The pair of molecules / ions with the same bo | vedclass

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(B) To determine the bond order,we use the formula: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $

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$O_2, C_2$

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(B) To determine the bond order,we use the formula: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $B_2$ ($10$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$. Bond order = $\frac{1}{2} (6 - 4) = 1$.For $C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. Bond order = $\frac{1}{2} (8 - 4) = 2$.For $O_2$ ($16$ electrons): Bond order = $2$.For $O_2^{+}$ ($15$ electrons): Bond order = $2.5$.For $O_2^{-}$ ($17$ electrons): Bond order = $1.5$.For $H_2^{+}$ ($1$ electron): Bond order = $0.5$.For $Li_2$ ($6$ electrons): Bond order = $1$.Re-evaluating the options:$A$: $B_2$ ($BO$=$1$),$C_2$ ($BO$=$2$)$B$: $O_2$ ($BO$=$2$),$C_2$ ($BO$=$2$)$C$: $O_2^{+}$ ($BO$=$2.5$),$O_2^{-}$ ($BO$=$1.5$)$D$: $H_2^{+}$ ($BO$=$0.5$),$Li_2$ ($BO$=$1$)The correct pair with the same bond order is $O_2$ and $C_2$.

The pair of molecules / ions with the same bond order value is

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