What is the percentage of s-character in the orbital occupied by lone pair of ${\mathrm{NH}}_3$? It is given that : $\cos {107}^{\circ }=-0.292$

The percentage of s-character in the orbital occupied by lone pair of ${\mathrm{NH}}_3$ is $32.2\mathrm{\%}$.

$\begin{array}{l}\cos {107}^{\circ }=\frac{s}{s-1}\\ -0.292=\frac{s}{s-1}\end{array}$

$\Rightarrow s=0.226$ $(\mathrm{in} \mathrm{bond} \mathrm{pair})$

 $\Rightarrow S$ in lone pair $=1-0.226\times 3=0.322\mathrm{\%}$

$\Rightarrow =32.2\mathrm{\%}$

Hence, option B is correct.