Answer:
B.O. of \[{{N}_{2}}(3)>B.O.\] of \[N_{2}^{+}\,(2\cdot
5)\,\] but B.O. of \[O_{2}^{+}\,(2\cdot 5)>B.O.\] of \[{{O}_{2}}\,(2)\].
Greater the bond order, greater is the bond dissociation energy.
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