Lewis Structures, Formal Charge & Resonance Notes

Lewis Structures, Formal Charge & Resonance

At the heart of chemistry lies a deceptively simple question: how do atoms share, transfer, or hold onto electrons when they form chemical bonds? In 1916, the American physical chemist Gilbert Newton Lewis (1875–1946) published a landmark paper that answered this question with an elegant visual language — one that remains, over a century later, the first tool every chemistry student learns for understanding molecular structure.

Lewis structures are a dot-and-line notation that maps the valence electrons of atoms onto molecular architectures. But drawing them correctly requires two companion skills: formal charge analysis (which tells us whether a proposed structure is reasonable) and resonance theory (which tells us when a single Lewis structure is insufficient to describe reality).

Together, these three ideas — Lewis structures, formal charge, and resonance — form a unified framework that bridges the gap between isolated atoms and the molecules they build.

Part I: Lewis Structures

1.1 Before Lewis: The Electron and the Chemical Bond

By the early 1900s, J.J. Thomson had discovered the electron (1897), and Rutherford had proposed the nuclear atom (1911). Chemists knew that atoms contained negatively charged electrons and positively charged nuclei, but the connection between this atomic structure and chemical bonding remained unclear.

Several ideas were in the air:

1.2 Lewis's 1916 Paper and the Cubic Atom

Lewis's original 1916 paper ("The Atom and the Molecule", *Journal of the American Chemical Society*) proposed a model in which:

Lewis's cubic model was geometrically restrictive, but the electron-pair bonding concept was revolutionary. In 1923, Lewis published his textbook *Valence and the Structure of Atoms and Molecules*, which refined the theory and introduced the dot notation we use today.

1.3 Langmuir's Contribution

Irving Langmuir (1881–1957), working at General Electric, independently developed many of the same ideas and was instrumental in popularizing Lewis's notation. Langmuir coined the term "covalent bond" and championed the octet rule. He received the Nobel Prize in Chemistry in 1932.

2.1 What Are Valence Electrons?

Valence electrons are the electrons in the outermost shell (highest principal quantum number, *n*) of an atom. They are the electrons available for bonding and chemical reactions.

The number of valence electrons for a main-group element equals its group number in the periodic table:

For transition metals, the situation is more complex because *d* electrons can participate in bonding. Lewis structures are primarily designed for main-group elements.

2.2 Lewis Dot Symbols for Atoms

The Lewis dot symbol for an atom represents its valence electrons as dots placed around the element symbol. The first four electrons are placed singly on four sides (top, right, bottom, left), then pairing begins:

More precisely:

The number of unpaired electrons in an atom's Lewis symbol often corresponds to the number of bonds it typically forms:

3.1 Statement

The octet rule, formulated by Lewis and refined by Langmuir, states:

Atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with 8 electrons in their valence shell (or 2 for hydrogen and helium).

This is essentially a statement that atoms strive for the electron configuration of the nearest noble gas.

3.2 Why Eight?

The number 8 arises from the quantum mechanical structure of atoms:

3.3 The Duet Rule for Hydrogen and Helium

Hydrogen and helium have only the 1*s* orbital in their valence shell. They follow the duet rule — they are stable with 2 electrons (analogous to the electron configuration of helium, 1s²).

3.4 Exceptions to the Octet Rule

The octet rule is a powerful guideline, but it is not a law of nature. There are several categories of exceptions:

a) Incomplete Octets (Fewer Than 8 Electrons)

Some elements, particularly those in groups 1–3, can be stable with fewer than 8 electrons:

       F
       |
   F — B — F         (B has 6 electrons, not 8)

BF₃ is electron-deficient and a strong Lewis acid — it readily accepts a lone pair from Lewis bases like NH₃.

  H — Be — H          (Be has 4 electrons)

Why are incomplete octets stable for B and Be?

These atoms are small and have low-lying empty orbitals. Having fewer bonds with lower formal charge is more energetically favorable than forcing an octet with high formal charges.

b) Expanded Octets (More Than 8 Electrons)

Elements in period 3 and below can accommodate more than 8 electrons because they have accessible *d* orbitals (or, more accurately in modern theory, because the larger atomic size allows more atoms to be accommodated around the central atom):

       Cl
       |
  Cl — P — Cl
       |
      Cl
       |
      Cl         (P has 10 electrons — trigonal bipyramidal)

  S surrounded by 6 F atoms (octahedral)

Important caveat: The role of *d* orbitals in expanded octets has been questioned by modern computational chemistry. Studies by Magnusson (1990) and others suggest that *d* orbital participation is minimal and that the expanded octet description is best understood through hypervalent bonding models involving ionic character and multi-center bonding. However, for the purposes of drawing Lewis structures, the expanded octet notation remains a useful practical tool.

c) Odd-Electron Species (Radicals)

Some molecules have an odd number of electrons, making a complete octet impossible for at least one atom:

  :N=Ö:  with one unpaired electron    (or :N̈—Ö· with a lone pair on N)

NO is a free radical — it has one unpaired electron. It is highly reactive and plays crucial roles in biology (vasodilator, neurotransmitter) and atmospheric chemistry.

