Chapter 15: Problem 3

Give reasons for the following. (a) \(\mathrm{SF}_{6}\) is known but \(\mathrm{OF}_{6}\) is not known. (b) \(\mathrm{H}_{2} \mathrm{~S}\) is more volatile than \(\mathrm{H}_{2} \mathrm{O}\). (c) Oxygen exists as \(\mathrm{O}_{2}\) but sulphur exists as \(\mathrm{S}_{8}\) (d) \(\mathrm{H}_{6} \mathrm{TeO}_{4}\) is known but the corresponding acids for the other Group 16 elements are not known.

Short Answer

Expert verified

\(\mathrm{SF}_{6}\) exists due to sulfur's ability to form six stable bonds, while \(\mathrm{OF}_{6}\) does not exist because oxygen can only form two stable bonds. \(\mathrm{H}_{2}\mathrm{S}\) is more volatile than \(\mathrm{H}_{2}\mathrm{O}\) because it has weaker intermolecular forces. Oxygen exists as \(\mathrm{O}_{2}\) due to the ability of oxygen to form pi bonds, whereas sulfur exists as \(\mathrm{S}_{8}\) because it has a more stable eight atom structure. \(\mathrm{H}_{6}\mathrm{TeO}_{4}\) exists while other similar molecules do not due to the size and electronegativity of tellurium.

Step by step solution

\(\mathrm{SF}_{6}\) versus \(\mathrm{OF}_{6}\)

Fluorine is the most electronegative element, and will form six stable bonds with sulfur, itself being in the third period of the periodic table, having d-orbitals available. Oxygen, however, is also very electronegative - even more so than sulfur - and as it's in the second period, it does not have d-orbitals. Therefore, oxygen cannot form more than two bonds stably, making \(\mathrm{OF}_{6}\) non-existent.

\(\mathrm{H}_{2}\mathrm{S}\) versus \(\mathrm{H}_{2}\mathrm{O}\)

The volatility of a molecule depends on the strength of the forces between the molecules. In \(\mathrm{H}_{2}\mathrm{O}\), the oxygen atom forms two polar covalent bonds with hydrogen, resulting in a significant difference in electronegativity and creating strong hydrogen bonds. On the other hand, \(\mathrm{H}_{2}\mathrm{S}\) has weaker dipole-dipole interactions and lacks hydrogen bonding. Therefore, \(\mathrm{H}_{2}\mathrm{S}\) is more easily vaporized and is more volatile than \(\mathrm{H}_{2}\mathrm{O}\).

\(\mathrm{O}_{2}\) versus \(\mathrm{S}_{8}\)

This is determined by the atom's ability to form pi bonds. Oxygen atoms can form two pi bonds, which allows them to exist as diatomic molecules (\(\mathrm{O}_{2}\)). Sulphur atoms have heavier atomic masses and larger atomic sizes, making pi bonding less efficient. Therefore, sulphur exists as an eight-atom ring structure \(\mathrm{S}_{8}\), a more stable configuration.

\(\mathrm{H}_{6}\mathrm{TeO}_{4}\) versus Corresponding Group 16 Acids

The reason that \(\mathrm{H}_{6}\mathrm{TeO}_{4}\) exists while corresponding acids for other Group 16 elements do not is due to the size and electronegativity of the central atom. The central atom in \(\mathrm{H}_{6}\mathrm{TeO}_{4}\) is Tellurium, a large atom with lower electronegativity, which allows for stable bonding. On the contrary, other Group 16 elements are smaller and have higher electronegativities, making stable bonding less likely.

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Short Answer

Step by step solution

\(\mathrm{SF}_{6}\) versus \(\mathrm{OF}_{6}\)

\(\mathrm{H}_{2}\mathrm{S}\) versus \(\mathrm{H}_{2}\mathrm{O}\)

\(\mathrm{O}_{2}\) versus \(\mathrm{S}_{8}\)

\(\mathrm{H}_{6}\mathrm{TeO}_{4}\) versus Corresponding Group 16 Acids

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