What is the order of stability of $N_{2}$ and its ions?

N_{2} > N_{2}^{+} = N_{2}^{-} > N_{2}^{2 -}

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N_{2}^{+} > N_{2}^{-} > N_{2} > N_{2}^{2 -}

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N_{2}^{-} > N_{2}^{+} > N_{2} > N_{2}^{2 -}

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N_{2}^{2 -} > N_{2}^{-} = N_{2}^{+} > N_{2}

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Solution

The correct option is A $N_{2} > N_{2}^{+} = N_{2}^{-} > N_{2}^{2 -}$
Stability of a molecule is related with bond order like this:-
$S t a b i l i t y α B o n d o r d e r$
and to calculate bond order, formula is = $\frac{N u m b e r o f e l e c t r o n s i n B M O - N u m b e r o f e l e c t r o n s i n A B M O}{2}$
$N_{2}$ has highest bond order So the correct option is [A]

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