OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 7
1 mark · Easy difficulty · Multiple Choice
Identify the best explanation for the increasing trend in boiling points from HCl to HI among hydrogen halides.
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Trend in Boiling Points of Hydrogen Halides
This question assesses your understanding of intermolecular forces vs intramolecular bonds, the factors governing the strength of London dispersion forces (induced dipole–dipole interactions), and the distinction between permanent dipoles, hydrogen bonds, and covalent bonds in Group 7 hydrides.
Identifying the Origin of the Boiling Point Trend (HCl → HI)
Data: HCl (188 K) < HBr (206 K) < HI (238 K)
✅ Correct Answer: C
The strength of the induced dipole–dipole interactions (London forces) increases.
- Descending from HCl to HI, the number of electrons per molecule increases significantly (18 → 36 → 54 electrons).
- More electrons lead to greater fluctuations in electron density and larger temporary dipoles.
- This causes stronger induced dipole–dipole forces, which require more thermal energy to overcome during boiling.
💡 Key Knowledge
- Intramolecular vs Intermolecular: Boiling a simple covalent molecular substance breaks only weak intermolecular forces, never strong covalent bonds.
- London Forces: Present in all molecular substances. Their magnitude depends on the total number of electrons and the surface contact area.
- Hydrogen Bonding Criteria: Requires hydrogen bonded directly to a highly electronegative atom with a lone pair (only N, O, or F). Therefore, HCl, HBr, and HI do not form hydrogen bonds.
- Electronegativity Trend: Electronegativity decreases down Group 7 (Cl > Br > I). Hence, dipole–dipole attractions actually weaken from HCl to HI.
❌ Why Other Options Are Incorrect
- A is incorrect: Boiling does not break covalent bonds inside the molecule. Furthermore, H–X covalent bond strength actually decreases down Group 7 as atomic radius increases.
- B is incorrect: A classic distractor! Despite containing hydrogen and a halogen, HCl, HBr, and HI do not have hydrogen bonds (only HF in Group 7 exhibits hydrogen bonding).
- D is incorrect: Chlorine is more electronegative than bromine and iodine. The H–Cl bond is the most polar, meaning permanent dipole–dipole attractions decrease towards HI, rather than increase. London forces easily outweigh this decrease.
🧠 Exam Technique: 2-Step Elimination
- Step 1: Eliminate intramolecular traps. When asked about boiling or melting points of simple molecular compounds, immediately cross out any option mentioning covalent bonds (eliminates A).
- Step 2: Check for genuine hydrogen bonding. Look for H attached directly to N, O, or F. If none are present, hydrogen bonding is impossible (eliminates B).
- Step 3: Decide between London forces and permanent dipoles. Notice the massive increase in total electrons (Cl = 17, Br = 35, I = 53). London forces dominate the trend, making C the definitive answer.
Topics
Module 2: Foundations in chemistry · Module 3: Periodic table and energy · 2.2 Electrons, bonding and structure · 3.1 The periodic table
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.