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(B) The total number of electrons in $F_2$ is $18$. According to Molecular Orbital Theory,the electronic configuration of $F_2$ is $\sigma 1s^2, \sigm

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$O_2^{2-}$

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(B) The total number of electrons in $F_2$ is $18$. According to Molecular Orbital Theory,the electronic configuration of $F_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order $= \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = 1$. Now,checking the options: $O_2^+$ has $15$ electrons,bond order $= 2.5$. $O_2^{2-}$ has $18$ electrons,bond order $= \frac{1}{2} (10 - 8) = 1$. $O_2$ has $16$ electrons,bond order $= 2$. $N_2^+$ has $13$ electrons,bond order $= 2.5$. Thus,$O_2^{2-}$ has the same bond order as $F_2$.

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

The species,which has the bond order same as that of $F_2$ molecule is

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