Question

Consider the following statements:
Statement I: $H_2Se$ is more acidic than $H_2Fe$
Statement II: $H_2Se$ has higher bond dissociation enthalpy
In light of the above statements, choose the correct option.

• A. Statement-I is true & Statement-II is false
• B. Statement-I is false & Statement-II is true
• C. Both Statement-I & Statement-II are true
• D. Both Statement-I & Statement-II are false

Hint:

When comparing the acidities of compounds, consider the size of the central atom and the bond dissociation energies. Larger atoms tend to form weaker bonds with hydrogen, increasing acidity.

Answer 1

The Correct Option is A

Let's evaluate both statements one by one:
Statement I: \( H_2Se \) is more acidic than \( H_2Fe \):** \( H_2Se \) is more acidic than \( H_2Fe \) because selenium is in the same group as oxygen but below sulfur in the periodic table. As we move down a group, the bond strength between the hydrogen and the element (e.g., Se-H) decreases, making it easier to lose the proton (H\(^+\)) and thus increasing acidity. On the other hand, in \( H_2Fe \), iron has a much stronger bond with hydrogen, making it less acidic. Hence, Statement I is **true**.
Statement II: \( H_2Se \) has higher bond dissociation enthalpy:** Bond dissociation enthalpy increases with increasing bond strength.
Since Se-H bonds in \( H_2Se \) are weaker than Fe-H bonds in \( H_2Fe \), \( H_2Se \) has lower bond dissociation enthalpy. Thus, Statement II is **false**.
Therefore, the correct answer is (1) Statement-I is true & Statement-II is false.