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

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

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

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

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

answer is B.

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Detailed Solution

N atom in ${NO}_{2}^{+}$ is sp hybridized whereas in ${NO}_{2}^{-}$ and ${NO}_{3}^{-}$ , it is sp² hybridized.

Bond order = $\frac{No . of bond between two atoms in all possible structures}{No . of resonating structures}$

Bond order of ${NO}_{2}^{+}$ = 2

Bond order of ${NO}_{2}^{-}$ :

B.O = 2+12=1.5

Bond order of ${NO}_{3}^{-}$ :

B.O = 2+1+13=1.33

Since $Bond length \propto \frac{1}{Bond order}$

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

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