The stability order of $O_{2}$ and its ions is:

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

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

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

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

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Solution

The correct option is A $O_{2}^{2 +} > O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$
$O_{2}^{2 t} = σ 15^{2} . σ * 15^{2} σ * 25^{2} π 2 P_{z}^{2} (π 2 P_{x}^{2} = π 2 P_{y}^{2})$
no. of fond $= \frac{f o n d i n g e l e m e n t n o n f o n d i n g e l e m e n t}{2}$
$= \frac{10 - 4}{2} = 3$
$O_{2}^{+} \Rightarrow σ 15^{2} σ * 15^{2} σ * 25^{2} σ * 25^{2} π 2 P_{z}^{2} (π 2 P_{x}^{2} = π 2 P_{y}^{2}) π * 2 P_{x}^{1}$
no. of fond $= \frac{10 - 5}{2} = 2.5$
$O_{2} \Rightarrow σ 15^{2} σ * 15^{2} σ * 25^{2} σ * 25^{2} π 2 P_{z}^{2} (π 2 P_{x}^{2} = π 2 P_{y}^{2}) (π * 2 P_{x}^{1} = π * 2 P_{y}^{1})$
no. of fond $= \frac{10 - 6}{2} = 2$
$O_{2}^{-} \Rightarrow$ no. of fond $= \frac{10 - 7}{2} = 1.5$
$O_{2}^{2 -} \Rightarrow$ no. of fond $= \frac{10 - 8}{2} = 1$
Stability order
$O_{2}^{2 +} > O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$

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