See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: C-O Bond Lengths in Compounds **Analyze each compound:** **(A) $CO_2$:** Linear structure $O=C=O$ - Both C-O bonds are double bonds - All C-O bonds have **identical length** ✓ **(B) $HCOOH$:** Formic acid structure $$\begin{array}{c} O \\ \| \\ H-C \\ \: \: \: \: \: \| \\ \: \: \: \: OH \end{array}$$ - One C=O (carbonyl, shorter ~1.20 Å) - One C-O (single bond to -OH, longer ~1.35 Å) - **Two different C-O bond lengths** ✗ **(C) $CO_3^{2-}$:** Carbonate ion (trigonal planar) - Resonance structures make all three C-O bonds equivalent - All bonds are identical intermediate length (~1.29 Å) - All C-O bonds have **same length** ✓ **(D) $HCOO^-$:** Formate ion (conjugate base of formic acid) - Resonance delocalization between two C-O bonds - Both C-O bonds are equivalent (order = 1.5) - All C-O bonds have **same length** ✓ **Answer:** **(B) $HCOOH$** — has mixed C=O (double) and C-O (single) bonds with different lengths, unlike the others which show bond equivalence through resonance or identical multiple bonds.