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(A) The bond order is calculated using the formula: $Bond \ order = \frac{1}{2} 	imes (N_b - N_a)$, where $N_b$ is the number of bonding electrons an

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$N_2^+$ and $O_2^+$

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(A) The bond order is calculated using the formula: $Bond \ order = \frac{1}{2} 	imes (N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.(1) For $N_2^+$ ($13$ electrons): $Bond \ order = \frac{1}{2} 	imes (9 - 4) = 2.5$.For $O_2^+$ ($15$ electrons): $Bond \ order = \frac{1}{2} 	imes (10 - 5) = 2.5$.Since both have a bond order of $2.5$, this pair is correct.(2) For $F_2$ ($18$ electrons): $Bond \ order = 1$.For $Ne_2$ ($20$ electrons): $Bond \ order = 0$.(3) For $O_2$ ($16$ electrons): $Bond \ order = 2$.For $B_2$ ($10$ electrons): $Bond \ order = 1$.(4) For $C_2$ ($12$ electrons): $Bond \ order = 2$.For $N_2$ ($14$ electrons): $Bond \ order = 3$.

Which of the following pairs have identical values of bond order?

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