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Bond Order and Enthalpy

The correct o...

Question

The correct order of bond lengths is:

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is C $O_{2}^{2 -} > O_{2}^{-} > O_{2}^{+}$
$Molecular orbital configuration for O_{2}^{+} is (σ_{1 s})^{2} (σ_{1 s}^{})^{2} (σ_{2 s})^{2} (σ_{2 s}^{})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (π_{2 p_{x}}^{})^{1} (π_{2 p_{y}}^{})^{0} (σ_{2 p_{z}}^{})^{0}$
$For O_{2}^{+}, bond order = \frac{6 - 1}{2} = 2.5$

$Molecular orbital configuration for O_{2}^{-} is$
$(σ_{1 s})^{2} (σ_{1 s}^{})^{2} (σ_{2 s})^{2} (σ_{2 s}^{})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (π_{2 p_{x}}^{})^{2} (π_{2 p_{y}}^{})^{1} (σ_{2 p_{z}}^{})^{0}$
$For O_{2}^{-}, bond order = \frac{6 - 3}{2} = 1.5$

$Molecular orbital configuration for O_{2}^{2 -}$
$(σ_{1 s})^{2} (σ_{1 s}^{})^{2} (σ_{2 s})^{2} (σ_{2 s}^{})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (π_{2 p_{x}}^{})^{2} (π_{2 p_{y}}^{})^{2} (σ_{2 p_{z}}^{*})^{0}$
$For O_{2}^{2 -}, bond order = \frac{6 - 4}{2} = 1.0$

Thus, the order of bond orders is:
$O_{2}^{+} > O_{2}^{-} > O_{2}^{2 -}$

We know that, $Bond order \propto \frac{1}{bond length}$

So, the correct order of bond lengths will be:
$O_{2}^{2 -} > O_{2}^{-} > O_{2}^{+}$

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Similar questions

Q. What is the stability order of the following species?

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