AQA A-Level Chemistry Paper 3, 2019: Question 17

Periodic Trends: Properties of Hydrogen Halides (HCl to HI)

What this question tests

This question assesses your understanding of Group 7 (halogens) chemistry, specifically the physical and chemical trends across the hydrogen halides (HCl, HBr, HI). Key concepts include intermolecular forces vs. boiling points, covalent bond strength (bond enthalpy), electronegativity trends, and redox properties (reducing ability of halide species).

Question 17

Multiple Choice: Identifying the Incorrect Trend

AQA Chemistry A-Level • Inorganic Chemistry • Group 7 Trends

✅ Correct Answer: A

Statement A is NOT correct: "The boiling points decrease."

From HCl to HI, the boiling points actually increase:

HCl: -85 °C (188 K)
HBr: -66 °C (207 K)
HI: -35 °C (238 K)

Because the question asks which statement is not correct, option A is the correct choice.

💡 Key Knowledge: Why Boiling Points Increase

Intermolecular Forces: HCl, HBr, and HI do not exhibit hydrogen bonding (only HF does). They experience permanent dipole-dipole forces and van der Waals (induced dipole-dipole) forces.
Dominant Factor: Down the group from Cl to I, the number of electrons per molecule increases significantly (HCl: 18 e⁻, HBr: 36 e⁻, HI: 54 e⁻).
Stronger London/van der Waals forces: More electrons lead to a larger, more polarisable electron cloud, producing stronger induced dipole-dipole interactions that require more thermal energy to overcome.

🧠 Systematic Breakdown of All Options

A (Incorrect statement — Correct choice): Boiling points increase from HCl to HI due to increasing strength of van der Waals forces, which outweigh the decrease in permanent dipole-dipole attractions.
B (Correct statement): H-X bond dissociation energy decreases. As the halogen atom gets larger down the group, atomic radius increases, the H-X bond length becomes longer, and orbital overlap weakens. The shared pair of electrons is further from the halogen nucleus, so less energy is needed to break the bond.
C (Correct statement): Polarity of the H-X bond decreases. Electronegativity decreases down Group 7 (Cl: 3.0 > Br: 2.8 > I: 2.5). The difference in electronegativity between H (2.1) and X decreases, making the bond less polar.
D (Correct statement): They are more easily oxidised in aqueous solution. Down Group 7, the halide ions (Cl⁻, Br⁻, I⁻) become progressively stronger reducing agents. Iodide ions lose electrons most readily due to greater ionic radius and increased electron shielding:
2I⁻(aq) → I₂(aq) + 2e⁻ (readily oxidised)

❌ Common Errors & Traps

Missing the negative stem: Overlooking the word "not" in the question and picking a statement that is scientifically true (e.g. B, C, or D).
Confusing covalent bonds with intermolecular forces: Thinking that because the H-X covalent bond weakens down the group (Statement B), the boiling point must also decrease. Boiling involves breaking intermolecular forces, not covalent bonds!
Overemphasising dipole moments: Assuming permanent dipole-dipole forces determine the boiling point trend. Van der Waals forces increase far more dramatically down the group and completely override the slight drop in dipole strength.

📐 Exam Strategy for Group 7 Hydrides

HF is the anomaly: Remember the classic V-shaped boiling point graph. HF has a very high boiling point due to hydrogen bonding.
Trend from HCl to HI: After the drop from HF to HCl, boiling points monotonically rise from HCl to HBr to HI.
Reducing power trend: Remember reactions with concentrated H₂SO₄: Cl⁻ cannot reduce sulfur (+6), Br⁻ reduces sulfur to SO₂ (+4), and I⁻ reduces sulfur all the way to S (0) and H₂S (-2). This confirms that HI / I⁻ is the most easily oxidised.

Mark Scheme Allocation: 1 mark for selecting option A. No partial marks awarded.

AQA A-Level Chemistry Paper 3, 2019: Question 17

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