Which member in each pair has the stronger intermolecular dispersion forces: (a) \(\mathrm{Br}_{2}\) or \(\mathrm{O}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH},(\mathrm{c}) \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl} ?\)

Short Answer

Expert verified
In each pair, the molecule with the stronger intermolecular dispersion forces are: (a) \(\mathrm{Br}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\), and (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\). This is due to their larger size, electron cloud, and polarizability compared to their counterparts.

Step by step solution

01

Pair (a) comparison: \(\mathrm{Br}_{2}\) vs \(\mathrm{O}_{2}\)

Since both molecules are homonuclear diatomic molecules, the primary difference between them is their size and electron cloud. A larger electron cloud is more easily polarizable, which leads to stronger dispersion forces. Bromine is larger than oxygen and has more electrons, so its electron cloud is larger. Thus, \(\mathrm{Br}_{2}\) has stronger dispersion forces than \(\mathrm{O}_{2}\).
02

Pair (b) comparison: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) vs \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\)

In this case, both molecules have similar functional groups (Sulfur-containing). The major difference between them is the length of their carbon chains. A longer carbon chain will lead to a larger surface area and polarization, thus resulting in stronger dispersion forces. Therefore, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) has stronger dispersion forces than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\).
03

Pair (c) comparison: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) vs \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\)

For this pair, both molecules have chlorine atoms, so the primary difference in their dispersion forces comes from the size and shape of the molecules. The molecule \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) has a more compact shape and two methyl groups, which make it more polarizable. Therefore, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) has stronger dispersion forces than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\).

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