Suggest, giving reasons, which substance in each pair is likely to have the higher boiling point: (a) \(\mathrm{H}_{2} \mathrm{~S}\) or \(\mathrm{H}_{2} \mathrm{O}\); (b) \(\mathrm{NH}_{3}\) or \(\mathrm{PH}_{3}\); (c) \(\mathrm{KBr}\) or \(\mathrm{CH}_{3} \mathrm{Br}\); (d) \(\mathrm{CH}_{4}\) or \(\mathrm{SiH}_{4}\) *

Intermolecular forces are the forces of attraction or repulsion which act between neighboring particles (atoms, molecules, or ions). These forces are responsible for the various physical properties of compounds, including boiling points. The stronger these interactions are, the more energy will be required to overcome them, resulting in a higher boiling point. Fundamental types of intermolecular forces include ionic bonds, hydrogen bonds, and London dispersion forces, each with varying strengths and implications on boiling points.

For instance, the comparison between \(\text{H}_2\text{S}\) and \(\text{H}_2\text{O}\) hinges on the presence of stronger hydrogen bonds in water due to high electronegativity differences, hence a higher boiling point.

Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom is bound to a highly electronegative atom (such as oxygen, nitrogen, or fluorine) and is attracted to another electronegative atom. This form of bonding is significantly stronger than other dipole-dipole interactions and is a key factor in determining the physical properties of molecules.

In the solution of comparing \(\text{NH}_3\) to \(\text{PH}_3\), \(\text{NH}_3\) exhibits hydrogen bonding because nitrogen forms a strong partial charge with hydrogen, resulting in a high boiling point. \(\text{PH}_3\), however, lacks these strong interactions, which explains its lower boiling point.

Ionic bonds are formed when an electron is transferred from one atom to another, resulting in the creation of ions that are electrically charged. These ions then attract each other to form a solid structure. Ionic compounds, such as salts, have high melting and boiling points because the ionic bonds are much stronger than other intermolecular forces, requiring significant amounts of energy to dissociate.

In our exercise, \(\text{KBr}\) is an ionic compound and has a substantially higher boiling point compared to \(\text{CH}_3\text{Br}\), which is a molecular compound with weaker intermolecular forces.

Electronegativity refers to the ability of an atom to attract shared electrons when forming a compound. The greater the difference in electronegativity between atoms in a bond, the more polar the bond becomes. This polarity can lead to stronger intermolecular attractions, such as dipole-dipole interactions or hydrogen bonds. As a result, molecules with highly electronegative atoms generally display higher boiling points due to these stronger intermolecular forces.

This property is essential when analyzing why \(\text{H}_2\text{O}\) has a higher boiling point than \(\text{H}_2\text{S}\), as oxygen is more electronegative than sulfur, leading to strong hydrogen bonds in water.

London dispersion forces are the weakest intermolecular forces and arise from temporary dipoles created when electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. These forces may be weak individually, but when many atoms are involved, they can exert a substantial collective force. Substances with larger molar masses and more electrons will typically have stronger London dispersion forces and therefore higher boiling points.

This concept is illustrated in the comparison between \(\text{CH}_4\) and \(\text{SiH}_4\). Despite both being non-polar and having dispersion forces, the larger molar mass and electron cloud of \(\text{SiH}_4\) results in higher boiling point than that of \(\text{CH}_4\).