Explain the important aspects of resonance with reference to the CO₃²⁻ ion.

Exercises · 4.11

CO_3^{2-}

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Concept understanding — Lewis Dot Structures

Why do we need Lewis dot structures?

Atoms are held together in molecules by chemical bonds — but what exactly is a bond? In the early 20th century, Gilbert N. Lewis realised that the key lies in the valence electrons (the outermost electrons). He noticed that atoms of noble gases (like Ne, Ar) are extremely stable and unreactive, and they all have 8 electrons in their outermost shell (except helium, which has 2). This led to the octet rule: atoms tend to gain, lose, or share electrons to achieve a full outer shell of 8 electrons (or 2 for hydrogen).

Lewis dot structures are simply a shorthand picture of this idea. They show:

Which atoms are connected to which
How many valence electrons each atom contributes
How those electrons are arranged as bonding pairs (shared) or lone pairs (unshared)

The precise statement

A Lewis dot structure (or electron dot structure) represents the valence electrons of an atom or molecule using dots placed around the element's symbol. Each dot stands for one valence electron. Shared pairs (bonds) are shown as lines, and unshared pairs as pairs of dots.

For a single atom, you write the element symbol and place dots on its four sides (top, bottom, left, right) — up to 8 dots. The order of filling doesn't matter for the final picture, but conventionally you place one dot on each side first, then pair them up.

For example:

Carbon (group 14, 4 valence electrons): $\cdot \text{C} \cdot$ (four single dots)
Oxygen (group 16, 6 valence electrons): $\cdot \ddot{\text{O}} \cdot$ (two single dots and two pairs)

How to draw a Lewis structure for a molecule

Here's the step-by-step method you'll use in exams:

Count total valence electrons — add up valence electrons from all atoms. For ions, add 1 electron for each negative charge, subtract 1 for each positive charge.
Identify the central atom — usually the least electronegative element (not hydrogen or fluorine). Place it in the centre.
Connect atoms with single bonds — each bond uses 2 electrons. Subtract these from your total.
Complete octets of outer atoms — place remaining electrons as lone pairs on terminal atoms (except hydrogen, which only needs 2).
Place leftover electrons on the central atom — if any remain.
If the central atom has fewer than 8 electrons, form multiple bonds — move lone pairs from outer atoms to create double or triple bonds until the central atom has an octet.

Watch out

A common mistake: forgetting that hydrogen only needs 2 electrons (a duet), not 8. Never put more than 2 electrons around H.

A concrete example: water (H₂O)

Total valence electrons: O has 6, each H has 1 → $6 + 1 + 1 = 8$ electrons.
Central atom: oxygen (least electronegative after H).
Connect: O—H bonds (2 bonds × 2 electrons = 4 electrons used).
Remaining: $8 - 4 = 4$ electrons → place as two lone pairs on oxygen.
Check: O has 2 bonds (4 electrons) + 2 lone pairs (4 electrons) = 8. Each H has 1 bond (2 electrons) = 2. Done.

The structure: $\text{H} - \ddot{\text{O}} - \text{H}$

What the structure tells you

Once drawn, a Lewis structure reveals:

Bond order (single, double, triple)
Lone pairs (which affect molecular shape and reactivity)
Formal charge (a bookkeeping tool to check which structure is most stable) …

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Resonance describes how the carbonate ion's structure is a hybrid of three equivalent forms, each with one C=O double bond and two C–O single bonds in different positions. The real ion has all three C–O bonds identical and intermediate in character, with the negative charge distributed equally over all three oxygens.

Why resonance exists

A single Lewis structure sometimes cannot capture the true electron distribution in a molecule. When we try to draw $CO_3^{2-}$ , we face a choice: which oxygen gets the double bond? The answer is that no single oxygen is special—the molecule doesn't "pick" one structure. Instead, the actual ion is a blend, or resonance hybrid, of all valid Lewis structures. This isn't about the molecule flipping between forms; it's about our notation being inadequate to show delocalized electrons in one picture.

Drawing the resonance structures of $CO_3^{2-}$

1. Count valence electrons

Carbon contributes 4, each oxygen contributes 6, and the $2-$ charge adds 2 more:

4 + 3(6) + 2 = 24 \text{ valence electrons}

2. Sketch the skeleton

Carbon is the central atom (less electronegative), bonded to three oxygens in a trigonal planar arrangement.

3. Distribute electrons to satisfy octets

If we place single bonds to all three oxygens (using 6 electrons), we have 18 electrons left. Putting lone pairs on the oxygens and forming one C=O double bond to satisfy carbon's octet, we get:

\begin{array}{ccc} \text{Structure I} & \text{Structure II} & \text{Structure III} \\[0.5em] \begin{array}{c} \phantom{-}O^- \\ \phantom{-}| \\ \phantom{-}C=O \\ \phantom{-}| \\ \phantom{-}O^- \end{array} & \begin{array}{c} \phantom{-}O \\ \phantom{-}|| \\ \phantom{-}C-O^- \\ \phantom{-}| \\ \phantom{-}O^- \end{array} & \begin{array}{c} \phantom{-}O^- \\ \phantom{-}| \\ \phantom{-}C-O^- \\ \phantom{-}|| \\ \phantom{-}O \end{array} \end{array}

Each structure has one C=O double bond (bond order 2) and two C–O single bonds (bond order 1), with the double bond in a different position. These three structures are equivalent by symmetry—they have the same energy.

Tip

The double-headed arrow $\leftrightarrow$ between resonance structures means "contributes to the hybrid," NOT a chemical equilibrium. The ion does not oscillate; it exists as the average of all forms simultaneously.

Key aspects of resonance in $CO_3^{2-}$

4. The resonance hybrid

The actual carbonate ion is a resonance hybrid—a weighted average of all three structures. Because the three forms are equivalent, each contributes equally (⅓ each). The result:

All three C–O bonds are identical in length and strength
Each bond has character intermediate between single and double: bond order = $\frac{1 + 2 + 1}{3} = \frac{4}{3} \approx 1.33$
The $2-$ charge is delocalized equally over all three oxygens, so each carries $-\frac{2}{3}$ formal charge

5. Experimental evidence

X-ray crystallography confirms that all three C–O bond lengths in $CO_3^{2-}$ are identical at about 129 pm—shorter than a typical C–O single bond (~143 pm) but longer than a C=O double bond (~120 pm). This is direct proof of resonance.

6. Stability through delocalization …

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Q.Explain the important aspects of resonance with reference to the CO32−CO_3^{2-}CO32−​ ion.

Why resonance exists

Drawing the resonance structures of CO32−CO_3^{2-}CO32−​

Key aspects of resonance in CO32−CO_3^{2-}CO32−​

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Q.Explain the important aspects of resonance with reference to the $CO_3^{2-}$ ion.

Drawing the resonance structures of $CO_3^{2-}$

Key aspects of resonance in $CO_3^{2-}$

Unlock everything free for 14 days