Question

Bond energy of $ F_{2},Cl_{2},Br_{2},I_{2} $ follows the order:

• A. $ Br_{2} > F_{2} > Cl_{2} > I_{2} $
• B. $ F_{2} > Cl_{2} > Br_{2} > I_{2} $
• C. $ I_{2} > F_{2} > Br_{2} > Cl_{2} $
• D. $ F_{2} > I_{2} > Br_{2} > Cl_{2} $

Answer 1

Answer: (B)

Bond energy of $F_{2}, C l_{2}, B r_{2}$ and $I_{2}$ are as:
$F-F \rightarrow 150.6\, mol ^{-1}(k J)$
$C l-C l \rightarrow 242.7 \,mol ^{-1}(k J) $
$B r-B r \rightarrow 188.3 \,mol ^{-1}(k J)$
$I - I \rightarrow 150.6\, mol ^{-1}(k J)$
Bond energy of $F_{2}$ is lower because of repulsion between non bonded electrons of small sized $F$ atom.
So the order is $F_{2} < C l_{2}>B r_{2}>I_{2}$