Suggest reasons why the $B-F$ bond lengths in $BF $$_{3}(130 \,pm )$ and $B F_{4}^{-}(143\, pm )$ differ.
Suggest reasons why the $B-F$ bond lengths in $BF $$_{3}(130 \,pm )$ and $B F_{4}^{-}(143\, pm )$ differ.
The $B-F$ bond length in $BF $$_{3}$ is shorter than the $B-F $ bond length in $B F_{4}^{-} $. $BF _{3}$ is an electron- deficient species. With a vacant $p$ -orbital on boron, the fluorine and boron atoms undergo $p$$n$$-pn$ back-bonding to remove this deficiency. This imparts a double bond character to the $B-F$ bond.
This double-bond character causes the bond length to shorten in $BF_3$ $(130\, pm ) .$ However, when $BF$ $_{3}$ coordinates with the fluoride ion, a change in hybridisation from $s p^{2}$ (in $BF _{3}$ ) to $s p^{3}\left(\text { in } B F_{4}^{-}\right)$ occurs. Boron now forms $4 \sigma$ bonds and the double-bond character is lost. This accounts for a $B-F$ bond length of $143\, pm$ in $B F_{4}^{-}$ ion.
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$(I)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}Si - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
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\end{array}$ $(II)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}C - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{(C{H_3})}_3}}
\end{array}$
Compare $\pi - $ bond strength between $B$ and $N$ given in two compounds
$(I)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}Si - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Si{{(C{H_3})}_3}}
\end{array}$ $(II)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}C - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{(C{H_3})}_3}}
\end{array}$