Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Key Idea For ${\mathrm{sp}}^2$ hybridisation, there must be $3\sigma$-bonds or $2\sigma$-bonds along with a lone pair of electrons. (i) ${\mathrm{NO}}_2^{-}\Rightarrow 2\sigma +1\mathrm{lp}=3$, ie, ${\mathrm{sp}}^2$ hybridisation (ii) ${\mathrm{NH}}_3\Rightarrow 3\sigma +1\mathrm{lp}=4$, ie, ${\mathrm{sp}}^3$ hybridisation (iii) ${\mathrm{BF}}_3\Rightarrow 3\sigma +0\mathrm{lp}=3$, ie, ${\mathrm{sp}}^2$ hybridisation (iv) ${\mathrm{NH}}_2^{-}\Rightarrow 2\sigma +2\mathrm{lp}=4$, ie, ${\mathrm{sp}}^3$ hybridisation (v) ${\mathrm{H}}_2\mathrm{O}\Rightarrow 2\sigma +2\mathrm{lp}=4$, ie, ${\mathrm{sp}}^3$ hybridisation Thus, among the given pairs, only ${\mathrm{BF}}_3$ and ${\mathrm{NO}}_2^{-}$ have ${\mathrm{sp}}^2$ hybridisation.