  :O—Ṅ=O:   with one unpaired electron on N

NO₂ dimerizes to N₂O₄ to pair up its unpaired electrons:

  O₂N—NO₂

d) Paramagnetic Molecules

Molecular oxygen (O₂) has 12 valence electrons. A Lewis structure predicts all electrons are paired:

:O=O:           (all electrons paired → diamagnetic?)

Yet experimentally, O₂ is paramagnetic — it is attracted to a magnetic field, indicating two unpaired electrons. This is a famous failure of Lewis theory and is correctly explained by molecular orbital (MO) theory, which shows that the two highest-energy electrons in O₂ occupy degenerate π* antibonding orbitals with parallel spins (Hund's rule).

This is one of the most important limitations of Lewis structures.

4.1 Single Bonds

A single bond consists of one shared pair of electrons (2 electrons). It is represented by a single line (—) between two atoms.

Example — Hydrogen (H₂):

H ·  +  · H   →   H : H   or   H—H

Each hydrogen now has 2 electrons (duet rule satisfied).

Example — Water (H₂O):

        H
        |
   H — O:         (O has 2 bonding pairs + 2 lone pairs = 8 electrons)
        (2 lone pairs on O)

4.2 Double Bonds

A double bond consists of two shared pairs of electrons (4 electrons). It is represented by a double line (=).

Example — Oxygen (O₂):

:O = O:           (each O has 2 bonding pairs + 2 lone pairs = 8 electrons)

Example — Carbon dioxide (CO₂):

:O = C = O:       (C has 4 bonding pairs; each O has 2 bonding + 2 lone pairs)

4.3 Triple Bonds

A triple bond consists of three shared pairs of electrons (6 electrons). It is represented by a triple line (≡).

Example — Nitrogen (N₂):

:N ≡ N:           (each N has 3 bonding pairs + 1 lone pair = 8 electrons)

This is the strongest bond in chemistry among homonuclear diatomics (bond energy = 945 kJ/mol), which explains why N₂ is so unreactive.

Example — Carbon monoxide (CO):

:C ≡ O:           (with lone pairs, see formal charge discussion below)

4.4 Coordinate (Dative) Bonds

A coordinate bond (also called a dative bond or coordinate covalent bond) forms when both electrons in the shared pair come from the same atom. After formation, a coordinate bond is identical to a regular covalent bond.

Example — Ammonium ion (NH₄⁺):

        H
        |
   H — N — H    ← The fourth N—H bond is a coordinate bond
        |          (both electrons from N's lone pair)
        H

The nitrogen in NH₃ donates its lone pair to H⁺:

   H                H
    \                \
     N:  +  H⁺  →   N—H   ⁺
    /                /
   H                H

Example — Hydronium ion (H₃O⁺):

        H
        |
   H — O — H    ⁺   (one O—H bond is a coordinate bond)

4.5 Bond Order, Bond Length, and Bond Strength

The bond order is the number of shared electron pairs between two bonded atoms:

General trends:

Step 1: Count Total Valence Electrons

Add up the valence electrons from each atom. For polyatomic ions, add electrons for negative charges or subtract electrons for positive charges:

Example — Carbonate ion (CO₃²⁻):

Step 2: Identify the Central Atom

The central atom is typically:

Exception: Hydrogen is always terminal (it can only form one bond). In oxyacids (H₂SO₄, HNO₃, etc.), hydrogen is bonded to oxygen, and oxygen bridges to the central atom.

Step 3: Draw Single Bonds from the Central Atom to Each Surrounding Atom

Each single bond uses 2 electrons. Subtract these from the total.

Step 4: Distribute Remaining Electrons as Lone Pairs

Starting with the outer (terminal) atoms, give each enough electrons to complete its octet (or duet for H). Then place any remaining electrons on the central atom.

Step 5: If the Central Atom Still Lacks an Octet, Form Multiple Bonds

Convert lone pairs from terminal atoms into bonding pairs (double or triple bonds) between the terminal and central atoms until the central atom has an octet.

Step 6: Verify

Count all electrons to ensure you've used exactly the number calculated in Step 1. Check that each atom (except H) has an octet.

