question_answer

Why \[{{(Si{{H}_{3}})}_{3}}N\] is a weaker base than \[{{(C{{H}_{3}})}_{3}}N\]? or Why \[N{{(C{{H}_{3}})}_{3}}\] is more basic than \[N(SiM{{e}_{3}})\]?

Answer:

                This is because the lone pair of electrons of N (present in the 2 p orbital) is transferred to the empty d-orbital of Si (\[p\pi -d\pi \] overlapping) and hence it is not available for protonation. As a result, \[N{{(SiM{{e}_{3}})}_{3}}\] is a weaker base than \[N{{(C{{H}_{3}})}_{3}}\] or \[N{{(Si{{H}_{3}})}_{3}}\] is a stronger base than \[N{{(C{{H}_{3}})}_{3}}\].