Bond order of O2, O2+, O2- and O22- is in order

Bond order of $O_{2}, O_{2}^{+}, O_{2}^{-} and O_{2}^{2 -}$ is in order

A
$O_{2}^{-} < O_{2}^{2 -} < O_{2} < O_{2}^{+}$
B
$O_{2}^{2 -} < O_{2}^{-} < O_{2} < O_{2}^{+}$
C
$O_{2}^{+} < O_{2} < O_{2}^{-} < O_{2}^{2 -}$
D
$O_{2} < O_{2}^{+} < O_{2}^{-} < O_{2}^{2 -}$

Solution:

	Electrons	Bond order
$O_{2}$	16	$\frac{N_{b} - N_{a}}{2} = \frac{10 - 6}{2} = 2$
$O_{2}^{+}$	15	$\frac{10 - 5}{2} = 2.5$
$O_{2}^{-}$	17	$\frac{10 - 7}{2} = 1.5$
$O_{2}^{2 -}$	18	$\frac{10 - 8}{2} = 1$

Thus, BO is in order

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

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