Chemistry - CHEMICAL BONDING Question with Solution | TestHub

(a)$\begin{array}{cccccc}& {\text{CH}}_4& \to & {\text{CH}}_3& -& {\text{CH}}_3\\ & \text{4 bp}+\text{0 lp}& & \text{4 bp}& & \text{4 bp}\\ \text{Hybridisation}& {\text{sp}}^3& & {\text{sp}}^3& & {\text{sp}}^3\\ \text{Structure}& \text{Tetrahedral}& & \text{Tetrahedral}& & \text{Tetrahedral}\end{array}$


(b)$\begin{array}{cccc}& {\text{NH}}_3& \to & {\text{NH}}_4^{+}\\ & 3\text{bp}+1\text{lp}& & 4\text{bp}\\ \text{Hybridisation}& {\text{sp}}^3& & {\text{sp}}^3\end{array}$
Structure              pyramidal        tetrahedral


(c)$\begin{array}{cccc}& {\text{BF}}_3& \to & {\text{BF}}_4^{-}\\ & 3\text{bp}& & 4\text{bp}\\ \text{Hybridisation}& {\text{sp}}^{\text{2}}& & {\text{sp}}^3\\ \text{Structure}& \text{trigonal}& & \text{tetrahedral planar}\end{array}$


(d)$\begin{array}{cccc}& {\text{H}}_2\text{O}& \to & {\text{H}}_3{\text{O}}^{+}\\ & 2\text{bp}+2\text{lp}& & 3\text{bp}+1\text{lp}\\ \text{Hybridisation}& {\text{sp}}^3& & {\text{sp}}^3\\ \text{Structure}& \text{angular}& & \text{Pyramidal}\end{array}$

Thus conversion of$\mathrm{B}{\mathrm{F}}_3$into${\mathrm{B}\mathrm{F}}_4^{-}$involves change in both hybridisation and shape.