Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The lone pair on N in $\mathrm{N}({\mathrm{CH}}_3{)}_3$ is readily available, making it more basic. In $\mathrm{N}({\mathrm{SiH}}_3{)}_3$, the lone pair is delocalized into empty d-orbitals of Si via p$\pi$-d$\pi$ bonding, reducing its basicity and leading to a planar geometry. Therefore, $\mathrm{N}({\mathrm{CH}}_3{)}_3$ is more basic.