If the sum of bond orders of O_2^{-} and O_2^ | vedclass

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

Vedclass · Accepted Answer

$1.20$

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(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $\frac{N_b - N_a}{2}$,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 $O_2^{-}$ ($17$ electrons): Bond order = $1.5$.For $O_2^{2-}$ ($18$ electrons): Bond order = $1.0$.Sum of bond orders of $O_2^{-}$ and $O_2^{2-}$ = $1.5 + 1.0 = 2.5$. Thus,$x = 2.5$.For $O_2^{2+}$ ($14$ electrons): Bond order = $3.0$.We need to express the bond order of $O_2^{2+}$ $(3.0)$ in terms of $x$ $(2.5)$.$3.0 = k 	imes 2.5 \implies k = \frac{3.0}{2.5} = 1.2$.Therefore,the bond order of $O_2^{2+}$ is $1.2x$.

If the sum of bond orders of $O_2^{-}$ and $O_2^{2-}$ is $x$,then bond order of $O_2^{2+}$ will be (in $x$)

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