Consider the compounds \(\mathrm{Cl}_{2}, \mathrm{HCl}, \mathrm{F}_{2},\) NaF, and HF. Which compound has a boiling point closest to that of argon? Explain.

To solve this problem, we first need to understand the interactions that affect boiling points. Boiling points are generally determined by the strength of the intermolecular forces in a given substance. The stronger the forces, the higher the boiling point. The major types of intermolecular forces are: 1. London dispersion forces 2. Dipole-dipole interactions 3. Hydrogen bonding In addition, we should also note that argon is a noble gas with only London dispersion forces.

Now let's analyze the types of intermolecular forces present in each compound: 1. \(\mathrm{Cl}_{2}\) - As a diatomic nonpolar molecule, it exhibits only London dispersion forces. 2. \(\mathrm{HCl}\) - This polar molecule exhibits dipole-dipole interactions and London dispersion forces. 3. \(\mathrm{F}_{2}\) - As a diatomic nonpolar molecule, it also exhibits only London dispersion forces. 4. NaF - This ionic compound has very strong ionic bonds. 5. HF - This compound has hydrogen bonding due to the F-H bond, in addition to London dispersion forces and dipole-dipole interactions.

We can now compare the boiling points of the given compounds based on the intermolecular forces present: 1. \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) have the weakest interactions, thus having lower boiling points. 2. \(\mathrm{HCl}\) and HF have stronger intermolecular forces (dipole-dipole and hydrogen bonding), resulting in higher boiling points. 3. NaF has the highest boiling point due to the strong ionic bonds. Out of all the compounds, \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) have interactions similar to argon, which has only London dispersion forces. Since fluorine and chlorine are both in the same group as argon in the periodic table, their forces will have similar strengths, making their boiling points quite close.

By comparing the intermolecular forces and boiling points of the given compounds, we can conclude that both \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) have boiling points similar to that of argon due to their similar London dispersion forces. However, \(\mathrm{Cl}_{2}\) exhibits a higher boiling point, which is closer to argon's boiling point. Thus, the compound with a boiling point closest to that of argon is \(\mathrm{Cl}_{2}\).