The bond order of $O_{2}^{+}$ is

1

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1.5

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2.5

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3

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Solution

The correct option is C $2.5$
The electronic configuration of $O_{2}^{+}$ according to MOT is $(σ_{1 s})^{2} (σ_{1 s}^{})^{2} (σ_{2 s})^{2} (σ_{2 s}^{})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} = (π_{2 p_{y}})^{2} (π_{2 p_{x}}^{})^{1} = (π_{2 p_{y}}^{})^{0} (σ_{2 p_{z}}^{*})^{0}$
Bond order is $= \frac{10 - 5}{2} = 2.5$

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Similar questions

What is meant by the term bond order? Calculate the bond order of: $N_{2}$ , $O_{2}$ , $O_{2}^{+}$ and $O_{2}^{-}$ .

O_{2}, O_{2}^{+}, O_{2}^{-}

O_{2}^{2 -}

O_{2}, O_{2}^{+}, O_{2}^{-}

O_{2}^{2 -}

N_{2}^{+}

N_{2}

O_{2}^{+}

O_{2}

The correct statement regarding O $_{2}$ $^{+}$ is

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