A) \[N{{H}_{3}}\]
B) \[{{H}_{2}}O\]
C) \[BC{{l}_{3}}\]
D) None of these
Correct Answer: D
Solution :
\[\text{BC}{{\text{l}}_{\text{3}}}\] has bond angle equal to \[\text{12}{{\text{0}}^{o}}\] (trigonal planar). \[\text{N}{{\text{H}}_{\text{3}}}\] and \[{{\text{H}}_{\text{2}}}\text{O}\] have \[\text{s}{{\text{p}}^{3}}\] hybridization but due to the presence of lone pair of electrons, they have bond angle less than 1 \[{{109}^{o}}28\] \[(N{{H}_{3}}-{{107}^{o}},{{H}_{2}}O-{{104.5}^{o}}),\] \[\text{As}{{\text{H}}_{\text{3}}}\] (\[s{{p}^{3}}\] hybrid) has smaller bond angle than \[\text{N}{{\text{H}}_{\text{3}}}\] due to less electronegativity of As than N.
You need to login to perform this action.
You will be redirected in
3 sec
