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Chapter 12: Problem 7

The compounds \(\mathrm{Br}_{2}\) and \(\mathrm{ICl}\) have the same number of electrons, yet \(\mathrm{Br}_{2}\) melts at \(-7.2^{\circ} \mathrm{C}\), whereas \(\mathrm{ICl}\) melts at \(27.2^{\circ} \mathrm{C}\). Explain.

Short Answer

Expert verified
Despite having the same number of electrons, \(\mathrm{ICl}\) melts at a higher temperature than \(\mathrm{Br}_{2}\) because \(\mathrm{ICl}\) has stronger dipole-dipole attractions due to the polar nature of its bond, whereas \(\mathrm{Br}_{2}\) has weaker London dispersion forces due to the nonpolar nature of its bond. These stronger intermolecular forces in \(\mathrm{ICl}\) require more energy to break, resulting in a higher melting point than \(\mathrm{Br}_{2}\).

Step by step solution

01

Understanding the Molecules

First, understand the structure of the molecules in question. \(\mathrm{Br}_{2}\) is an element and as a diatomic molecule, it is composed of two bromine atoms bonded together. On the other hand, \(\mathrm{ICl}\) is a compound formed by iodine and chlorine atoms. The important thing to note here is that \(\mathrm{ICl}\) has polar covalent bond whereas \(\mathrm{Br}_{2}\) has nonpolar covalent bond.
02

Understanding Bond Polarity

The electronegativity difference between iodine and chlorine in \(\mathrm{ICl}\) makes it a polar molecule. This means there are dipole-dipole attractions in addition to the London dispersion forces. The bond polarity creates an additional intermolecular force of attraction between the molecules, causing it to have a higher melting point than a similar molecule (\(\mathrm{Br}_{2}\)) without this type of bond.
03

Understanding Melting Points

The melting point of a substance is based on the strength of the bonds or forces that hold the particles of the substance together. The stronger these forces or bonds, the more energy is required to break them and change the substance from a solid to a liquid. It thus requires more energy to break the stronger dipole-dipole attractions in \(\mathrm{ICl}\) than the weaker London forces in \(\mathrm{Br}_{2}\). This greater energy requirement corresponds to a higher melting point. So, despite having the same number of electrons, \(\mathrm{ICl}\) melts at a higher temperature than \(\mathrm{Br}_{2}\) due to the polar nature of its bond.

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