Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Calculate the electron pair.

$\mathrm{E}.\mathrm{P}=\frac{\mathrm{Valence} \mathrm{electron} \mathrm{of} \mathrm{central} \mathrm{atom} + \mathrm{monovalent} \mathrm{atom} + \mathrm{negative} \mathrm{charge}- \mathrm{positive} \mathrm{charge}}{2}$

For ${\mathrm{NO}}_3^{-}$

$\mathrm{E}.\mathrm{P}=\frac{5+1}{2}= 3$

It means there is the requirement of the three orbitals for the hybridisation.

Therefore the hybridisation will be ${\mathrm{sp}}^2$
$\mathrm{s}{\mathrm{p}}^2$ (triangular planer)