The increasing order of bond order of O_2, O_ | vedclass

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(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$.For $O_2$ $(16 \ e^-)$: $BO = \frac{10

Vedclass · Accepted Answer

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

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(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$.For $O_2$ $(16 \ e^-)$: $BO = \frac{10-6}{2} = 2$.For $O_2^+$ $(15 \ e^-)$: $BO = \frac{10-5}{2} = 2.5$.For $O_2^-$ $(17 \ e^-)$: $BO = \frac{10-7}{2} = 1.5$.For $O_2^{2-}$ $(18 \ e^-)$: $BO = \frac{10-8}{2} = 1$.Thus,the increasing order of bond order is $O_2^{2-} < O_2^- < O_2 < O_2^+$.

The increasing order of bond order of $O_2, O_2^+, O_2^-$,and $O_2^{2-}$ is:

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