Question

The bond dissociation energy of $B-F$ in $B{F}_3$ is 646 $kJ$ $mo{l}^{-1}$ whereas that of $C-F$ in $C{F}_4$ is 515 $kJmo{l}^{-1}$. 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 $0$ bond between $B$ and $F$ in $B{F}_3$ as compared 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 $B{F}_3$ than that between $C$ and $F$ in $C{F}_4$

Answer 1

The correct option is 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 $C{F}_4$

Boron in $B{F}_3$ has a vacant p–orbital, allowing back bonding while carbon in $C{F}_4$ has no vacant orbital, so no back bonding is feasible. Thus, B–F bond is stronger than $C{F}_4$.Due to small size and higher electronegativity, carbon has a strong tendency to form multiple bonds either with itself or with the other atoms of similar size such as oxygen and nitrogen.

$d\pi -p\pi$ bonding is the formation of a π molecular orbital by the overlap of a d orbital on one atom with a $p$ or $p^{\ast }$ orbital on another atom.