The correct answer is 2. C > A > B > D > E.
Group 16 elements, also known as chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). These elements can form hydrides of the type \(H_2E\), where E represents the central atom from this group.
The thermal stability of these hydrides is influenced by the strength of the \(H-E\) bond. As you move down the group, several key factors change:
1. Atomic Size: The atomic size increases as you move down the group. Larger atoms have more diffuse electron clouds, making it easier for them to accommodate additional electrons. This results in weaker bonding between hydrogen and the central atom.
2. Electronegativity: Electronegativity is a measure of an element's ability to attract electrons. As you move down the group, electronegativity decreases. Oxygen is the most electronegative element in this group, while polonium is the least electronegative. Lower electronegativity means the central atom is less effective at attracting and holding onto electrons, leading to weaker bonds with hydrogen.
3. Bond Strength: Weaker \(H-E\) bonds result in lower bond dissociation energies. A lower bond dissociation energy means it takes less energy to break the \(H-E\) bond and release hydrogen gas. Hydrides with weaker bonds are less thermally stable because they decompose more readily at lower temperatures.
Based on these factors, we can understand the trend in thermal stability of the group 16 hydrides:
\(H_2O\) (Water): Oxygen is the most electronegative element and the smallest in this group, resulting in a strong \(H-O\) bond. This makes water the most thermally stable hydride in the group.
\(H_2S\) (Hydrogen Sulphide): Sulphur is larger than oxygen and less electronegative, resulting in a weaker \(H-S\) bond compared to \(H-O\). Thus, \(H_2S\) is less thermally stable than \(H_2O\).
\(H_2Se\) (Hydrogen Selenide): As you move down the group, the trend continues. Selenium is larger and less electronegative than sulfur, leading to an even weaker \(H-Se\) bond compared to \(H-S\).
\(H_2Te\) (Hydrogen Telluride): Tellurium is larger and less electronegative than selenium, resulting in a weaker \(H-Te\) bond compared to \(H-Se\). Thus, \(H_2Te\) is less thermally stable than \(H_2Se\).
\(H_2Po\) (Hydrogen Polonide): Polonium, being the largest and least electronegative element in the group, forms the weakest \(H-Po\) bond. Consequently, \(H_2Po\) is the least thermally stable hydride in the group.
So, the trend in thermal stability of group 16 hydrides is a direct consequence of changes in atomic size and electronegativity as you move down the group, which affects the strength of the \(H-E\) bonds. |
