Decreasing order of stability of O2,O2-,O+2 and O2-2 is [2015]

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Solution

Q.
Decreasing order of stability of $O_{2}, O_{2}^{-}, O_{2}^{+}$ and $O_{2}^{2 -}$ is [2015]

1 $O_{2}^{2 -} > O_{2}^{-} > O_{2} > O_{2}^{+}$

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

3 $O_{2} > O_{2}^{-} > O_{2}^{+} > O_{2}^{2 -}$

4 $O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$

Ans.
(4)

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

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

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

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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