Question

In which case, geometry of the molecule is pyramidal?

• A. $N(C{H}_3{)}_3$
• B. $N(Si{H}_3{)}_3$
• C. Both (A) and (B)
• D. None of these

Answer 1

The correct option is A

$N(C{H}_3{)}_3$

In $N(Si{H}_3{)}_3$, due to back-bonding from $N$ to $Si$, the lone pair of $N$ is out of hybridization and is changed into $\pi$ -coordinate bond. So, hybridization changes from $s{p}^3$ to $s{p}^2$ and shape changes from pyramidal to planar. For trimethylamine, the nitrogen's lone pair is localized on the $N$ atom since the carbons, already with octets, do not have a suitable $d$-orbital available for back-bonding.