Correct order of bond length is-

$C O_{3}^{2 -} > C O_{2} > C O$

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$C O_{2} > C O > C O_{3}^{2 -}$

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$C O > C O_{2} > C O_{3}^{2 -}$

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None of these

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Solution

The correct option is A
$C O_{3}^{2 -} > C O_{2} > C O$

Let's take a look at the structure of $C O_{3}^{2 -}, C O_{2} a n d C O$

$C O_{2} \Rightarrow . . O . . = C = . . O . .$

$C O \Rightarrow: C \equiv O :$

So, bond order will be:
$C O_{3}^{2 -} \Rightarrow 1.33$
$C O_{2} \Rightarrow 2$
$C O \Rightarrow 3$

We know,
$C O > C O_{2} > C O_{3}^{2 -}$ (Bond Order)

We know
$B o n d o r d e r \propto \frac{1}{B o n d L e n g t h}$

$\Rightarrow C O_{3}^{2 -} > C O_{2} > C O$ (Bond Length)

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