Answer:
In \[N{{(C{{H}_{3}})}_{3}},\,N\] is \[s{{p}^{3}}\]-hybridized
with one lone pair of electrons and hence \[N{{(C{{H}_{3}})}_{3}}\] is
pyramidal. In \[N{{(Si{{H}_{3}})}_{3}},\]Si is \[s{{p}^{3}}\]-hybridized and
forms three \[\sigma \]-bonds with H and fourth with the N-atom. Further due to
the presence of d-orbitals, Si has a strong tendency to form \[p\pi -d\pi \]double
bond with N-atom. For this to occur, N must be \[s{{p}^{2}}\]-hybridized so
that the p-orbital on N containing the lone pair of electrons can overlap with
the empty d-orbital on Si as shown in Fig. 11.14, page 11/26. As a result, \[N{{(Si{{H}_{3}})}_{3}}\]
is planar.
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