A) \[NO_{2}^{-}>NO_{2}^{+}>N{{O}_{2}}\]
B) \[NO_{2}^{+}>NO_{2}^{-}>N{{O}_{2}}\]
C) \[N{{O}_{2}}>NO_{2}^{+}>NO_{2}^{-}\]
D) \[NO_{2}^{+}>N{{O}_{2}}>NO_{2}^{-}\]
Correct Answer: D
Solution :
As the number of lone pairs of electrons increases, bond angle decreases. Thus, the order of bond angle is \[\underset{(2\,bp\,+0\,lp)}{\mathop{NO_{2}^{+}}}\,>\underset{(2\text{ }bp\text{ }+1\text{ }unpaired\text{ }electron)}{\mathop{N{{O}_{2}}}}\,\] \[>\underset{(2\,bp\,\,+\,1\,\,lp)}{\mathop{N{{O}_{2}}}}\,\]
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