Worked Examples

Example 1: Water (H₂O)

Step 1: Total valence electrons

Step 2: Central atom = O (less electronegative than... well, O is the only non-hydrogen)

Step 3: Draw single bonds

H — O — H        (2 bonds × 2 e⁻ = 4 e⁻ used, 4 remaining)

Step 4: Distribute remaining electrons

        :  
   H — O — H      (4 remaining electrons → 2 lone pairs on O)
        :

Step 5: Check octets

Result:

   H — Ö — H

Bent molecular geometry, 104.5° bond angle.

Example 2: Carbon Dioxide (CO₂)

Step 1: Total valence electrons

Step 2: Central atom = C (less electronegative)

Step 3: Single bonds

O — C — O          (2 bonds × 2 e⁻ = 4 e⁻ used, 12 remaining)

Step 4: Distribute remaining electrons

:Ö: — C — :Ö:     (12 e⁻ distributed: 6 on each O → 3 lone pairs each)

But wait — C only has 4 electrons (2 from each bond). It needs 4 more.

Step 5: Form double bonds

Convert one lone pair from each O into a bonding pair:

:O = C = O:         (each O now has 2 lone pairs + 2 bonding pairs)

Step 6: Verify

Result:

:O = C = O:

Linear geometry, 180° bond angle.

Example 3: Nitrate Ion (NO₃⁻)

Step 1: Total valence electrons

Step 2: Central atom = N

Step 3: Single bonds

   O
   |
O—N—O              (3 bonds × 2 e⁻ = 6 e⁻ used, 18 remaining)

Step 4: Distribute remaining electrons

   :Ö:
   |
:Ö—N—Ö:            (18 e⁻: 6 on each O → 3 lone pairs)

N has only 6 electrons (3 bonds). Needs 2 more.

Step 5: Form a double bond

   :O:
   ‖
:O—N—O:            (convert one lone pair from any O into a double bond)

Now N has 8 electrons (1 double bond + 2 single bonds = 4 + 2 + 2 = 8).

Result (one resonance structure):

       :O:
       ‖
  :O — N — O: ⁻

But as we will discuss in Part III (Resonance), the double bond is not localized to one N—O bond. It is delocalized over all three N—O bonds. The true structure is a resonance hybrid of three equivalent structures.

Example 4: Sulfuric Acid (H₂SO₄)

Step 1: Total valence electrons

Step 2: Central atom = S (less electronegative than O; H is always terminal)

Step 3: Draw the connectivity. H₂SO₄ has the structure where two H atoms are bonded to O atoms, and the O atoms are bonded to S:

     O
     ‖
HO — S — OH
     ‖
     O

Step 4-5: After single bonds (S—O—H × 2, S—O × 2) and lone pairs:

       :O:
       ‖
  :O — S — O:     
       ‖
       :O:

(with each O having lone pairs to complete octets)

Step 6: S has 12 electrons (expanded octet) — allowed for period 3 elements.

Example 5: Phosphorus Pentachloride (PCl₅)

Step 1: Total valence electrons

Step 2: Central atom = P

Step 3-5:

       Cl
       |
  Cl — P — Cl
       |
      Cl
       |
      Cl

P has 10 electrons (5 bonds) — expanded octet. Each Cl has 3 lone pairs + 1 bond = 8.

Total: 5 bonds (10e⁻) + 15 lone pairs (30e⁻) = 40 ✓

Trigonal bipyramidal geometry.

Example 6: Xenon Tetrafluoride (XeF₄)

Step 1: Total valence electrons

Step 2: Central atom = Xe

Step 3-5:

       F
       |
   F — Xe — F       (Xe also has 2 lone pairs)
       |
       F

Xe has 4 bonds + 2 lone pairs = 12 electrons (expanded octet). Square planar geometry.

6.1 Hydrogen Bonding Partners (Oxyacids)

In oxyacids like H₂SO₄, H₃PO₄, and HClO₄, the hydrogen atoms are bonded to oxygen atoms, not directly to the central atom. The structure is:

    O                 O
    ‖                 ‖
HO — S — OH     HO — P — OH
    ‖                 ‖
    O                 OH

The general formula is: central atom bonded to O, and some of those O atoms have H attached.

6.2 Polyatomic Ions as Units

When drawing Lewis structures for salts (e.g., NH₄Cl), draw the polyatomic ion (NH₄⁺) separately from the counterion (Cl⁻):

        H
        |
   H — N — H   ⁺      :Cl:⁻
        |
        H

6.3 Molecules with Multiple Central Atoms

For larger molecules like ethane (C₂H₆), ethylene (C₂H₄), or acetylene (C₂H₂), there are multiple central atoms:

Ethane (C₂H₆):

   H   H
    \ /
     C — C
    / \ / \
   H   H   H

All single bonds, each C has 8 electrons.

Ethylene (C₂H₄):

   H     H
    \   /
     C = C
    /   \
   H     H

Double bond between carbons, each C has 8 electrons.

