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Q.
The percentage of $s$-character in the hybrid orbitals of nitrogen in $NO^+_2, NO^-_3$ and $NH_4^+$ respectively are
KCETKCET 2020Chemical Bonding and Molecular Structure
Solution:
In $NO _{2}{ }^{+}$, the steric number of $N$ is two (two sigma bonds). Thus, it is sp hybridized. Hence, the percentage s-character is $=1 / 2 \times 100=50 \%$.
In $O _{3}^{-}$, the steric number of $N$ is three (three sigma bonds). Thus, it is $sp ^{2}$ hybridized. Hence the percentage s-character is $=1 / 3 \times 100=33.33 \%$.
In $NH _{4}^{+}$, the steric number of $N$ is four (four sigma bonds). Thus, it is $sp ^{3}$ hybridized.
Hence, the percentage s-character is $=1 / 4 \times 100=25 \%$