Question

The percentage of p-character in each orbital of the central atom used for bonding in $N{H}_3$ is:

• A. 25%
• B. 75%
• C. >75%
• D. 33.3%

Answer 1

The correct option is C >75%

As we know that the hybridization of $N$ in $N{H}_3$ is $s{p}^3,$ the percent $p$ character must be $3/4\times 100=75\%$.

This happens when the structure doesn't have any lone pairs and it is a simple tetrahedral structure. But, here it has a pyramidal shape.

And due to lone pair-bond pair repulsion, the bond angle decreases from an ideal 109.5 degree to 107 degree.

As bond angle decreases, % s character also decreases. But % p character increases. As a result of which here % p will be more than 75% and % s will be less than 25%.