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Chemistry Question on Molecular Orbital Theory

The correct order of C−OC-O bond length among CO,CO32−CO,CO_3^{2-}, CO2CO_2 is

A

CO2<CO32−<COCO_2 < CO_3^{2-} < CO

B

CO<CO32−<CO2CO < CO_3^{2-} < CO_2

C

CO32−<CO2<COCO_3^{2-} < CO_2 < CO

D

CO<CO2<CO32−CO < CO_2 < CO_3^{2-}

Answer

CO<CO2<CO32−CO < CO_2 < CO_3^{2-}

Explanation

Solution

A bond length is the average distance between the centres of nuclei of two bonded atoms.
A multiple bond (double or triple bonds) is always shorter than the corresponding single bond.
The C-atom in CO32−CO_3^{2-} is sp2sp^2 hybridised as shown:
The C-atom in CO2CO_2 is spsp hybridised with bond distance carbon-oxygen is 122 pm122\, pm.
O=C=O↔+O≡C−O−↔O = C = O \leftrightarrow ^+ O \equiv C - _O^- \leftrightarrow
\hspace37mm O−−C≡O+_O^- - C \equiv _O^+
The C-atom in CO is sp hybridised with C-O bond distance is 110 pm.
       :C≡O+:\, \, \, \, \, \, \, :C \equiv O^+ :
So, the correct order is
      CO<CO2<CO32−\, \, \, \, \, \, CO < CO_2 < CO_3^{2-}