The correct arrangement of the species in the decreasing order of the bond length between
carbon and oxygen in them is

• A

  \( \mathrm{CO}_{2}, \mathrm{HCO}_{2}^{-}, \mathrm{CO}, \mathrm{CO}_{3}^{2-} \)
• B

  \( \mathrm{CO}, \mathrm{CO}_{3}{ }^{2-}, \mathrm{CO}_{2}, \mathrm{HCO}_{2} \)
• C

  \( \mathrm{CO}, \mathrm{CO}_{2}, \mathrm{HCO}_{2}^{-}, \mathrm{CO}_{3}^{2-} \)
• Correct answer
  D

  \( \mathrm{CO}_{3}{ }^{2-}, \mathrm{HCO}_{2}{ }^{-}, \mathrm{CO}_{2}, \mathrm{CO} \)

Solution

We know Bond order \( \propto \frac{1}{\text { Bond length }} \)
CO molecule have \( 3 \) bonds between \( C \) and \( O \).
\[ C=O \]
\( \mathrm{CO}_{2} \) molecule have \( 2 \) double bonds between \( \mathrm{C} \) and \( \mathrm{O} \)
\[ O=C=O \]
\( \mathrm{CO}_{3}{ }^{2-} \) ion have \( 3 \) partial double bonds between \( \mathrm{C} \) and \( \mathrm{O} \), due to resonance.

\( \mathrm{HCO}_{2}{ }^{-} \)ion also show resonance, but due to the presence of hydrogen atom attached to carbon atom it has less
resonance hybrid, therefore, \( \mathrm{HCO}_{2}{ }^{-} \)ion have greater bond order than \( \mathrm{CO}_{3}{ }^{2-} \)

Hence, the correct decreasing order of bond length is \( \mathrm{CO}_{3}{ }^{2-}, \mathrm{HCO}_{2}{ }^{-}, \mathrm{CO}_{2}, \mathrm{CO} \)

Hence, the correct answer is option: 'D'.