(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6)

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(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6) = 2.0$.For $O_2^{-}$ ($17$ electrons): $B.O. = \frac{1}{2} (10 - 7) = 1.5$.For $O_2^{+}$ ($15$ electrons): $B.O. = \frac{1}{2} (10 - 5) = 2.5$.For $O_2^{2-}$ ($18$ electrons): $B.O. = \frac{1}{2} (10 - 8) = 1.0$.Comparing these values,the bond order is maximum in $O_2^{+}$.

Class 11 Chemistry Chemical Bonding and Molecular Structure Molecular orbital theory

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The bond order is maximum in

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$O_2$
B
$O_2^{-}$
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$O_2^{+}$
D
$O_2^{2-}$

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