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According to $MO$ theory which of the following lists ranks the nitrogen species in terms of increasing bond order

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$N_{2}^{-} < N_{2} < N_{2}^{2 -}$

$N_{2}^{2 -} < N_{2}^{-} < N_{2}$

$N_{2} < N_{2}^{2 -} < N_{2}^{-}$

$N_{2}^{-} < N_{2}^{2 -} < N_{2}$

answer is B.

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Detailed Solution

$N_{2} (14 e^{-}) - σ 1 s^{2} σ * 1 s^{2} σ 2 s^{2} σ * 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2}$

Bond order of $N_{2} = \frac{10 - 4}{2} = 3$

Similarly, $N_{2}^{-} = \frac{10 - 5}{2} = 2.5, N_{2}^{2 -} = \frac{10 - 6}{2} = 2.0$ .

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