Question

What is the order of stability of $N_2$ and its ions?

• A. $N_{2} > N_{2}^{+} = N_{2}^{-} > N_{2}^{2-}$
• B. $N^{+}_{2} > N_{2}^{-} > N_{2} > N_{2}^{2-}$
• C. $N^{-}_{2} > N_{2}^{+} > N_{2} > N_{2}^{2-}$
• D. $N^{2-}_{2} > N_{2}^{-} = N_{2}^{+} > N_{2}$

Answer 1

Answer: (A)

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

Answer 2

Bond order of $N_{2} = 3$, $N_{2}^{+} = 2.5$, $N_{2}^{-}= 2.5$ and $N_{2}^{2-}$ is $2$. Higher the bond order, more is the stability.