question_answer

The bond dissociation energy of B-F in \[B{{F}_{3}}\]is \[646\,kJ\,mo{{l}^{-\,1}}\]. The correct reason For higher B-F bond dissociation energy as compared to that of C-F is

A) smaller size of B-atom as compared to that of C-atom.

B) stronger or-bond between B and F in \[B{{F}_{3}}\] to that between C and F in \[C{{F}_{4}}.\]

C) significant \[p\pi -p\pi \] interaction between B and F in \[B{{F}_{3}}\]whereas there is no possibility of such interaction between C and F in \[C{{F}_{4}}.\]

D) lower degree of \[p\pi -p\pi \] interaction between B and F in \[B{{F}_{4}}\] than that between C and F in \[C{{F}_{4}}.\]

Correct Answer: C

Solution :

[c] In \[B{{F}_{3}}\] there is significant \[p\pi -p\pi \] interaction (back bonding) between unshared p-orbital (having no electron) of boron and the lone pair of electron over fluorine in 2p-orbital. This provides extra stability and thus bond dissociation energy of \[B-F\] in \[B{{F}_{3}}\] is more than \[C-F\] in \[C{{F}_{4}},\] where no such back bonding exist.