Question

Structure of \[{\rm{N}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}\] and \[{\rm{N}}{\left( {{\rm{Si}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}\]are different it is due to the fact that:
(A) Silicon also use d- orbital for multiple bonding
(B) In case of \[{\rm{N}}{\left( {{\rm{Si}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}\], lone pair of \[{\rm{N}}\]-atom, is transferred to the empty d-orbital of silicon\[{\rm{(}}\,{\rm{P\pi - d\pi }}\,{\rm{overlapping}}\,{\rm{)}}\]
(C) Both (A) and (B)
(D) None of the above

Answer 1

Hint:: As we know that nitrogen is the first member of its family and has p-orbital electrons and s-orbital in its outermost shell. Silicon is the second member of its family and has p-orbital electrons and s-orbital in its outermost shell

Complete answer:
Nitrogen is the first member of nitrogen family and has \[{\rm{2p}}\,{\rm{ - orbital}}\] and \[{\rm{2s}}\,{\rm{ - orbital}}\] in its outermost shell so, it can donate its electron to the empty orbitals as it is a Lewis base.
Silicon is the second member of carbon family and has filled\[{\rm{3p}}\,{\rm{ - orbital}}\] and \[{\rm{3s}}\,{\rm{ - orbital}}\] in its outermost shell and also empty d-orbital.
So, it uses its d- orbital to form bonding with nitrogen. Nitrogen donates its electron from p- orbital. Therefore, nitrogen forms \[{\rm{P\pi - d\pi }}\,{\rm{bonding}}\,\] with silicon atoms.
Therefore, both the options (option A and option B) are correct.

Hence, the correct option is (C).

Note:: \[{\rm{N}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}\]is a good Lewis base but due to empty d-orbital of silicon \[{\rm{N}}{\left( {{\rm{Si}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}\]is neither good Lewis base nor good Lewis acid. This is due to the fact that nitrogen donates its lone pair to d-orbital of silicon hence deficiency of electrons in structure of \[{\rm{N}}{\left( {{\rm{Si}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}\]is reduced and also basicity of this structure is also reduced.