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NCERT Exemplar · Q48

Q.Explain why CO3^2- ion cannot be represented by a single Lewis structure. How can it be best represented?

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The carbonate ion CO32−\text{CO}_3^{2-} cannot be captured by a single Lewis structure because experiments show all three C–O bonds are identical in length and strength, yet any one Lewis diagram forces us to draw one double bond and two single bonds. The ion is best represented as a resonance hybrid of three equivalent structures.

Why a single Lewis structure fails

When you draw the Lewis structure for CO32−\text{CO}_3^{2-}, you place carbon at the center bonded to three oxygen atoms, with two extra electrons accounting for the 2−2- charge. Following the octet rule, you end up with something like this:

O=C−O−−O−\text{O}=\text{C}{-}\text{O}^-{-}\text{O}^-

One oxygen forms a double bond with carbon, while the other two carry negative charges and form single bonds. The structure obeys all the rules: carbon has eight electrons, each oxygen has eight electrons, and the total charge is 2−2-.

But here's the problem: this structure predicts that one C–O bond should be shorter and stronger (the double bond, roughly 120 pm) while the other two should be longer and weaker (single bonds, roughly 143 pm). Experimental measurements tell a different story. All three C–O bonds in carbonate are identical, each measuring about 129 pm—intermediate between a single and double bond. No single Lewis structure can account for this symmetry.

Watch out

A common mistake is thinking the "real" structure rapidly flips between the three forms. The electrons are not hopping around; the ion exists in one state that is a blend of all three descriptions simultaneously.

The concept of resonance

The resolution lies in recognizing that our Lewis dot notation is a limited tool. The actual electron distribution in CO32−\text{CO}_3^{2-} does not correspond to any one Lewis diagram. Instead, the true structure is a resonance hybrid—a weighted average of multiple valid Lewis structures.

For carbonate, we can draw three equivalent structures by placing the double bond on each of the three oxygens in turn:

Structure I:

O=C−O−−O−\text{O}=\text{C}{-}\text{O}^-{-}\text{O}^-

Structure II:

O−−C=O−O−\text{O}^-{-}\text{C}=\text{O}{-}\text{O}^-

Structure III:

O−−O−−C=O\text{O}^-{-}\text{O}^-{-}\text{C}=\text{O}

Each structure is equally valid and contributes equally to the hybrid. The double-headed arrow ↔\leftrightarrow connects them to show they are resonance forms, not separate species in equilibrium.

What the resonance hybrid means physically

In the actual carbonate ion, the "extra" π\pi-bonding pair of electrons is delocalized across all three C–O bonds. Instead of one full double bond, each C–O bond has partial double-bond character—specifically, each bond order is 43\frac{4}{3} (one σ\sigma bond plus one-third of a π\pi bond shared among three positions). This delocalization:

  • Makes all three bonds identical in length and energy. …

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