The decreasing order of $N - O$ bond lengths in $N O_{2}^{+}$ , $N O_{2}^{-}$ and $N O_{3}^{-}$ is:

N O_{3}^{-} > N O_{2}^{+} > N O_{2}^{-}

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N O_{3}^{-} > N O_{2}^{-} > N O_{2}^{+}

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N O_{2}^{+} > N O_{3}^{-} > N O_{2}^{-}

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N O_{2}^{-} > N O_{3}^{-} > N O_{2}^{+}

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Solution

The correct option is B $N O_{3}^{-} > N O_{2}^{-} > N O_{2}^{+}$
By finding out the bond order of the molecules, we can predict the bond length order.
For $N O_{2}^{+}$ ,
B.O=2
$B o n d o r d e r = \frac{N o . o f b o n d s}{N o . o f r e s o n a t i n g s t r u c t u r e s}$ ,
For $N O_{2}^{-}$ ,
B.O= $\frac{3}{2} = 1.5$
For $N O_{3}^{-}$ ,
B.O= $\frac{4}{3} = 1.33$
The order of bond order is:
$N O_{2}^{+} > N O_{2}^{-} > N O_{3}^{-}$
$B o n d o r d e r \propto \frac{1}{B o n d l e n g t h}$
Thus, the order of bond length is just reverse.

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