Question

The bond dissociation energy of $B-F$ in $B{{F}_{3}}$ is $646\text{ }kJmo{{l}^{-1}}$ whereas that of $C-F$ in $C{{F}_{4}}$ is $515\text{ }kJmo{{l}^{-1}}$. 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 $\sigma $- bond between $B$ and $F$ in $B{{F}_{3}}$ as compared to 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

Hint: Think about the electronic configurations of both boron and carbon and how they are hybridized to form bonds with fluorine. Take note of any empty orbitals.

Complete step by step solution:
In $B{{F}_{3}}$ boron is $s{{p}^{2}}$ hybridized since it needs to accommodate electrons from 3 fluorine atoms. The configuration of boron will be:
i) Ground state

ii) Hybridized state

Here, we can see that boron undergoes $s{{p}^{2}}$ hybridization and 1 $2p$ orbital remains empty.
Now, let us look at carbon, and the hybridization it undergoes to form 4 $\sigma $- bonds with fluorine. The electronic configuration of carbon is:


Here, we can see that carbon undergoes $s{{p}^{3}}$ hybridization to accommodate fluorine.
Now to compare both these structures, we need to take into consideration the concept of back-bonding. Back-bonding occurs when lone pairs of an atom interact with empty orbitals of the atom bonded with it.
In this example, the lone pairs on the fluorine atom can undergo $p\pi -p\pi $ interaction with the empty $2p$ orbital that is present in boron. Such a phenomenon will not occur in carbon as all the orbitals are hybridized and filled. This back-bonding has a $\pi $- bond nature as a sigma bond is already present between boron and fluorine and makes the $B-F$ bond harder to break. This results in a higher energy of dissociation of the bond than other bonds like $C-F$ that do not have back-bonding.

Hence, the answer 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 $C{{F}_{4}}$’

Note: We know that size decreases along a period so carbon is smaller than boron and will have a stronger bond in absence of back-bonding. The $\sigma $- bond in both the molecules is of the same type and any change in bond dissociation energy will not be seen due to that.