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 $N_2^{+}\ge$ 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.