Question

The geometry of $N(C{H}_3{)}_3$ and $N(Si{H}_3{)}_3$ respectively, is:

• A. linear, planar
• B. pyramidal, pyramidal
• C. planar, pyramidal
• D. pyramidal, planar

Answer 1

Answer: (D)

pyramidal, planar

In the case of $N(C{H}_3{)}_3$, geometry is pyramidal and in case of $N(Si{H}_3{)}_3$, geometry is planar.

In trimethyl amine, N atom is $s{p}^3$ hybridized which results in pyramidal geometry. There is no pi bonding because C has no low-lying d-orbital.

In trisilyl amine, N atom is $s{p}^2$ hybridized which results in planar geometry. There is $p\pi -d\pi$ bonding because Si has vacant d-orbital.

Filled $2p$ orbital of N overlaps with vacant $3d$ orbital of Si to from $p\pi -d\pi$ bond.

Hence, option D is correct.