Question

The boiling point of bromine is 59 °C. Which of the following best predicts the boiling point of iodine monochloride, a polar compound?
Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Lower than 59 °C because ionic bonding in bromine is stronger than covalent bonding in iodine monochloride.

Lower than 59 °C because dipole-dipole interactions in iodine monochloride are weaker than in bromine.

Higher than 59 °C because ionic bonding in iodine monochloride is stronger than H-bonding in bromine.

Answer 1

Answer:

Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Explanation:

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Answer 2

Answer:

Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Explanation:

Intermolecular forces are broadly classified into four types:

1) Ionic

2) Hydrogen bonding

3) Dipole-dipole

4) London dispersion

The strongest is the ionic interaction while the weakest are the dispersion forces. Bromine ($Br_{2}$) is a non-polar molecule due to the absence of a permanent dipole. The only force of attraction in non-polar molecules are the weak dispersion forces.

In contrast iodine monochloride ($ICl$) is a polar molecule which are held together dipole-dipole forces that are stronger than the London dispersion forces.

Hence, the boiling point of ($ICl$) would be higher than that of  ($Br_{2}$)