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(C) The bond order $(BO)$ of a species is calculated as $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding orbitals and $N_a

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$O_2^{+}$

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(C) The bond order $(BO)$ of a species is calculated as $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.$(a)$ $O_2$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. $BO = \frac{10 - 6}{2} = 2.0$.$(b)$ $F_2$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$. $BO = \frac{10 - 8}{2} = 1.0$.$(c)$ $O_2^{+}$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1$. $BO = \frac{10 - 5}{2} = 2.5$.$(d)$ $F_2^{-}$: Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2, \sigma^* 2p_z^1$. $BO = \frac{10 - 9}{2} = 0.5$.Thus,$O_2^{+}$ has the highest bond order.

List-$I$ (Molecule / ion)	List-$II$ (Bond order)
$1$. $NO$	$i$. $1.5$
$2$. $CO$	$ii$. $2.0$
$3$. $O_2^{-}$	$iii$. $2.5$
$4$. $O_2$	$iv$. $3.0$

Among the following,the species with the highest bond order is

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Match List-$I$ with List-$II$ and select the correct option:
List-$I$ (Molecule / ion) List-$II$ (Bond order)
$1$. $NO$ $i$. $1.5$
$2$. $CO$ $ii$. $2.0$
$3$. $O_2^{-}$ $iii$. $2.5$
$4$. $O_2$ $iv$. $3.0$

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