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(B) The bond length is inversely proportional to the bond order $(B.O.)$.$1$. For $O_2$ (dioxygen molecule): The bond order is $2.0$.$2$. For $Na_2O_2

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$Na_2O_2$

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(B) The bond length is inversely proportional to the bond order $(B.O.)$.$1$. For $O_2$ (dioxygen molecule): The bond order is $2.0$.$2$. For $Na_2O_2$ (contains peroxide ion,$O_2^{2-}$): The bond order is $1.0$.$3$. For $KO_2$ (contains superoxide ion,$O_2^-$): The bond order is $1.5$.Comparing the bond orders: $2.0 (O_2) > 1.5 (O_2^-) > 1.0 (O_2^{2-})$.Since bond length is inversely proportional to bond order,the species with the lowest bond order will have the maximum bond length.Therefore,$O_2^{2-}$ (in $Na_2O_2$) has the maximum bond length.

Column $I$	Column $II$
$(A)$ $B_2$	$(p)$ Paramagnetic
$(B)$ $N_2$	$(q)$ Undergoes oxidation
$(C)$ $O_2^{-}$	$(r)$ Undergoes reduction
$(D)$ $O_2$	$(s)$ Bond order $\geq 2$
	$(t)$ Mixing of '$s$' and '$p$' orbitals

Which of the following species shows the maximum bond length?

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