Concept:Bond energy decreases as the size of the bonding atoms increases, because a longer bond distance results in weaker electrostatic attraction between the nuclei and the shared electrons.
Formula:$$ \text{Bond Energy Trend: HF} > \text{HCl} > \text{HBr} > \text{HI} $$
Solution:- Iodine is the largest halogen in the given options, located at the bottom of Group 7.
- When Iodine bonds with Hydrogen, the resulting H-I bond is very long.
- Because the shared electron pair is so far from the Iodine nucleus (and shielded by many inner electron shells), the bond is extremely weak.
- Therefore, HI requires the least amount of energy to break (lowest bond energy).
Why other options are incorrect:- Option A, Option C, Option D: Fluorine, Chlorine, and Bromine are smaller atoms, forming shorter and progressively stronger bonds.
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