Question
The decreasing values of bond angles from $$N{H_3}\left( {{{106}^ \circ }} \right)\,{\text{to}}\,Sb{H_3}\left( {{{101}^ \circ }} \right)$$ down group-15 of the periodic table is due to
A.
decreasing $$lp{\text{ - }}bp$$ repulsion
B.
decreasing electronegativity
C.
increasing $$bp{\text{ - }}bp$$ repulsion
D.
increasing p - orbital character in $$s{p^3}$$
Answer :
decreasing electronegativity
Solution :
The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.
\[\begin{matrix}
N{{H}_{3}} \\
{{107}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
P{{H}_{3}} \\
{{94}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
AS{{H}_{3}} \\
{{92}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
Sb{{H}_{3}} \\
{{91}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
Bi{{H}_{3}} \\
{{90}^{\circ }} \\
\end{matrix}\]
NOTE : This can also be explained by the fact that as
the size of central atom increases $$s{p^3}$$ hybrid orbital becomes more distinct with increasing size of central atom i.e, pure $$p$$ - orbitals are utilized in $$M - H$$ bonding