Question

Which of the following molecules are planar in shape?

• A. $N(Si{H}_3{)}_3$
• B. $N(C{H}_3{)}_3$
• C. $P(Si{H}_3{)}_3$
• D. Both (a) and (c)

Answer 1

The correct option is A $N(Si{H}_3{)}_3$

a) In $N(Si{H}_3{)}_3$, $N$ has one lone pair and three bond pairs. Hence, it should posses tetrahedral geometry and pyramidal shape. But it forms a planar structure because of the presence of back bonding between the filled 2p orbital of the central atom $\left(N\right)$ and the empty 3d orbital of $Si$.

b) $N(C{H}_3{)}_3$ has tetrahedral geometry and pyramidal shape due to the absence of back bonding as $C$ does not have vacant d-orbital.

c) In $P(Si{H}_3{)}_3$, the shape is pyramidal as there is very less extent or no back bonding between $P$ and $Si$. This is due to the inefficient overlaping of orbitals.