Compare acidity of group 16 hydrides

The hydrides are: $H_2O,\, H_2S,\, H_2Se,\, H_2Te$

Down the group, acidity increases because the $E-H$ bond ($E =$ chalcogen) becomes weaker as atomic size increases.

Thus, $H_2O < H_2S < H_2Se < H_2Te$ in acidic strength.

Therefore, $H_2Te \text{ is more acidic than } H_2Se$

So Statement I: "$H_2Se$ is more acidic than $H_2Te$" is false.

Compare bond enthalpy of dissociation

Down group 16, size of the central atom increases, overlap between $1s$ orbital of H and the valence orbital of the chalcogen decreases, so the $H-E$ bond becomes weaker.

Hence bond enthalpy decreases: $H-O > H-S > H-Se > H-Te$

Therefore, $\text{bond enthalpy}(H-Se) > \text{bond enthalpy}(H-Te)$

So Statement II: "$H_2Se$ has higher bond enthalpy for dissociation than $H_2Te$" is true.

Evaluate the options

• A: Both true  Incorrect
• B: I true, II false  Incorrect
• C: Both false  Incorrect
• D: I false, II true $\checkmark$ Correct

Conclusion

The correct option is: $\boxed{D}$

Comparison with stored answer

Stored correct answer: D

My derived answer also is D, so they agree.