Acetylene (C₂H₂):

H — C ≡ C — H

Triple bond between carbons, each C has 8 electrons.

Part II: Formal Charge

Often, multiple Lewis structures can be drawn for the same molecule, all satisfying the octet rule but differing in where double bonds, lone pairs, or charges are placed. How do we decide which is the best (most stable, most likely to represent reality)?

Formal charge is the primary tool for making this decision.

The formal charge on an atom in a Lewis structure is the charge it would have if all bonding electrons were shared equally between the bonded atoms:

Where:

Equivalently, since B/2 counts the number of bonds (each bond contributes 1 electron "belonging" to each atom under equal sharing):

Or even more simply:

where "dots" = lone pair electrons and "lines" = number of bonds.

Intuition

Formal charge answers the question: "If we pretend the bonding electrons are shared equally, does this atom have more or fewer electrons than it started with?"

Example 1: Carbon Dioxide (CO₂)

Structure: :O = C = O:

All formal charges are zero. This is the best Lewis structure for CO₂.

Example 2: Carbon Monoxide (CO)

CO has 10 valence electrons. One valid Lewis structure is:

:C ≡ O:

Formal charges: C = −1, O = +1

This seems counterintuitive — oxygen is more electronegative, yet it bears the positive formal charge. However, formal charge is not the same as actual charge (discussed below). This structure is actually the best representation because:

Example 3: Nitrate Ion (NO₃⁻)

Three resonance structures exist (all single bonds and one double bond):

Structure A (double bond to top O):

       :O:⁻
       |
  :O = N — O: ⁻   

Wait, let me be more careful.

For NO₃⁻ (24 electrons total):

Structure with one N=O double bond:

       :O:
       ‖
  :O — N — O: ⁻

Sum of formal charges: +1 + 0 + (−1) + (−1) = −1 ✓ (matches the ion charge)

Example 4: Ammonium Ion (NH₄⁺)

        H
        |
   H — N — H  ⁺
        |
        H

Sum: +1 + 0(×4) = +1 ✓

Example 5: Ozone (O₃)

Ozone has 18 valence electrons. Two major resonance structures:

Structure A:

:O = O — O: ⁻     (double bond on left)

Sum: 0 + 1 + (−1) = 0 ✓

When multiple Lewis structures are possible, the best structure is the one that satisfies these criteria, applied in this priority order:

Rule 1: Minimize Formal Charges

The best structure has formal charges closest to zero on all atoms. A structure where every atom has FC = 0 is ideal.

Rule 2: No Like Charges on Adjacent Atoms

Avoid placing positive charges next to positive charges, or negative charges next to negative charges. Like charges repel.

Rule 3: Negative Formal Charges on More Electronegative Atoms

If formal charges are unavoidable:

This is intuitive: electronegative atoms "want" more electron density.

Rule 4: Avoid Excessive Formal Charges

Structures with formal charges of ±2 or higher are generally very unfavorable unless no alternatives exist.

Rule 5: Respect the Octet Rule for C, N, O, F

For second-period elements (C, N, O, F), do not exceed an octet. Expanded octets are only allowed for period 3 and beyond (P, S, Cl, etc.).

Summary of Priorities (in order of importance):

Formal charge and oxidation state are two different bookkeeping systems for assigning charges to atoms:

Neither formal charge nor oxidation state represents the actual charge on an atom, which is determined by electron density distributions (computable by quantum mechanical methods). In reality, atoms in molecules have partial charges that are typically much smaller than either formal charge or oxidation state would suggest.

Case Study: Carbon Monoxide (CO)

Three possible Lewis structures for CO:

Structure A: :C ≡ O: (triple bond, each atom has 1 lone pair)

Structure B: :C = O: (double bond, C has 2 lone pairs, O has 2 lone pairs)

C: 4 + O: 6 = 10. :C = O: has 2 lone pairs on C (4e⁻) + 2 bonds (4e⁻) + 2 lone pairs on O (4e⁻) = 12. Too many electrons.

Actually :C = O: with only 1 lone pair on C and 2 on O: 2 + 4 + 4 = 10. ✓

FC: C = 4 − 2 − 2 = 0, O = 6 − 4 − 2 = 0

Hmm, but C only has 6 electrons (1 lone pair + 2 bonds = 2 + 4 = 6). Incomplete octet.

Structure C: :C — O: (single bond, C has 3 lone pairs, O has 3 lone pairs)

OK let me redo this properly.

CO has 10 valence electrons.

Structure A: :C ≡ O:

Structure B: :C = Ö:

So Structure B uses too many electrons. That rules it out.

Structure C: :Ċ — Ö:

Way too many electrons. Ruled out.

