The increasing order of the bond order of $O_{2}$ , $O_{2}^{-}$ , $O_{2}^{+}$ and $O_{2}^{2 -}$ ?

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Solution

Bond order:
Every molecule has a bonding hierarchy.
Each molecule has a different bond arrangement
The bond order can be calculated as half the difference between the number of electrons in nonbonding (bonding-free) and bonding-filled molecular orbitals.
The formula for calculating bond order:
$B . O . = \frac{N_{B} - N_{A}}{2} $
The number of electrons in the antibonding and bonding molecular orbitals, respectively, is $N_{A}, N_{B}$
Step 1: Calculating bond order of given species:
$B . O for O_{2} = \frac{1}{2} \times [10 - 6] B . O for O_{2} = 2$
$B . O for O_{2}^{-} = \frac{1}{2} \times [10 - 7] B . O for O_{2}^{-} = 1.5$
$B . O for O_{2}^{+} = \frac{1}{2} \times [10 - 5] B . O for O_{2}^{+} = 2.5$
$B . O for O_{2}^{2 -} = \frac{1}{2} \times [10 - 8] B . O for O_{2}^{2 -} = 1$
Step 2: Arranging in increasing order of bond length
$O_{2}^{2 -}$ < $O_{2}^{-}$ < $O_{2}$ < $O_{2}^{+}$

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The increasing order of the bond order of $O_{2}$ , $O_{2}^{-}$ , $O_{2}^{+}$ and $O_{2}^{2 -}$ ?