JEE Main · 2019 · 12 Apr 2019 · Shift 1 · p-Block Elements

In $(\mathrm{SiH}{}_3){}_3\mathrm{N}$ the lone pair on nitrogen is drawn into $p\pi \text{–}d\pi$ back-bonding with the empty $d$ orbitals of silicon. That delocalisation flattens the nitrogen to a planar, $s{p}^2$-like geometry and, by tying up the lone pair, leaves it far less available for donation. In $(\mathrm{CH}{}_3){}_3\mathrm{N}$ carbon has no such $d$ orbitals, so no back-bonding occurs and the lone pair stays fully available. Hence $(\mathrm{SiH}{}_3){}_3\mathrm{N}$ is planar and less basic than $(\mathrm{CH}{}_3){}_3\mathrm{N}$.