So the only valid Lewis structure with octets is :C ≡ O: — the triple bond structure with FC(C) = −1, FC(O) = +1. This is the correct Lewis structure for CO.

Case Study: Nitrogen Dioxide (NO₂)

NO₂ has 17 valence electrons (odd-electron molecule).

Structure A:

:O = N — O·

(N has 1 double bond + 1 single bond + 1 unpaired electron = 7 electrons on N... not quite 8)

Let me be more careful:

Actually, for NO₂ with 17 electrons:

Structure A:

:O = N̈ — Ö:

That's N with 1 lone pair, 1 double bond, 1 single bond: 2 + 4 + 2 = 8 on N ✓

O(double): 4 lone e⁻ + 4 bonding e⁻ = 8 ✓

O(single): 6 lone e⁻ + 2 bonding e⁻ = 8 ✓

Total lone pair e⁻: 2 + 4 + 6 = 12

Total bonding e⁻: 4 + 2 = 6

Total: 18. But we need 17.

Hmm, since it's a radical (odd electron), one atom must have an unpaired electron.

Let me try:

:O = N — Ö·

N: 1 lone pair + 2 bonds (double + single) = 2 + 4 + 2 = 8 ✓

O(double): 4 lone e⁻ + 4 bonding = 8 ✓

O(single): 6 lone e⁻ + 1 unpaired + 2 bonding = 9... that's 9 electrons on O, which exceeds octet for a second-period element.

Let me try:

·O — N = O:

where the dot on the left O is an unpaired electron.

Actually, let me count properly for each structure.

17 valence electrons to distribute.

Structure: O=N—Ö·

Let me just count electrons around each atom properly:

Total: 8 + 8 + 7 = 23... no that counts bonding electrons twice.

Total unique electrons = non-bonding (4 + 2 + 5 = 11) + bonding (6) = 17 ✓

OK this works. The unpaired electron is on the terminal O that's single-bonded. But it could also be on N:

Structure: ·Ö—N=O:

with the unpaired electron on the singly-bonded O. FC:

All FC = 0! This is actually a good structure.

But the real NO₂ has the unpaired electron delocalized (resonance between the two equivalent structures).

OK, I think I've spent enough time working out the details of individual examples. Let me move on and write the comprehensive text without getting bogged down in every electron count.

Formal charge is not just an exercise in bookkeeping — it has real predictive power:

a) Predicting Molecular Stability

Structures with all zero formal charges are generally the most stable. This explains why:

b) Predicting Reactivity

Atoms with non-zero formal charges are often the reactive sites in a molecule:

c) Predicting Acid-Base Behavior

In oxyacids, formal charge helps explain acid strength:

For oxyacids with the general structure (HO)_m XO_n:

Example:

Part III: Resonance

Consider the carbonate ion, CO₃²⁻. If we draw a Lewis structure with one C=O double bond and two C—O single bonds, we get a perfectly valid Lewis structure with reasonable formal charges. But there's a problem: experimental evidence shows that all three C—O bonds in CO₃²⁻ are identical — they all have the same bond length (129 pm), which is intermediate between a C—O single bond (143 pm) and a C=O double bond (120 pm).

No single Lewis structure can account for this. The double bond can be placed on any of the three oxygen atoms, giving three equivalent structures. The real ion is a composite of all three.

This is the resonance problem, and resonance theory is its solution.

2.1 Definition

Resonance is a concept that describes molecular species that cannot be adequately represented by a single Lewis structure. Instead, the true electronic structure is described as a weighted average (superposition) of two or more contributing structures (also called resonance structures or canonical forms).

Key points:

2.2 The Resonance Hybrid

The resonance hybrid is the actual molecule. It is represented by a double-headed arrow (↔) connecting the contributing structures:

For the carbonate ion:

     :O:              :O:⁻             :O:⁻
      ‖                 |                 |
 :O — C — O: ⁻  ↔  O = C — O: ⁻  ↔  :O — C = O
                         ⁻                 ⁻
   Structure I      Structure II      Structure III

The hybrid has:

2.3 The Double-Headed Arrow (↔) vs. Equilibrium Arrow (⇌)

This distinction is critically important:

A common student mistake is confusing these two symbols. Resonance structures are NOT in equilibrium with each other. They do not represent different molecules.

2.4 Analogy

Think of the resonance hybrid as a mule — a mule is not a horse one moment and a donkey the next. It is a single, distinct animal that has characteristics of both. Similarly, the resonance hybrid is a single molecular species with characteristics of all its contributing structures.

Another analogy: if you describe a rhinoceros as a "cross between a tank and a unicorn," the rhinoceros is real and the tank and unicorn are the imaginary descriptions. The resonance structures are like the tank and unicorn — useful fictions.

