Q29 / 150 · MHT CET 2026 (16 April Shift 1) · Chemistry
Identify increasing order of acidity for following diprotic acids in aqueous solutions.
- Correct answerA$\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}$
- B$\text{H}_2\text{Se} < \text{H}_2\text{S} < \text{H}_2\text{Te}$
- C$\text{H}_2\text{Te} < \text{H}_2\text{S} < \text{H}_2\text{Se}$
- D$\text{H}_2\text{Se} < \text{H}_2\text{Te} < \text{H}_2\text{S}$
Solution
The given compounds are hydrides of Group 16 elements.
As we move down the group from Sulphur (S) to Tellurium (Te), the atomic size of the central atom increases.
This leads to an increase in the H-E bond length (where E is S, Se, Te) and a corresponding decrease in the bond dissociation enthalpy.
Consequently, the H-E bond becomes weaker, making the release of an $\text{H}^{+}$ ion easier.
Therefore, the acidic strength increases down the group.
The increasing order of acidity is $\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}$.
Answer: $\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}$
As we move down the group from Sulphur (S) to Tellurium (Te), the atomic size of the central atom increases.
This leads to an increase in the H-E bond length (where E is S, Se, Te) and a corresponding decrease in the bond dissociation enthalpy.
Consequently, the H-E bond becomes weaker, making the release of an $\text{H}^{+}$ ion easier.
Therefore, the acidic strength increases down the group.
The increasing order of acidity is $\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}$.
Answer: $\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}$
Hence, the correct answer is option: 'A'.