Identify the pair in which the difference in | vedclass

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(B) The bond order is calculated as: $	ext{Bond order} = \frac{1}{2} [N_b - N_a]$,where $N_b$ is the number of bonding electrons and $N_a$ is the num

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$O_2^{2-}, O_2^{2+}$

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(B) The bond order is calculated as: $	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 anti-bonding electrons. For $O_2$ ($16$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 6] = 2.0$. For $O_2^{-}$ ($17$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 7] = 1.5$. For $O_2^{+}$ ($15$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 5] = 2.5$. For $O_2^{2-}$ ($18$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 8] = 1.0$. For $O_2^{2+}$ ($14$ electrons): $	ext{Bond order} = \frac{1}{2} [10 - 4] = 3.0$. Calculating the differences: $(A)$ $|2.5 - 1.5| = 1.0$ $(B)$ $|3.0 - 1.0| = 2.0$ $(C)$ $|3.0 - 2.0| = 1.0$ $(D)$ $|3.0 - 2.5| = 0.5$ The maximum difference is $2.0$ for the pair $O_2^{2-}$ and $O_2^{2+}$.

Identify the pair in which the difference in bond order value is maximum.

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