Rule 1: Only Electrons Move, Not Atoms

In resonance, only π (pi) electrons and lone pair electrons can be redistributed. The positions of atoms (the σ-bond framework) never changes between resonance structures.

This means:

Rule 2: The Skeleton (σ Framework) Must Be Identical

All resonance structures must have the same connectivity of atoms. If two structures have different atom arrangements, they are different molecules, not resonance structures.

Rule 3: All Resonance Structures Must Have the Same Number of Electrons

The total number of valence electrons must be conserved across all contributing structures.

Rule 4: All Atoms (Except H) Should Satisfy the Octet Rule

Structures where all atoms have complete octets are more stable than those with incomplete octets (though exceptions exist for elements with expanded octets).

Rule 5: Evaluate Stability Using These Criteria

Among valid resonance structures, the most stable (lowest energy) contributors are weighted more heavily in the hybrid:

a) Minimize formal charges:

Structures with fewer formal charges are more stable. A structure with FC = 0 on all atoms is preferred.

b) Keep negative charges on more electronegative atoms:

If formal charges are unavoidable, place FC = −1 on O, N, F (electronegative atoms) rather than on C or H.

c) Avoid like charges on adjacent atoms:

Positive next to positive, or negative next to negative, is destabilizing.

d) Avoid charge separation:

A structure with +1 on one atom and −1 on another is less stable than one with no formal charges, even if the total charge is the same.

e) Preserve aromaticity (for cyclic π systems):

Aromatic resonance structures (following Hückel's rule: 4n+2 π electrons in a cyclic, planar, conjugated system) are particularly stable.

f) Expanded octets are allowed for period 3+ elements:

Structures with 10 or 12 electrons around S, P, Cl, etc. are acceptable if they reduce formal charges.

Rule 6: Equivalent Resonance Structures Contribute Equally to the Hybrid

When all contributing structures are identical in energy (as in CO₃²⁻ or C₆H₆), each contributes equally. When structures have different energies, the lowest energy structure contributes most to the hybrid.

Organic chemists use curved arrows (also called "electron-pushing" arrows) to show how electrons flow from one resonance structure to another:

The Rules of Curved Arrows:

Example: Carbonate Ion

     :O:⁻                  :O:
      |          →          ‖
 :O — C = O         :O = C — O: ⁻
      (structure I)       (structure II)

Curved arrow: Start at the lone pair on the singly-bonded O⁻, draw to the C—O bond position → converts that O's lone pair into a C—O double bond, while simultaneously the adjacent C=O double bond breaks and becomes a lone pair on that oxygen.

More precisely:

Curved Arrows Represent Electron Redistribution, Not Physical Motion

The arrows do not mean that electrons physically move from one location to another in the actual molecule. In the hybrid, the electrons are already delocalized. The arrows simply show the logical relationship between two formal representations.

Example 1: Ozone (O₃)

Ozone has 18 valence electrons and two equivalent resonance structures:

:O = O — O: ⁻     ↔     ⁻:O — O = O:
  Structure I              Structure II

Formal charges in Structure I:

Hybrid:

Experimental evidence:

Example 2: Benzene (C₆H₆)

Benzene is the most famous example of resonance. The two Kekulé structures are:

    H               H                  H               H
    |               |                  |               |
    C       C       C                  C       C       C
   / \     / \     / \       ↔       / \     / \     / \
  C    C  C    C  C    C            C    C  C    C  C    C
   \  / \  / \  / \  /               \  / \  / \  / \  /
    C       C       C                  C       C       C
    |               |                  |               |
    H               H                  H               H

Or more simply:

   ⬡ (with alternating double bonds)   ↔   ⬡ (with alternating double bonds, shifted by one)

Key features of the hybrid:

The circle-in-hexagon notation represents the delocalized π system:

    C — C
   / ⬡ \
  C     C
   \   /
    C — C

Benzene also has additional (less important) resonance structures involving charge separation (Dewar structures):

    C⁺—C           ⁻C—C⁺
   /     \        /      \
  C       C⁻  ↔ C        C
   \     /        \      /
    C—C             C—C

These contribute less to the hybrid because they involve charge separation, but they are part of the full resonance picture.

Example 3: Carboxylate Ion (RCO₂⁻)

The carboxylate ion has two equivalent resonance structures:

         :O:              :O:⁻
          ‖                 |
   R — C — O: ⁻     ↔    R — C = O

Hybrid:

Experimental evidence:

Example 4: Nitrate Ion (NO₃⁻) — Complete Analysis

Three equivalent resonance structures:

    :O:              :O:⁻             :O:⁻
     ‖                 |                 |
:O — N — O: ⁻  ↔  O = N — O: ⁻  ↔  :O — N = O
                         ⁻                 ⁻
  Structure I      Structure II      Structure III

Hybrid:

Experimental evidence:

Example 5: Amide Ion Resonance in Peptide Bonds

The peptide bond in proteins exhibits resonance, making it one of the most biologically significant examples:

        O          O⁻
        ‖          |
 —C — N — H  ↔  —C = N⁺ — H
        |              |
        R              R

Consequences of this resonance:

Example 6: Allyl System (C₃H₅⁺, C₃H₅·, C₃H₅⁻)

The allyl cation (CH₂=CH—CH₂⁺) has two resonance structures:

CH₂ = CH — CH₂⁺    ↔    ⁺CH₂ — CH = CH₂

The positive charge is delocalized over the terminal carbons.

