Which of the following options represents the correct bond order?

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

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

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

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

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Solution

The correct option is C $O_{2}^{-} < O_{2} < O_{2}^{+}$
$Bond Order = \frac{Bonding electrons - Antibonding electrons}{2}$

$O_{2}^{-} \to$ Total electron = 8 + 9 = 17
Configuration:
$σ (1 s)^{2} σ^{} (1 s)^{2} σ (2 s)^{2} σ^{} (2 s)^{2} σ (2 p_{z})^{2} π (2 p_{x})^{2} π (2 p_{y})^{2} π^{} (2 p_{x})^{2} π^{} (2 p_{y})^{1}$
$Bond Order = \frac{10 - 7}{2} = 1.5$

$O_{2} \to$ Total electron = 8 + 8 = 16
Configuration:
$σ (1 s)^{2} σ^{} (1 s)^{2} σ (2 s)^{2} σ^{} (2 s)^{2} σ (2 p_{z})^{2} π (2 p_{x})^{2} π (2 p_{y})^{2} π^{} (2 p_{x})^{1} π^{} (2 p_{y})^{1}$
$Bond Order = \frac{10 - 6}{2} = 2$

$O_{2}^{+} \to$ Total electron = 8 + 7 = 15
Configuration:
$σ (1 s)^{2} σ^{} (1 s)^{2} σ (2 s)^{2} σ^{} (2 s)^{2} σ (2 p_{z})^{2} π (2 p_{x})^{2} π (2 p_{y})^{2} π^{*} (2 p_{x})^{1}$
$Bond Order = \frac{10 - 5}{2} = 2.5$

Hence, the correct bond order in the following species is:
$O_{2}^{+} > O_{2} > O_{2}^{-} .$

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