Increasing- order of bond length in $NO, {NO}^{+}$ and ${NO}^{-}$

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${NO}^{+} < {NO}_{} < {NO}^{-}$

$NO > {NO}^{-} > N^{+}$

$NO < {NO}^{+} < {NO}^{-}$

$NO < {NO}^{+} = {NO}^{-}$

answer is B.

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Detailed Solution

Electronic configuration of No:

$\begin{array}{l} σ 1 s^{2} σ 1 s^{2} σ 2 s^{2} σ 2 s^{2} π 2 P_{x}^{2} = π 2 P_{y}^{2} σ 2 P_{z}^{2} . \\ π 2 P_{x} = π 2 P_{y} \end{array}$

Bond order $= \frac{N_{b} - N a}{2} = \frac{10 - 5}{2} = 2.5$

Electronic configuration of ${NO}^{-}$ :

$\begin{array}{l} σ 1 s^{2} σ 1 s^{2} σ 2 s^{2} σ 2 s^{2} π 2 P_{x}^{2} = π 2 P_{y}^{2} σ 2 P_{z}^{2} \\ π 2 P_{x} = π 2 P_{y} \end{array}$

Bond order $= \frac{N_{b} - N a}{2} = \frac{10 - 6}{2} = 2$

Here the

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

So order of bond length in increasing order will be

${NO}^{+} < {NO}^{2} < {NO}^{-}$

Hence correct option is B.

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