The allyl anion (CH₂=CH—CH₂⁻) has:

CH₂ = CH — CH₂⁻    ↔    ⁻CH₂ — CH = CH₂

The negative charge is delocalized over the terminal carbons.

These allylic systems are crucial in organic chemistry because the resonance stabilization makes allylic intermediates more stable than simple alkyl cations or anions.

Example 7: Phenoxide Ion

The phenoxide ion (C₆H₅O⁻) shows how the negative charge on oxygen is delocalized into the benzene ring:

    O⁻             O              O              O
    |              ‖              |              |
   ⬡        ↔     ⬡⁻       ↔    ⬡⁻       ↔    ⬡⁻
                       (with negative charge on ortho/para positions)

This resonance stabilization makes phenol (pKa ≈ 10) much more acidic than cyclohexanol (pKa ≈ 16) — the phenoxide ion is stabilized by charge delocalization into the ring.

6.1 Definition

The resonance energy (or delocalization energy) is the difference in energy between the actual molecule (the resonance hybrid) and the most stable contributing resonance structure:

Since the hybrid is always more stable than any individual contributing structure, the resonance energy is always positive (stabilizing).

6.2 Measured Resonance Energies

Trend: The more resonance structures that contribute significantly, and the more equivalent they are, the greater the resonance stabilization.

6.3 Resonance and Chemical Stability

Resonance energy explains many experimental observations:

The bond order in a resonance hybrid is calculated as:

Alternatively:

Examples:

Misconception 1: "The molecule switches between resonance structures."

Reality: The molecule exists in a single, unchanging state — the resonance hybrid. It does not oscillate between different structures. The resonance structures are a human invention to describe a single reality that cannot be captured by one Lewis structure.

Misconception 2: "Resonance structures are real."

Reality: Individual resonance structures do not exist. Only the hybrid exists. The contributing structures are mathematical components of a quantum mechanical description, not physical entities.

Misconception 3: "Resonance requires identical structures."

Reality: Resonance structures can be (and often are) different in energy. The most stable structure contributes the most to the hybrid. Equivalent structures contribute equally, but non-equivalent structures can also participate in resonance — they simply contribute less.

Misconception 4: "Double-headed arrows mean equilibrium."

Reality: The double-headed arrow (↔) represents resonance (a single molecule described by multiple structures). Equilibrium arrows (⇌) represent a chemical reaction between different molecules.

Misconception 5: "More resonance structures always mean more stability."

Reality: What matters is the quality of the resonance structures, not the quantity. Two high-quality (low-energy, all-octet, minimal-charge) resonance structures contribute more stabilization than ten poor ones. For example, both benzene and the hypothetical molecule 1,3,5-cyclohexatriene could be said to have two Kekulé structures, but the point is that benzene's hybrid is lower in energy than either structure alone.

Misconception 6: "Resonance is the same as tautomerism."

Reality: Tautomers are different molecules in equilibrium (e.g., keto-enol tautomerism: the keto form and enol form are different compounds with different atomic positions). Resonance structures represent the same molecule with different electron arrangements (same atomic positions).

10.1 Conjugated Systems

A conjugated system is a molecule with alternating single and double bonds (or lone pairs adjacent to π bonds), allowing electron delocalization over a continuous chain or ring.

Examples:

10.2 1,3-Butadiene

CH₂ = CH — CH = CH₂

Resonance structures:

CH₂ = CH — CH = CH₂    ↔    ⁺CH₂ — CH = CH — CH₂⁻    ↔    ⁻CH₂ — CH = CH — CH₂⁺
   Structure I                   Structure II                    Structure III

Structures II and III involve charge separation and contribute much less than Structure I. But the minor contribution still has effects:

10.3 Color and Conjugation

The absorption of visible light by molecules is directly related to resonance and conjugation. As the extent of conjugation increases:

This is why:

Part IV: Connecting All Three Concepts

When analyzing a molecule, the three concepts work together in a systematic workflow:

Step 1: LEWIS STRUCTURE
   ↓ Draw the structure using valence electrons, octet rule
Step 2: FORMAL CHARGE
   ↓ Calculate FC on each atom; evaluate structure quality
Step 3: RESONANCE
   ↓ If multiple valid structures exist, identify resonance contributors
   ↓ Calculate bond orders, charge distribution
Result: A complete electronic description of the molecule

Let's apply all three concepts to the sulfate ion.

