A) \[NH_{4}^{+}>N{{H}_{3}}>NH_{2}^{-}\]
B) \[N{{H}_{3}}>NH_{2}^{-}>NH_{4}^{+}\]
C) \[NH_{2}^{-}<N{{H}_{3}}>NH_{4}^{+}\]
D) \[N{{H}_{3}}>NH_{4}^{+}>NH_{2}^{-}\]
Correct Answer: A
Solution :
The correct order of decreasing bond angle is \[NH_{4}^{+}>N{{H}_{3}}>NH_{2}^{-}\] This is because all of them involves \[s{{p}^{3}}\] hybridisation. The number of lone pair of electrons present on N-atom are 0, 1, 2 respectively. Greater the number of lone pairs, greater the repulsion on the bond pairs and hence smaller is the bond angle.
You need to login to perform this action.
You will be redirected in
3 sec
