Question

Consider the following statements:
Statement-I: H$_2$Se is more acidic than H$_2$Te.
Statement-II: H$_2$Se has higher bond dissociation enthalpy than H$_2$Te.
In light of the above statements, choose the correct option:

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

Hint:

When comparing the acidity of hydrides in Group 16, remember that as you move down the group, the bond dissociation enthalpy decreases and acidity increases due to the larger atomic size.

Answer 1

The Correct Option is D

Step 1: Analyzing Statement-I (Acidity of H\(_2\)Se and H\(_2\)Te)
The acidity of hydrides of Group 16 elements increases as we move down the group because the bond between hydrogen and the chalcogen becomes weaker due to increasing atomic size.
This results in easier dissociation of the proton (H\(^+\)) and thus stronger acidity.
Therefore, H\(_2\)Te (hydride of tellurium) should be more acidic than H\(_2\)Se (hydride of selenium) since Te is larger than Se. Thus, Statement-I is false, because H\(_2\)Te is more acidic than H\(_2\)Se.
Step 2: Analyzing Statement-II (Bond Dissociation Enthalpy)
Bond dissociation enthalpy is the energy required to break a bond.
As the size of the central atom increases, the bond dissociation enthalpy generally decreases.
For H\(_2\)Se and H\(_2\)Te, since Te is larger than Se, the H-Te bond will be weaker than the H-Se bond, meaning H\(_2\)Se will have a higher bond dissociation enthalpy compared to H\(_2\)Te. Thus, Statement-II is also false, because H\(_2\)Se has a higher bond dissociation enthalpy than H\(_2\)Te.
Step 3: Conclusion
Both statements are false.
Therefore, the correct option is (4).