Edexcel A-Level Chemistry Paper 3, June 2024: Question 2

✅ Model Answer

• Step 1: Random movement of electrons creates a temporary (instantaneous) uneven distribution of electron density within a molecule.
• Step 2: This produces an instantaneous (or temporary) dipole in the first molecule.
• Step 3: This temporary dipole induces a dipole in a neighbouring/adjacent molecule, leading to an electrostatic attraction.

🧠 Exam Technique & Mark Breakdown

To secure all 3 marks, you must follow the logical sequence of events:

• Mark 1: Mentioning electron movement and uneven/asymmetrical distribution.
• Mark 2: Stating the formation of an instantaneous/temporary dipole.
• Mark 3: Explaining the induction of a dipole in a neighbouring molecule.

❌ Common Errors & Pitfalls

• Permanent Dipoles: Never mention permanent dipoles when discussing non-polar alkanes. This loses marks instantly.
• Atom vs Electron: Vague references to "atoms moving" instead of electron clouds shifting will fail to score Mark 1.

💡 Key Knowledge

London forces exist between all molecules (polar or non-polar). They are weaker than permanent dipole-dipole interactions and hydrogen bonds, but their cumulative effect increases significantly with larger molecular size and greater surface contact area.

✅ Correct Answer

Compound A has the highest boiling temperature because it is unbranched, meaning it has a greater surface area (or more points of contact between molecules), resulting in stronger London forces that require more energy to overcome.

🧠 Exam Technique & Mark Breakdown

Full marks require linking structure directly to intermolecular force strength:

• Mark 1: Identifying compound A and noting it has less/least branching or greater London forces.
• Mark 2: Explaining that increased surface area or closer molecular packing allows for more points of contact.

❌ Common Errors & Pitfalls

• Covalent Bonds: Never state that covalent bonds are broken during boiling. Examiners penalise this heavily.
• Incorrect Reason: Do not claim compound A has "more electrons"—all isomers A–D share the exact same molecular formula ( C₆H₁₄ ) and therefore the same total number of electrons!

💡 Key Knowledge

As branching increases, molecules become more spherical, reducing the surface contact area between adjacent molecules. This decreases the magnitude of London forces and lowers the boiling temperature.

✅ Correct Answer Summary

• CCl₄ (Tetrachloromethane): Not deflected (non-polar molecule due to symmetrical shape, despite polar C-Cl bonds).
• CHCl₃ (Trichloromethane): Deflected (polar molecule due to non-symmetrical shape and polar C-Cl/C-H bonds).
• C₆H₁₄ (Hexane): Not deflected (non-polar molecule with non-polar bonds).

🧠 Exam Technique & Mark Breakdown

Marks are awarded across three analytical criteria:

• Mark 1: Stating correct deflection outcomes for the three liquids.
• Mark 2: Referencing bond polarity (difference in electronegativities creating polar bonds).
• Mark 3: Explaining overall molecular polarity/symmetry (dipoles cancelling out vs. a net dipole moment).

❌ Common Errors & Pitfalls

• Assuming Polar Bonds = Polar Molecule: Students often see polar C-Cl bonds in CCl₄ and incorrectly assume the liquid is polar. You must account for 3D molecular geometry and symmetry where bond dipoles cancel out.

💡 Key Knowledge

Charged rods attract polar molecules because the permanent dipole aligns with the electrostatic field, pulling the stream sideways. Non-polar liquids ( CCl₄ and C₆H₁₄ ) experience no net electrostatic attraction and flow straight down.