Question

Given below are two statements: 
Statement I: $ H_2Se $ is more acidic than $ H_2Te $ 
Statement II: $ H_2Se $ has higher bond enthalpy for dissociation than $ H_2Te $
In the light of the above statements, choose the correct answer from the options given below.

• A. Statement I is false but Statement II is true
• B. Statement I is true but Statement II is false
• C. Both Statement I and Statement II are false
• D. Both Statement I and Statement II are true

Hint:

- Remember the periodic trend: acidity of hydrides increases down the group - Bond enthalpy decreases with increasing atomic size in a group - Larger atoms form weaker bonds, making their hydrides more acidic

Answer 1

The Correct Option is A

To address the question, each statement will be analyzed individually within the context of chemistry, focusing on the properties of chalcogen group hydrides.

1. Statement I: \( H_2Se \) is more acidic than \( H_2Te \)

The acidity of hydrides within Group 16 (chalcogens) increases down the group. The acidity trend for \( H_2\text{X} \) hydrides (where \( \text{X} \) is a chalcogen) is \( H_2O<H_2S<H_2Se<H_2Te \). Consequently, \( H_2Te \) is more acidic than \( H_2Se \) owing to larger atomic size and weaker bond strength, which promotes proton release.

Therefore, Statement I is false.

1. Statement II: \( H_2Se \) has a higher bond enthalpy for dissociation than \( H_2Te \)

Bond enthalpy generally decreases down a group as atomic size increases, leading to weaker bonds. Thus, \( H_2Se \), positioned higher in the group than \( H_2Te \), possesses a greater bond enthalpy (stronger bonds) compared to \( H_2Te \).

Consequently, Statement II is true.

Based on the preceding analysis, the correct conclusion is that Statement I is false and Statement II is true.