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Problems · Problem 4.3

Q.Explain the structure of the CO32−CO_3^{2-} ion in terms of resonance.

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The carbonate ion CO32−CO_3^{2-} has three equivalent resonance structures with delocalized π\pi bonding, giving each C–O bond a bond order of 43\frac{4}{3} and explaining its trigonal planar geometry with 120° bond angles.

The carbonate ion is a textbook example of why we need resonance theory. If you try to draw a single Lewis structure, you'll find yourself forced to choose which oxygen gets the double bond—but experiments show all three C–O bonds are identical in length and strength. That's the signature of resonance: the true structure is a hybrid, not any one drawing.

Why resonance matters here

Carbon has 4 valence electrons, each oxygen has 6, and the 2−2- charge adds 2 more, giving us 4+3(6)+2=244 + 3(6) + 2 = 24 valence electrons total. Carbon sits in the center (least electronegative), bonded to three oxygens. To satisfy the octet rule for carbon, we need at least one double bond—but there's no reason to prefer one oxygen over another. Nature doesn't pick favorites: the π\pi electrons spread out equally across all three bonds.

Step-by-step construction

1. Draw the skeleton and count electrons

Place carbon at the center with single bonds to three oxygens:

O−C−Owith one more OO - C - O \quad \text{with one more } O

That uses 3×2=63 \times 2 = 6 electrons, leaving 24−6=1824 - 6 = 18 to distribute.

2. Complete octets on the terminal atoms first

Each oxygen needs 6 more electrons (3 lone pairs) to complete its octet. That accounts for 3×6=183 \times 6 = 18 electrons—exactly what we have left. But now carbon has only 6 electrons (three single bonds), violating the octet rule.

3. Form a double bond to satisfy carbon's octet

Move one lone pair from any oxygen to form a C=OC=O double bond. Now carbon has 8 electrons. You can choose any of the three oxygens, giving three equivalent structures:

Structure I:O−−C=Owith O− belowStructure II:O=C−O−with O− belowStructure III:O−−C−O−with O double-bonded below\begin{array}{ccc} \text{Structure I:} & \quad & O^- - C = O \quad \text{with } O^- \text{ below} \\ \text{Structure II:} & \quad & O = C - O^- \quad \text{with } O^- \text{ below} \\ \text{Structure III:} & \quad & O^- - C - O^- \quad \text{with } O \text{ double-bonded below} \end{array}

Each structure has one C=OC=O double bond and two C−O−C-O^- single bonds. The negative charges sit on the oxygens with single bonds.

4. Recognize these as resonance structures

The three structures differ only in the placement of electrons, not atoms. The double-headed arrow ↔\leftrightarrow connects them:

O−−C..=O↔O=C..−O−↔(third equivalent form)O^- - \overset{..}{C} = O \quad \leftrightarrow \quad O = \overset{..}{C} - O^- \quad \leftrightarrow \quad \text{(third equivalent form)}

Important

Resonance structures are not in equilibrium. The ion doesn't flip between them. The true structure is a hybrid—a weighted average where each C–O bond has partial double-bond character.

5. Determine the bond order

In the hybrid, each C–O bond is identical. Across the three resonance structures, each oxygen participates in one double bond (bond order 2) and two single bonds (bond order 1 each) when you sum over all structures. Averaging:

Bond order per C–O=1×2+2×13=43≈1.33\text{Bond order per C–O} = \frac{1 \times 2 + 2 \times 1}{3} = \frac{4}{3} \approx 1.33

This intermediate bond order (between single and double) matches experimental bond lengths of about 129 pm—shorter than a typical C–O single bond (143 pm) but longer than a C=O double bond (120 pm).

6. Geometry and charge distribution

The ion is trigonal planar with sp2sp^2 hybridization on carbon and bond angles of 120°. The negative charge is delocalized equally over all three oxygens, so each carries −23-\frac{2}{3} of an electron's charge. This delocalization stabilizes the ion significantly compared to any single structure.

Tip

Whenever you see equivalent atoms around a central atom and not enough electrons to give them all double bonds, think resonance. The π\pi system will delocalize.

✓Final answer

The CO32−CO_3^{2-} ion is best described by three equivalent resonance structures, each with one C=O double bond and two C–O single bonds in different positions, resulting in a resonance hybrid with delocalized π\pi bonding, equal C–O bond lengths (bond order 43\frac{4}{3}), and trigonal planar geometry.

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