Decreasing order of stability of $O_{2}, O_{2}^{-}, O_{2}^{+} and O_{2}^{2 -} is$

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

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

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

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

answer is D.

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Detailed Solution

$O_{2} (16) : KK σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2} π^{*} 2 p_{x}^{1} = π^{*} 2 p_{y}^{1}$

$Bond order = \frac{1}{2} (8 - 4) = 2$

$O_{2}^{2 -} (18) : K K σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2} π^{*} 2 p_{x}^{2} = π^{*} 2 p_{y}^{2}$

$Bond order = \frac{1}{2} (8 - 6) = 1$

$O_{2}^{-} (17) : K K σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2} π^{*} 2 p_{x}^{2} = π^{*} 2 p_{y}^{1}$

$Bond order = \frac{1}{2} (8 - 5) = 1.5$

$O_{2}^{+} (15) : K K σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2} π^{*} 2 p_{x}^{1}$

$Bond order = \frac{1}{2} (8 - 3) = 2.5$

$As, bond order \propto stability$

The decreasing order of stability is

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

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