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Which of the following options represents the correct bond ord

Question

Which of the following options represents the correct bond order ?

A

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

B

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

C

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

D

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

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Text solutionVerified

Bond order of

O_{2}^{-}

O_{2}^{-} = 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{number of electrons in BMO - number of electrons ABMO}{2}

= \frac{10 - 7}{2} = \frac{3}{2} = 1.5

O_{2}^{+} = 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})

BO=

\frac{10 - 5}{2} = \frac{5}{2} = 2.5

O_{2}^{+} = 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})

BO=

\frac{10 - 6}{2} = \frac{5}{2} = 2.5

So, the correct sequence is

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

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Question Text	Which of the following options represents the correct bond order ?
Answer Type	Text solution:1
Upvotes	150