Question

The percentage of $s$-character in the hybrid orbitals of nitrogen in $N{O}_2^{+},N{O}_3^{-}$ and $N{H}_4^{+}$ 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)

In $N{O}_2{}^{+}$, 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 $s{p}^2$ hybridized. Hence the percentage s-character is $=1/3\times 100=33.33\%$.

In $N{H}_4^{+}$, the steric number of $N$ is four (four sigma bonds). Thus, it is $s{p}^3$ hybridized.

Hence, the percentage s-character is $=1/4\times 100=25\%$