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.