Question

Even though nitrogen in ammonia is in ${sp}^3$ hybridization, the bond angle deviate from ${109}^o{28}^{{}^{\prime }}$. Explain.

Answer 1

Ammonia $\left(N{H}_3\right)$

$N$ is $S{P}^3$ hybridised

$\therefore$ the bold angle ($\theta$)

should be ${109}^o{28}^{\prime }$

But due to lone pair Bond pair repulsion. It becomes less than ${109}^o{28}^{\prime }$