Question

Which from following molecules exhibits highest acidic nature?

• A. $H_2O$
• B. $H_2S$
• C. $H_2Se$
• D. $H_2Te$

Hint:

Logic Tip: For binary hydrides within a specific \textit{group}, bond strength (size of central atom) dominates over electronegativity in determining acidity. Larger atom $\rightarrow$ longer, weaker bond $\rightarrow$ stronger acid. (Note: For hydrides across a \textit{period}, electronegativity dominates).

Answer 1

The Correct Option is D

Concept:
The acidic character of hydrides of Group 16 elements ($H_2O, H_2S, H_2Se, H_2Te$) depends on how easily they can release a proton ($H^+$) in an aqueous solution. This ease of proton release is governed by the bond dissociation enthalpy of the $H-E$ bond (where E is the Group 16 element).
Step 1: Analyze the periodic trend for atomic size.
As we move down Group 16 in the periodic table: $$\text{O} \rightarrow \text{S} \rightarrow \text{Se} \rightarrow \text{Te}$$ The atomic size of the central atom increases significantly due to the addition of new electron shells.
Step 2: Relate atomic size to bond strength and acidity.
Because the size of the central atom ($E$) increases down the group, the bond length between Hydrogen and the central atom ($H-E$) also increases. A longer bond is a weaker bond, meaning the bond dissociation enthalpy decreases: $$\text{Bond strength: } H-O>H-S>H-Se>H-Te$$ Since the $H-Te$ bond is the weakest, it requires the least amount of energy to break, making it the easiest to release an $H^+$ ion.
Step 3: Determine the order of acidity.
The easier it is to lose a proton, the stronger the acid. Therefore, the acidic character increases down the group: $$H_2O<H_2S<H_2Se<H_2Te$$ $H_2Te$ exhibits the highest acidic nature among the given options.