Explain the observed trend in the boiling points of these compounds. $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{Te} & -2{ }^{\circ} \mathrm{C} \\ \mathrm{H}_{2} \mathrm{Se} & -41.5{ }^{\circ} \mathrm{C} \\ \mathrm{H}_{2} \mathrm{~S} & -60.7{ }^{\circ} \mathrm{C} \\ \mathrm{H}_{2} \mathrm{O} & 100^{\circ} \mathrm{C} \end{array} $$

Understanding the boiling points of compounds involves looking at periodic trends, which are patterns that exist within the periodic table and can predict various properties of the elements. For instance, if we consider the group of hydrogen chalcogenides presented in the exercise, a clear trend is observed.

As you move down the group on the periodic table—from oxygen (O) to tellurium (Te)—the atomic radius increases. This is because each element down the group has an additional electron shell compared to the one above it. Generally, with an increasing atomic radius and more electron shells, the atoms have a greater number of electrons that can participate in van der Waals forces. This effect causes an increase in the boiling point of the compounds as you go down the group, since more energy is required to break these intermolecular attractions during the phase change from liquid to gas.

However, this trend does not apply to water (H2O), which has a much higher boiling point than would be expected from its position on the periodic table. This anomaly is explained by hydrogen bonding, a stronger type of intermolecular force present in water.

The van der Waals forces are a type of intermolecular attraction that can affect the boiling points of compounds. They include attractions such as dipole-dipole interactions and London dispersion forces. The latter are especially significant in non-polar molecules, or molecules where the difference in electronegativity between the atoms is not enough to form significant dipoles.

London dispersion forces arise from the temporary fluctuations in electron distribution within molecules, which create temporary dipoles and induce dipoles in neighboring molecules. These forces are present in all molecules, whether polar or non-polar, but they are the only attractive forces in non-polar molecules. As the molecular weight of a compound increases, the number of electrons also increases, which leads to stronger London dispersion forces, and as a result, an increase in the boiling point of the compounds, as seen in the case of the hydrogen chalcogenides from H2S to H2Te.

Hydrogen bonding represents a special case of dipole-dipole interaction that occurs when a hydrogen atom covalently bonded to a highly electronegative atom, such as nitrogen, oxygen, or fluorine, is attracted to a lone pair of electrons on another electronegative atom. These bonds are much stronger than ordinary van der Waals forces and significantly influence boiling points.

In the case of water (H2O), each molecule can form up to four hydrogen bonds—with two bonds formed through its hydrogen atoms and two through the lone pairs on oxygen. This network of hydrogen bonds requires a substantial amount of energy to break, resulting in water's anomalously high boiling point compared to other group 16 hydrides.

Another key factor influencing boiling points is molecular weight. Molecules with a higher molecular weight typically have more electrons, creating stronger London dispersion forces through increased instantaneous dipole-inducing interactions. This correlation is evident as heavier molecules tend to have higher boiling points under similar circumstances.

However, it's crucial to recognize that molecular weight is just one aspect. The structure and shape of the molecule and the strength of intermolecular forces such as hydrogen bonding or ionic interactions can also play a significant role and sometimes can override the influence of molecular weight. As observed in the provided exercise, H2S, H2Se, and H2Te follow the expected trend, where increased molecular weight correlates with higher boiling points, while water, due to its capability for hydrogen bonding, diverges from this pattern and exhibits a much higher boiling point than what molecular weight alone would suggest.