Consider the ions/molecule O_{2}^{+}, O_{2}, | vedclass

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(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the nu

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$O_{2}^{2-} < O_{2}^{-} < O_{2} < O_{2}^{+}$

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(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.SpeciesBond Order$O_{2}^{2-}$$1.0$$O_{2}^{-}$$1.5$$O_{2}$$2.0$$O_{2}^{+}$$2.5$The increasing order of bond order is: $O_{2}^{2-} < O_{2}^{-} < O_{2} < O_{2}^{+}$.

Consider the ions/molecule $O_{2}^{+}, O_{2}, O_{2}^{-}, O_{2}^{2-}$. For increasing bond order,the correct option is ..... .

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