$N_{2}$ has greater bond energy than ${N_{2}}^{+}$ but $O_{2}$ has tower bond dissociation energy then ${O_{2}}^{+}$ . Explain.

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Solution

According to Molecular orbital theory: $N_{2}$ have 6 more electrons in a bonding atomic orbital so it's bond order is 3, whereas $N_{2}^{+}$ has only 5 in bonding so it's bond order is 2.5. Bond dissociation energy is directly proportional to Bond order of a molecule so $N_{2}$ has more bond dissociation energy.
$O_{2}^{+}$ has more bond dissociation energy than $O_{2}$ .Because the bond order in $O_{2}$ is 2 where as in $O_{2}^{+}$ is 2.5.If bond order is more energy required to break the bonds is more.

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