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Chemical Bonding: Question 9

Syllabus 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7

Multiple choice AS 1 mark

The boiling points of the Group 17 elements increase steadily down the group: F2\text{F}_2 boils at −188 ∘C-188\,^\circ\text{C}, Cl2\text{Cl}_2 at −34 ∘C-34\,^\circ\text{C}, Br2\text{Br}_2 at 59 ∘C59\,^\circ\text{C}, and I2\text{I}_2 at 184 ∘C184\,^\circ\text{C}.

Which of the following best explains this trend?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Identify what boiling actually overcomes

Boiling a simple molecular substance like a halogen does not break the covalent bonds within each molecule; it only overcomes the (much weaker) intermolecular forces between separate molecules. This immediately rules out any explanation based on the strength of the covalent X–X bond.

Step 2: Identify which intermolecular forces are present

Each halogen molecule (F2\text{F}_2, Cl2\text{Cl}_2, Br2\text{Br}_2, I2\text{I}_2) is made of two identical atoms, so there is no electronegativity difference within the molecule and therefore no permanent dipole. This rules out permanent dipole-dipole forces. There is also no hydrogen atom bonded to N, O or F in any of these molecules, so hydrogen bonding cannot occur.

This leaves van der Waals’ forces (induced dipole-induced dipole / London dispersion forces) as the only intermolecular force acting between halogen molecules. These arise from temporary, constantly-fluctuating dipoles caused by the movement of electrons, which induce corresponding dipoles in neighbouring molecules.

Going down Group 17, each successive halogen molecule has more electrons (F →\to Cl →\to Br →\to I, each with a larger, more diffuse electron cloud). More electrons means larger, more polarisable electron clouds, which produce stronger instantaneous and induced dipoles, and therefore stronger van der Waals’ forces between molecules.

Stronger intermolecular forces require more energy to overcome, so the boiling point increases steadily from F2\text{F}_2 up to I2\text{I}_2, consistent with the data given.

Why the other options are wrong

  • A: this confuses the intramolecular covalent bond with the intermolecular forces that boiling actually overcomes; the covalent bond is not broken on boiling.
  • C: electronegativity actually decreases down Group 17, and in any case, a diatomic molecule of two identical atoms cannot have a permanent dipole at all.
  • D: hydrogen bonding requires a hydrogen atom bonded directly to N, O or F; none of the halogen molecules contains any hydrogen atoms, so hydrogen bonding is impossible here.

Final answer

  • The trend is explained by increasing van der Waals’ (induced dipole-induced dipole) forces as molecule size and electron number increase down the group, option B.