The bond dissociation energy of B- F in BF_3 is 646 ,kJ , mol^ -1…

MCQ

The bond dissociation energy of $B- F$ in $BF_3$ is $646 \,kJ\, mol^{-1}$ whereas that of $C - F$ in $CF_4$ is $515\, kJ\, mol^{-1}.$ The correct reason for higher $B- F$ bond dissociation energy as compared to that of $C-F$ is

A
stronger $\sigma$ bond between $B$ and $F$ in $BF_3$ as compared to that between $C$ and $F$ in $CF_4.$
✓
significant $p\pi - p\pi $ interaction between $B$ and $F$ in $BF_3$ whereas there is no possibility of such interaction between $C$ and $F$ in $CF_4.$
C
lower degree of $p\pi - p\pi $ interaction between $B$ and $F$ in $BF_3$ than that between $C$ and $F$ in $CF_4.$
D
smaller size of $B-$ atom as compared to that of $C-$ atom.

✓

Answer

Correct option: B.

significant $p\pi - p\pi $ interaction between $B$ and $F$ in $BF_3$ whereas there is no possibility of such interaction between $C$ and $F$ in $CF_4.$

b
$\mathrm{BF}_{3}$ is a Lewis acid due to incomplete octet of Boron. Among $\mathrm{BF}_{3}, \mathrm{BCl}_{3}, \mathrm{BBr}_{3}$ and so on, B $\mathrm{F}_{3}$ is the weakest Lewis acid due to back bonding of lone pair electron of $F$ to $p$ -orbital of Boron.

B has vacant available p-orbital in $\mathrm{BF}_{3}$ and hence it involves p $\pi$ -p $\pi$ back bonding which is not possible in $\mathrm{CF}_{4}$ as $\mathrm{C}$ does not have any vacant orbital.

Hence, option $\mathrm{B}$ is correct.