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# Calculate the % s-character in the lone pair electron of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$, where the percentage s-character of bond pair is $22\%$ in each ${\rm{N - H}}$ bond.A. $60\%$B. $25\%$C. $44\%$D. $34\%$

Last updated date: 25th Jun 2024
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Hint: We know that hybridization is generally explained by the overlapping of two or more atomic orbitals and they usually form new atomic orbitals through the pairing of electrons.

We know that, the shape of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ molecule is distorted tetrahedral and the hybridization of this molecule is ${\rm{s}}{{\rm{p}}^{\rm{3}}}$. In the ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ molecule the number of ${\rm{N - H}}$ bond is three and it has one lone pair. The lone pair electrons are those electrons which are generally not shared with any other atom, and sometimes lone pair is termed as unshared electrons or non-bonding electrons.
The number of ${\rm{N - H}}$ bond present in ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ molecule is three. So, the percentage of one bond pair showing s-character is $22\%$, then three bonds showing s-character can be calculated as shown below.
$\begin{array}{c} 1\;{\rm{N - H}} = 22\% \\ 3\;{\rm{N - H}} = 22\% \times 3\\ = 66\% \end{array}$
The % of s-character in the lone pair of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ molecule is $100\% - 66\% = 34\%$.
Thus, the % s-character in the lone pair of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ molecule is $34\%$, when the percentage of s-character due to ${\rm{N - H}}$ bond is $22\%$.
Hence, the correct choice for this question is D that is $34\%$.
The hybridization of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ molecule is ${\rm{s}}{{\rm{p}}^{\rm{3}}}$ and the shape is distorted tetrahedral. So, the percentage character of any orbital can be determined with the help of shape and angle.