Question

What is the hybridization of $N$ in $N(C{H}_3{)}_3$ and $N(Si{H}_3{)}_3$?

• A. $s{p}^2\text{and}s{p}^2$
• B. $s{p}^3\text{and}s{p}^3$
• C. $s{p}^3\text{and}s{p}^2$
• D. $s{p}^2\text{and}s{p}^3$

Answer 1

The correct option is C $s{p}^3\text{and}s{p}^2$

In $N(C{H}_3{)}_3$ hybridisation and geometry will be $s{p}^3$ and pyramidal. There is no $\pi$ bonding takes place because of the absence of low lying d orbital.
In $N(Si{H}_3{)}_3$ hybridisation and geometry will be $s{p}^2$ and planar. Due to the presence of vacant d orbital $p\pi -d\pi$ back-bonding takes place.