Question

Bond energies in $NO$, $N{O}^{+}$ and $N{O}^{-}$ are such that:

• A. $NO>N{O}^{+}>N{O}^{-}$
• B. $N{O}^{-}>NO>N{O}^{+}$
• C. $N{O}^{+}>N{O}^{-}>NO$
• D. $N{O}^{+}>NO>N{O}^{-}$

Answer 1

The correct option is D $N{O}^{+}>NO>N{O}^{-}$

Greater the bond order, the higher is the energy required to break the bonds, since those molecules are more stable.

Bond order $\propto$ Bond Energy

BO of ${NO}^{+}=\frac{N_{BO}-N_{AO}}{2}=\frac{10-4}{2}=3$

BO of $NO=\frac{10-5}{2}=2.5$

BO of ${NO}^{-}=\frac{10-6}{2}=2$

Hence, ${NO}^{+}>NO>{NO}^{-}$.