Which member in each pair has the stronger intermolecu- lar dispersion forces? (a) \(\mathrm{H}_{2} \mathrm{O}\) or $\mathrm{CH}_{3} \mathrm{OH},\( (b) \)\mathrm{CBr}_{3} \mathrm{CBr}_{3}$ or (c) \(\mathrm{C}\left(\mathrm{CH}_{3}\right)_{4}\) or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ \(\mathrm{CCl}_{3} \mathrm{CCl}_{3}\)

Both \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) are polar molecules with hydrogen bonding and dipole-dipole interactions. However, we need to compare their dispersion forces. \(\mathrm{H}_{2} \mathrm{O}\) has a smaller size and less polarizable electron cloud compared to \(\mathrm{CH}_{3} \mathrm{OH}\), which has an additional \(\mathrm{CH}_3\) (methyl) group. The presence of the one more methyl group in methanol makes its electron cloud more polarizable; therefore, the dispersion forces are stronger in \(\mathrm{CH}_{3} \mathrm{OH}\).

Comparing \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) and \(\mathrm{CCl}_{3} \mathrm{CCl}_{3}\), we can see that both have a similar shape but have different halogens attached to the carbon atoms. Bromine atoms are larger and more polarizable than chlorine atoms, and therefore, the dispersion forces will be stronger in \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\).

Both \(\mathrm{C}\left(\mathrm{CH}_{3}\right)_{4}\) (neopentane) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (pentane) are nonpolar hydrocarbons. Neopentane has a compact, spherical shape, while pentane has a linear, extended shape. The extended shape of pentane allows it to have a larger surface area in contact with neighboring molecules, which leads to stronger dispersion forces compared to the more spherical neopentane. So, the molecules with stronger intermolecular dispersion forces in each pair are: - Pair (a): \(\mathrm{CH}_{3} \mathrm{OH}\) - Pair (b): \(\mathrm{CBr}_{3} \mathrm{CBr}_{3}\) - Pair (c): \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)