What is bond order in case of $O_{2}^{+}$ ?

2.5

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1.5

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Solution

The correct option is A 2.5
$O_{2}^{-} = 8 + 8 + 1 = 17 = σ 1 s^{2}, σ^{} s^{2}, σ 2 s^{2}, σ 2^{} s^{2}, σ 2 p_{z}^{2}, π 2 p x^{2}, π 2 p_{y}^{2}, π 2 p_{x}^{2 }, π 2 p_{y}^{ 1}$
Bond order $= \frac{No. of bonding electrons - No. of antibonding electrons}{2}$
$= \frac{10 - 7}{2} = \frac{3}{2} = 1.5$

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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 $O_{2}^{-}$ .

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

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