Question

The percentage of $s$-character in the hybrid orbitals of nitrogen in $NO^+_2, NO^-_3$ and $NH_4^+$ respectively are

• A. 33.3%, 50%, 25%
• B. 33.3%, 25%, 50%
• C. 50%, 33.3%, 25%
• D. 25%, 50%, 33.3%

Answer 1

Answer: (C)

50%, 33.3%, 25%

Answer 2

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 $=1 / 4 \times 100=25 \%$