Question

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

\(\mathrm{SF}_{6}\) 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.

Key concepts

Volatility of Compounds

Volatility refers to how easily a substance can turn into vapor. It depends on the forces between the molecules. The stronger these forces, the less volatile the substance. In the case of compounds like \( \text{H}_2\text{S} \) and \( \text{H}_2\text{O} \), the volatility difference is notable. Hydrogen sulfide (\( \text{H}_2\text{S} \)) is more volatile than water (\( \text{H}_2\text{O} \)).

• Hydrogen Bonding in Water: In water, there are strong hydrogen bonds. These bonds form because oxygen is highly electronegative, creating a strong attraction between hydrogen and oxygen atoms of different water molecules. This gives water a lower volatility.
• Weaker Interactions in Hydrogen Sulfide: In contrast, \( \text{H}_2\text{S} \) lacks these strong hydrogen bonds. It mainly has weaker dipole-dipole interactions, making it vaporize more easily, thus making it more volatile.

The strength of these intermolecular forces directly impacts the boiling point and evaporation rate, explaining why some compounds are more volatile than others.

Oxygen and Sulfur Compounds

Oxygen and sulfur, both belonging to group 16 of the periodic table, exhibit different properties in their common compounds due to their inherent atomic characteristics.

• Oxygen (\( \text{O}_2 \)): Oxygen prefers forming double bonds due to its small size and ability to form efficient pi bonds, leading to its diatomic molecular form \( \text{O}_2 \). These stable pi bonds allow oxygen to exist naturally as \( \text{O}_2 \).
• Sulfur (\( \text{S}_8 \)): On the other hand, sulfur forms a more complex structure. Sulfur atoms are larger, making pi bonding less effective. Instead, sulfur forms what is known as a puckered ring structure, or \( \text{S}_8 \). This eight-membered ring is the most stable form of sulfur.

These differences in structure significantly influence the chemical and physical properties of oxygen and sulfur compounds, including their reactivity and stability.

Electronegativity and Bonding

Electronegativity is a crucial concept in understanding the bonding characteristics of elements, including those in group 16 like oxygen and sulfur.

• Electronegativity Trend: Oxygen is more electronegative than sulfur. This means oxygen attracts electrons more strongly in compounds. As a result, oxygen can form strong, stable bonds with other elements, often resulting in polar compounds.
• Sulfur's Larger Size: Although sulfur is less electronegative, it can engage in multiple bonding possibilities due to its larger size, which allows the use of d-orbitals in bonding as seen in \( \text{SF}_6 \). The lack of available d-orbitals in oxygen means it can't form analogous compounds like \( \text{OF}_6 \).

The electronegativity influences the type and strength of bonds formed, dictating the compound's stability and existence.

Periodic Table Trends

Understanding periodic table trends helps us grasp why certain elements behave the way they do. In group 16, these trends are evident in the compounds they form.

• Atomic Size and Complexity: As you move down the group from oxygen to tellurium, the atomic size increases. This results in more complex bonding structures. For instance, tellurium can form \( \text{H}_6\text{TeO}_4 \), while other group 16 elements cannot form similar stable acids. This is because larger atoms like tellurium have less electronegativity and can hold more bonds due to a greater number of available orbitals.
• Bonding Capacity: These trends explain the observed bonding patterns, such as the formation of \( \text{SF}_6 \) but not \( \text{OF}_6 \), due to sulfur's ability to use d-orbitals in bonding.

Periodic trends in electronegativity and atomic size are pivotal in determining the nature and properties of compounds formed by group 16 elements, showcasing the broader patterns across the periodic table.