Arrange the following in order of decreasing $N - O bond length : {NO}_{2}^{+}, {NO}_{2}^{-}, {NO}_{3}^{-}$

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${NO}_{3}^{-} > {NO}_{2}^{+} > {NO}_{2}^{-}$

${NO}_{3}^{-} > {NO}_{2}^{-} > {NO}_{2}^{+}$

${NO}_{2}^{+} > {NO}_{3}^{-} > {NO}_{2}^{-}$

${NO}_{2}^{-} > {NO}_{3}^{-} > {NO}_{2}^{+}$

answer is B.

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Detailed Solution

N atom in ${NO}_{2}^{+} is sp hybridised while in {NO}_{2}^{-} and {NO}_{3}^{-}, it is {sp}^{2} hybridised .$

O = N = O Bond order = 2

Bond order= $\frac{2 + 1}{2} = 1.5$

Bond order $= \frac{2 + 1 + 1}{3} = 1.33$

Bond order $\propto \frac{1}{Bond Length}$

So, Bond length order is ${NO}_{3}^{-} > {NO}_{2}^{-} > {NO}_{2}^{+}$

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