The sum of bond order of O_2^{+},O_2^{-},O_2 | vedclass

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(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding e

Vedclass · Accepted Answer

$9$

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(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $O_2$ ($16$ electrons): Bond order = $2.0$.For $O_2^{+}$ ($15$ electrons): Bond order = $2.5$.For $O_2^{-}$ ($17$ electrons): Bond order = $1.5$.For $O_2^{2+}$ ($14$ electrons): Bond order = $3.0$.Sum of bond orders = $2.0 + 2.5 + 1.5 + 3.0 = 9.0$.

The sum of bond order of $O_2^{+}$,$O_2^{-}$,$O_2$ and $O_2^{2+}$ is equal to

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