AQA A-Level Chemistry AS Paper 1, June 2022: Question 4

✅ Correct Answer / Criteria

Two valid orientations are accepted by the mark scheme:

• Option A (Ethanol donor, NH₃ acceptor):
  • Show lone pair on :NH₃ pointing to H on ethanol.
  • Partial charges: O(δ−)—H(δ+)···:N(δ−)H₃.
  • Dotted line from lone pair on N to the H of ethanol.
  • 180° linear bond angle along the O—H···N group.
• Option B (NH₃ donor, Ethanol acceptor):
  • Show lone pair on ethanol's oxygen :O pointing to an H of NH₃.
  • Partial charges: ethanol O(δ−)···(δ+)H—N(δ−)H₂.
  • Dotted line from lone pair on O to the H of NH₃.
  • 180° linear bond angle along N—H···O.

🧠 Exam Technique: 3 Mandatory Criteria

1. M1: Correct partial charges (δ−, δ+, δ−) on the 3 atoms involved AND the lone pair on the electronegative acceptor atom (N or O).
2. M2: A clearly dashed/dotted line drawn directly from the lone pair to the δ+ hydrogen atom.
3. M3: Linearity: The covalent bond to H and the hydrogen bond must be drawn in a straight line (180° angle: O—H···:N or N—H···:O).

❌ Common Errors

• Omitting the lone pair on the acceptor atom (N or O).
• Drawing the hydrogen bond with a solid line instead of a dashed/dotted line.
• Non-linear arrangement (e.g., drawing the hydrogen bond at a sharp angle to the O—H bond).
• Putting partial charges on non-participating atoms like C or non-bonding H atoms incorrectly.

✅ Correct Answer

The (relative) tendency/power of an atom to attract a pair of electrons / electron density in a covalent bond.

💡 Key Knowledge

• Must specify an atom (or nucleus) — NOT an element or molecule.
• Must specify a pair of electrons / electron density.
• Must explicitly mention in a covalent bond.

❌ Common Errors

• Saying "attract a lone pair of electrons" (immediate zero).
• Referring to an "element's ability" rather than an "atom's ability".
• Forgetting to include the context: "in a covalent bond".

✅ Correct Answer

H and O  (or written as O—H )

📐 Step-by-Step Calculation

1. Find the difference in electronegativity (ΔEN) for candidate pairs:
   • C (2.5) & Br (2.8): ΔEN = 0.3
   • C (2.5) & O (3.5): ΔEN = 1.0
   • N (3.0) & H (2.1): ΔEN = 0.9
   • O (3.5) & H (2.1): ΔEN = 1.4 (Greatest difference)
2. Therefore, the O—H bond has the highest polarity.

🧠 Exam Technique

Always show you are picking the two atoms with the largest numerical difference. Ensure you name both atoms clearly: "H and O" or identify the bond as "O—H".

✅ Correct Answer

• Mark 1: The molecule is completely symmetrical / tetrahedral / has an even distribution of electron density.
• Mark 2: The dipoles cancel out.

💡 Key Knowledge

• A bond can be polar due to electronegativity differences, but a molecule's overall dipole depends on its 3D shape.
• CBr₄ has 4 bonding pairs and 0 lone pairs around carbon → regular tetrahedral shape with bond angles of 109.5°.
• Due to high spherical/spatial symmetry, individual bond dipoles point in opposing directions and sum vectorially to zero.

❌ Common Errors & Forbidden Phrases

• Do NOT say: "The polar bonds cancel out" — bonds are physical links, they cannot cancel; dipoles cancel!
• Do NOT say: "Partial charges cancel out" — this is rejected by AQA mark schemes.
• Stating only that it is symmetrical without explaining that the dipoles cancel.

✅ Correct Answer

• M1: CBr₄ has van der Waals’ forces between molecules.
• M2: CHBr₃ has van der Waals’ forces AND permanent dipole–dipole forces between molecules.
• M3: The van der Waals’ forces between CBr₄ molecules are stronger than the combined dipole–dipole and van der Waals' forces in CHBr₃ (because CBr₄ has more electrons / larger electron cloud / larger mass).

🧠 Exam Technique: Overcoming the Trap

Students instinctively think: "CHBr₃ has permanent dipoles, so it must have the higher boiling point!"

Reality: Van der Waals' forces depend heavily on the total number of electrons. CBr₄ has 146 electrons, whereas CHBr₃ has 112 electrons. The significantly larger electron cloud in CBr₄ produces much stronger induced dipoles, which outweigh the permanent dipole forces in CHBr₃.

❌ Critical Misconceptions

• Breaking covalent bonds: Never mention breaking C—Br or C—H bonds! If you mention breaking covalent bonds, M3 cannot be awarded. Boiling only overcomes intermolecular forces.
• Omitting van der Waals from CHBr₃: All molecular substances have van der Waals' forces. Do not claim CHBr₃ only has dipole-dipole forces.
• Failing to explicitly compare the forces between the named substances.