Answer:
In\[N{{H}_{3}}\], there is one
lone pair on N-atom which repels bond pairs whereas in \[{{H}_{2}}O\], there
are two lone pairs on O-atom which repel bond pairs. Thus, repulsion is more in\[{{H}_{2}}O\]
than in \[N{{H}_{3}}\] and hence, the bond angle in water is less than in\[N{{H}_{3}}\].
You need to login to perform this action.
You will be redirected in
3 sec