Step 1: Lewis Structure

Total valence electrons: S(6) + 4×O(6) + 2(charge) = 32

Draw with all single bonds first:

   :O:
   |
:O—S—O:
   |
   :O:

Each O has 3 lone pairs, S has 0 lone pairs. Count: 4 bonds (8e⁻) + 12 lone pairs (24e⁻) = 32 ✓

S has 8 electrons (4 bonds) — octet satisfied.

Step 2: Formal Charge (all-single-bond structure)

Sum: +2 + 4(−1) = −2 ✓

But FC(S) = +2 is quite high. Can we do better?

Alternative: Use double bonds

Since S is in period 3, it can have an expanded octet. Place double bonds to reduce formal charges:

Structure with four double bonds:

    :O:
     ‖
O = S = O
     ‖
    :O:

Sum: −2 + 4(0) = −2 ✓

FC(S) = −2 is unfavorable (S is not very electronegative).

Best structure: Two double bonds, two single bonds:

    :O:                 :O:⁻
     ‖                    |
O = S — O: ⁻    ↔    (and other resonance structures)
     ‖
    :O:⁻

Sum: 0 + 0 + 0 + (−1) + (−1) = −2 ✓

S has FC = 0, and negative charges are on the electronegative oxygen atoms. This is the best combination.

Step 3: Resonance

There are 6 resonance structures corresponding to the 6 ways to choose 2 of the 4 oxygen atoms for double bonds (C(4,2) = 6):

Each structure has:

Hybrid:

Experimental evidence:

ATOMIC STRUCTURE (electrons, orbitals)
         ↓
LEWIS STRUCTURES (electron bookkeeping, dot-and-line notation)
         ↓
FORMAL CHARGE (evaluating structure quality)
         ↓
RESONANCE (describing electron delocalization)
         ↓
BOND ORDER & CHARGE DISTRIBUTION (quantitative predictions)
         ↓
MOLECULAR PROPERTIES (bond lengths, reactivity, acidity, color)

Each concept builds on the previous one. Lewis structures give us a first approximation; formal charge tells us which approximation is best; and resonance tells us when we need to go beyond a single approximation.

Part V: Limitations of Lewis Theory

Despite its enormous utility, Lewis theory has well-known limitations:

As discussed earlier, the Lewis structure of O₂ predicts all electrons are paired (diamagnetic), but O₂ is experimentally paramagnetic. Molecular orbital theory correctly predicts two unpaired electrons.

Lewis structures tell us about connectivity but not about shape. VSEPR theory (Valence Shell Electron Pair Repulsion) and hybridization theory are needed to predict geometry.

For example, both CH₄ and NH₃ have similar Lewis structures (4 electron pairs around the central atom), but CH₄ is tetrahedral while NH₃ is trigonal pyramidal. Lewis structures alone cannot distinguish them.

Resonance theory is qualitative. To get quantitative descriptions of electron delocalization, bond orders, and charge distributions, one needs molecular orbital theory or computational quantum chemistry (density functional theory, ab initio methods).

Lewis structures are designed for main-group elements. Transition metal complexes with their partially filled *d* orbitals, variable oxidation states, and complex bonding patterns require ligand field theory or molecular orbital theory for adequate description.

The concept of *s*, *p*, and *d* orbitals with specific directional properties is not captured by Lewis's dot notation. Valence bond theory and hybridization (sp, sp², sp³, etc.) provide the bridge between Lewis structures and molecular geometry.

Molecules like diborane (B₂H₆) have bridging hydrogen atoms with 3-center 2-electron bonds that cannot be represented by conventional Lewis structures. Multi-center bonding models are required.

Conclusion

Lewis structures, formal charge, and resonance form an integrated framework that is the starting point for understanding chemical bonding. Gilbert N. Lewis's insight — that molecules are held together by shared electron pairs, and that atoms strive for stable electron configurations — has proven to be one of the most powerful simplifications in all of science.

Lewis structures give us a map of where electrons are (and are not) in a molecule. Formal charge tells us which map is the most reasonable when multiple maps are possible. Resonance tells us when no single map is sufficient and we need a composite picture.

No, these tools are not perfect — quantum mechanics provides a deeper and more accurate description. But as first approximations that can be sketched on paper without a computer, Lewis structures remain indispensable. They predict reactivity, explain acidity, guide synthesis, and provide the language in which organic chemists think and communicate. Over a century after their introduction, they continue to be the foundation upon which all deeper understanding of molecular structure is built.

Key References: