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Q. The Hybridisation of atomic orbitals of nitrogen in $\text{NO}_{2}^{+}$ $\text{, NO}_{3}^{-}$ , and $\text{NH}_{4}^{+ \, }$ are?

NTA AbhyasNTA Abhyas 2022

Solution:

The number of hybrid orbitals $=H=\frac{1}{2}\left(\right.V+M-C+A\left.\right)$
V = number of valence electrons
M= number of monovalent atoms(H, X=F,Cl,Br,I etc)
C = number of units of positive charge
A= number of units of negative charge
$\text{NO}_{2}^{+} \Rightarrow \text{H} = \frac{1}{2} \left(5 - 1\right) = 2,$ sp Hybridisation.
$\text{NO}_{3}^{-} \Rightarrow \text{H} = \frac{1}{2} \left(5 + 1\right) = 3,$ sp2 Hybridisation.
$\text{NH}_{4}^{+} \Rightarrow \text{H} = \frac{1}{2} \left(5 + 4 - 1\right) = 4,$ sp3 Hybridisation