Question

Which among the following hydrides has low bond dissociation energy of M–H bond (M = central atom)?

• A. H\(_2\)Se
• B. H\(_2\)S
• C. H\(_2\)Te
• D. H\(_2\)O

Hint:

Weaker bonds are formed when larger atoms are involved due to poor orbital overlap.

Answer 1

The Correct Option is C

Step 1: Understand bond dissociation energy trend.
Bond dissociation energy decreases as bond length increases.
Step 2: Analyze group 16 hydrides.
Down the group from oxygen to tellurium, atomic size increases and M–H bond length increases.
Step 3: Compare given hydrides.
\[ \text{O–H}>\text{S–H}>\text{Se–H}>\text{Te–H} \]
Thus, H\(_2\)Te has the weakest M–H bond.
Step 4: Conclusion.
H\(_2\)Te has the lowest bond dissociation energy.