Question

Bond order of $N_{2}^{+}$ and $N_{2}^{-}$ are same. Which relation is correct for $N_{2}^{+}$ and $N_{2}^{-}$ ?

• A. Bond energy of $N_{2}^{+}=$ bond energy of $N_{2}^{-}$
• B. Bond energy of $N_{2}^{+}>$ bond energy of $N_{2}^{-}$
• C. Bond energy of $N_{2}^{+} < $ bond energy of $N_{2}^{-}$
• D. Bond energy of $N_{2}^{+}\geq $ bond energy of $N_{2}^{-}$

Answer 1

Answer: (B)

By MOT, the bond orders of both are 2.5. Now acc to MOT only, the compound having more electrons in antibonding orbitals will form a weaker bond and a longer bond. If their bond orders are equal. $N^{2 -}$ has $5e^{-}$ in antibonding while $N^{2 +}$ has 4. So, $N^{2 +}$ will make a stronger bond and will have a shorter bond length.