The C–O bond length is inversely proportional to the bond order. A higher bond order means a stronger and shorter bond.

Step 1: Determine the bond order for CO.
In carbon monoxide (CO), there is a triple bond between carbon and oxygen (C≡O). The bond order is 3.

Step 2: Determine the bond order for CO₂.
In carbon dioxide (CO₂), there are two double bonds between carbon and oxygen (O=C=O). The bond order for each C=O bond is 2.

Step 3: Determine the bond order for CO₃²⁻.
In the carbonate ion (CO₃²⁻), there is resonance. The structure consists of one C=O double bond and two C–O single bonds, which are delocalized over all three C–O bonds. The average bond order for each C–O bond is $(1\times 2+2\times 1)/3=4/3=1.33$.

Step 4: Compare bond orders and determine bond lengths.
Bond orders: CO (3) > CO₂ (2) > CO₃²⁻ (1.33).
Since bond length is inversely proportional to bond order, the order of C–O bond length will be the reverse of the bond order:
CO < CO₂ < CO₃²⁻.

Option Analysis:

• A) CO < CO₃²⁻ < CO₂: Incorrect. This order incorrectly places CO₃²⁻ before CO₂.
• B) CO₃²⁻ < CO₂ < CO: Incorrect. This is the reverse of the correct order.
• C) CO < CO₂ < CO₃²⁻: Correct. This order correctly reflects the increasing bond length based on decreasing bond order.
• D) CO₂ < CO < CO₃²⁻: Incorrect. This order incorrectly places CO₂ before CO.

Correct Answer: (C)