03: Levels of Response Breakdown

AQA 6-Mark Criteria (AO1 = 1 mark, AO2 = 3 marks, AO3 = 2 marks)

Level	Marks	Criteria
Level 3	5–6	All 3 stages are covered and each stage is generally correct and virtually complete. Clear structure, no contradictory chemistry, accurate terminology.
Level 2	3–4	All 3 stages covered with minor gaps/errors OR 2 stages are generally correct and virtually complete.
Level 1	1–2	2 stages covered with gaps/inaccuracies OR 1 stage is generally correct and virtually complete.

💡 Examiner Rule of Thumb: Structure your answer clearly with three explicit headings matching the question bullets (Shape, Polarity, Intermolecular Forces). This guarantees the examiner can easily verify all 3 stages for Level 3!

Stage 1: Explaining the Tetrahedral Shape

Why both CCl₄ and CHCl₃ adopt tetrahedral geometry

✅ Model Answer Points

Around the central carbon atom, there are 4 bonding pairs of electrons and no lone pairs.
Electron pairs repel as far apart as possible (or repel equally) to reach a state of minimum repulsion.
This equal repulsion results in a tetrahedral arrangement with a bond angle of 109.5°.

💡 Key Knowledge: VSEPR Principles

Electron pairs in the outer shell of the central atom are negatively charged clouds that repel each other.

Both molecules have a Group 4 central carbon with 4 valence electrons forming 4 single covalent bonds.
Carbon valence count: 4 + 4 (from Cl/H atoms) = 8 electrons = 4 electron pairs.

🧠 Exam Technique

Always state both the number of bonding pairs and the number of lone pairs. Leaving out "no lone pairs" is one of the most common reasons students fail to make this stage "virtually complete".

❌ Common Errors to Avoid

Writing "4 bonds repel" instead of electron pairs repelling. (The mark scheme accepts bonds, but strictly rejects "atoms repel").
Forgetting to state that repulsion is minimised or that they repel as far apart as possible.

Stage 2: Molecular Polarity

Explaining why CCl₄ is non-polar while CHCl₃ is polar

✅ Model Answer Points

Chlorine is more electronegative than carbon, so all C–Cl bonds are polar (carrying a dipole C^δ+–Cl^δ–).
CCl₄ is non-polar: Because the molecule is symmetrical, the individual bond dipoles cancel out, leaving no overall permanent dipole.
CHCl₃ is polar: Because the molecule is asymmetrical (the C–H bond polarity differs from C–Cl), the bond dipoles do not cancel out, creating an overall permanent dipole.

💡 Key Knowledge: Polarity vs Symmetry

Dipole moment is a vector quantity: It has both magnitude and direction.
In CCl₄, all four vectors point toward identical vertices of a regular tetrahedron, pulling equally in all directions, so the net vector sum is zero ( μ = 0 ).
In CHCl₃, the dipole along C–H is negligible compared to the strong C–Cl dipoles, so the dipole vectors do not cancel.

🧠 Diagram Recommendation

If drawing 3D tetrahedral structures to support your answer:

Draw one central C atom with two normal lines (in the plane), one wedged bond (pointing out), and one dashed/hatched bond (pointing back).
Add partial charges: δ+ on C and δ– on each Cl atom.
Add dipole arrows pointing from C to Cl ( +---> ).

❌ Common Errors to Avoid

Claiming that CCl₄ has non-polar bonds. The bonds are polar; it is the molecule that is non-polar!
Saying "CCl₄ is symmetrical" without explaining the chemical consequence (that the dipoles cancel).

Stage 3: Intermolecular Forces & Boiling Points

Explaining why CCl₄ (77 °C) has a higher boiling point than CHCl₃ (61 °C)

✅ Model Answer Points

Identify forces in CCl₄: Only van der Waals (London dispersion) forces between molecules.
Identify forces in CHCl₃: Both permanent dipole-dipole forces and van der Waals forces between molecules.
Compare magnitude: The van der Waals forces in CCl₄ are stronger than the combined intermolecular forces (van der Waals + dipole-dipole) in CHCl₃.
Explain why: CCl₄ has more electrons (74 electrons vs 58 electrons) and a larger surface area / higher M_r, resulting in stronger induced dipoles requiring more thermal energy to overcome.

💡 Key Knowledge: Why vdW Can Beat Dipole-Dipole

Students often wrongly assume that dipole-dipole forces always lead to a higher boiling point than van der Waals forces alone.

Electron count CCl₄: 6 + (4 × 17) = 74 electrons.
Electron count CHCl₃: 6 + 1 + (3 × 17) = 58 electrons.
Because CCl₄ has 16 more electrons, its electron cloud is significantly more polarisable, producing much stronger temporary induced dipoles that outweigh the permanent dipole in CHCl₃.

❌ Fatal Error: Breaking Covalent Bonds

Zero credit is given for boiling points if you mention breaking covalent bonds!

Boiling is a physical change that overcomes intermolecular forces between molecules. Covalent bonds inside the molecule remain completely intact.

🧠 Examiner Distinction for 6/6 Marks

To secure full marks in Level 3, make sure you:

Explicitly state that CHCl₃ has both van der Waals and permanent dipole-dipole forces.
State clearly that the van der Waals forces in CCl₄ are stronger than the total/combined forces in CHCl₃.
Explicitly attribute the stronger van der Waals forces to more electrons (or higher M_r / larger molecule).

Full Mark (6/6) Exemplar Response

Shape:
Both CCl₄ and CHCl₃ have four bonding pairs of electrons and no lone pairs of electrons around the central carbon atom. Electron pairs repel each other equally as far apart as possible to minimise repulsion, giving both molecules a tetrahedral shape with bond angles of approximately 109.5°.

Polarity:
Chlorine is more electronegative than carbon, making the C–Cl bonds polar. CCl₄ is a symmetrical molecule, so the individual bond dipoles pull equally in opposite directions and cancel out; therefore, CCl₄ is non-polar. In contrast, CHCl₃ is asymmetrical because of the C–H bond, so the bond dipoles do not cancel out, meaning CHCl₃ has an overall permanent dipole and is polar.

Intermolecular Forces & Boiling Points:
Between non-polar CCl₄ molecules, only van der Waals forces exist. Between polar CHCl₃ molecules, both permanent dipole-dipole forces and van der Waals forces exist. However, CCl₄ has more electrons (74 e⁻) than CHCl₃ (58 e⁻), which makes the electron cloud in CCl₄ more polarisable. As a result, the van der Waals forces in CCl₄ are significantly stronger than the combined intermolecular forces in CHCl₃. More energy is required to overcome the intermolecular forces in CCl₄, giving it a higher boiling point (77 °C vs 61 °C).

Topics

Physical Chemistry · 3.1.3 Bonding

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.