The correct order of $C-O $ bond length among $CO,CO_3^{2-}$, $CO_2$ is
- $CO_2 < CO_3^{2-} < CO$
- $CO < CO_3^{2-} < CO_2$
- $CO_3^{2-} < CO_2 < CO$
- $CO < CO_2 < CO_3^{2-}$
The Correct Option is D
Solution and Explanation
A multiple bond (double or triple bonds) is always shorter than the corresponding single bond.
The C-atom in $CO_3^{2-}$ is $sp^2$ hybridised as shown:
The C-atom in $CO_2$ is $sp$ hybridised with bond distance carbon-oxygen is $122\, pm$.
$O = C = O \leftrightarrow ^+ O \equiv C - _O^- \leftrightarrow $
$ \hspace37mm$ $_O^- - C \equiv _O^+$
The C-atom in CO is sp hybridised with C-O bond distance is 110 pm.
$ \, \, \, \, \, \, \, :C \equiv O^+ :$
So, the correct order is
$ \, \, \, \, \, \, CO < CO_2 < CO_3^{2-}$
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