A) \[{{(C{{H}_{3}})}_{3}}N\]
B) \[{{(Si{{H}_{3}})}_{3}}N\]
C) \[P{{\left( C{{H}_{3}} \right)}_{3}}\]
D) \[P{{\left( Si{{H}_{3}} \right)}_{3}}\]
Correct Answer: B
Solution :
[b] In case of \[N{{(Si{{H}_{^{3}}})}_{3}},\] N atom is \[s{{p}^{2}}\] hybridised, the lone pair is present in 2p orbital and it is transferred to empty d orbital of Si forming \[d\pi -p\pi \] bond. Hence nitrogen with \[s{{p}^{2}}\] hybridization has trigonal planar shape.
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