Lewis Structures How atoms share electrons CHEM 1411 | Texas A&M University-Corpus Christi

Where we left off

Electron configurations describe electrons in isolated atoms.

But atoms don’t stay isolated.

When atoms bond, their valence electrons rearrange.

Lewis structures are the map.

Lewis Dot Symbols

Each dot = one valence electron.

HHeLiBNOFNeBe

Why do bonds form?

A shared electron is pulled by both nuclei at once.

That extra pull lowers energy, forming a bond.

Sharing electrons

One shared pair = single bond

Two shared pairs = double bond

Three shared pairs = triple bond

Single Bond

Single bond: one shared pair gives each hydrogen a duplet

HHHHHH

Duplet rule: Hydrogen and helium achieve stability with 2 valence electrons (one filled orbital).

Single Bond

One shared pair + three lone pairs on each fluorine

FFFFFF
FFFFFF

Not every electron pair is shared.

Octet rule: Main group atoms tend toward 8 valence electrons. Four orbitals, two electrons each.

Double Bond

Double bond: two shared pairs give each oxygen an octet

Octet rule: Main group atoms tend toward 8 valence electrons. Four orbitals, two electrons each.

Triple Bond

Triple bond: three shared pairs give each nitrogen an octet

NNNNNN
NNNNNN

Octet rule: Main group atoms tend toward 8 valence electrons. Four orbitals, two electrons each.

Ionic bonds

When the electronegativity difference is large enough, electrons transfer rather than share.

ClNaCl−Na+ClNa+−
NaClNaClNa+Cl−Na+Cl−+−

The bonding continuum

Most bonds are somewhere in between.

The electronegativity difference between two atoms determines where a bond falls on the spectrum.

Water (H2O)

Step 1: Total the number of valence electrons
Step 1: Total the number of valence electrons
H2O
O:
2 H:
Total ve:
2
6
8
Total ve:
8
Step 1: Total the number of valence electrons
O:
2 H:
2
6
H2O
Step 2: Place least electronegative atom (not H)
Placed ve:
0
8
Remaining ve:
Step 2: Place least electronegative atom (not H)
Placed ve:
0
8
Total ve:
8
Remaining ve:
Placed ve:
0
8
Total ve:
8
Remaining ve:
Step 2: Place least electronegative atom (not H)
Step 3: Place remaining atoms around central atom
Step 3: Place remaining atoms around central atom
Placed ve:
0
8
Total ve:
8
Remaining ve:
Placed ve:
0
8
Total ve:
8
Remaining ve:
Step 3: Place remaining atoms around central atom
Step 4: Draw a single bond between each terminal atom and the central atom.
Step 4: Draw a single bond between each terminal atom and the central atom.
Placed ve:
Total ve:
8
Remaining ve:
0
8
2 e–
2 e–
4
4
Placed ve:
4
4
Total ve:
8
Remaining ve:
2 e–
2 e–
Step 4: Draw a single bond between each terminal atom and the central atom.
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
Placed ve:
Total ve:
8
Remaining ve:
4
4
4 e–
0
0
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
4 e–
Placed ve:
0
0
Total ve:
8
Remaining ve:
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Octet and duplet patterns satisfied. Structure is good.

Formaldehyde (CH2O)

Step 1: Total the number of valence electrons
Step 1: Total the number of valence electrons
CH2O
O:
2 H:
Total ve:
2
6
12
C:
4
Total ve:
12
Step 1: Total the number of valence electrons
O:
2 H:
2
6
CH2O
C:
4
Step 2: Place least electronegative atom (not H)
Placed ve:
0
12
Remaining ve:
C
Step 2: Place least electronegative atom (not H)
Placed ve:
0
12
Total ve:
12
Remaining ve:
C
Placed ve:
0
12
Total ve:
12
Remaining ve:
Step 2: Place least electronegative atom (not H)
Step 3: Place remaining atoms around central atom
COHH
Step 3: Place remaining atoms around central atom
Placed ve:
0
12
Total ve:
12
Remaining ve:
COHH
Placed ve:
0
12
Total ve:
12
Remaining ve:
Step 3: Place remaining atoms around central atom
Step 4: Draw a single bond between each terminal atom and the central atom.
COHH
Step 4: Draw a single bond between each terminal atom and the central atom.
Placed ve:
Placed ve:
12
12
Total ve:
Total ve:
Remaining ve:
Remaining ve:
0
12
6
6
2 e–
2 e–
2 e–
COHH
Total ve:
12
Placed ve:
6
6
6
Remaining ve:
2 e–
2 e–
Step 4: Draw a single bond between each terminal atom and the central atom.
2 e–
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
COHH
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
Total ve:
12
Placed ve:
Remaining ve:
6
6
12
0
6 e–
COHH
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
Total ve:
12
Placed ve:
12
0
Remaining ve:
6 e–
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
COHH
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Octet and duplet patterns satisfied. Structure is good.

HCN

Your turn: CO2

Cyanate ion (OCN–)

Step 1: Total the number of valence electrons
Step 1: Total the number of valence electrons
OCN–
Negative charge:
C:
N:
O:
Total ve:
4
5
6
1
16
Total ve:
16
Step 1: Total the number of valence electrons
Negative charge:
C:
N:
O:
4
5
6
1
OCN–
Step 2: Place least electronegative atom (not H)
Placed ve:
0
16
Remaining ve:
C
Step 2: Place least electronegative atom (not H)
Total ve:
16
Placed ve:
Placed ve:
0
0
16
16
Remaining ve:
Remaining ve:
C
Step 2: Place least electronegative atom (not H)
Step 3: Place remaining atoms around central atom
Total ve:
16
Placed ve:
0
16
Remaining ve:
CON
Step 3: Place remaining atoms around central atom
Total ve:
16
Placed ve:
0
16
Remaining ve:
CON
Total ve:
16
Placed ve:
0
16
Remaining ve:
Step 3: Place remaining atoms around central atom
Step 4: Draw a single bond between each terminal atom and the central atom.
CON
Total ve:
16
Placed ve:
Remaining ve:
Step 4: Draw a single bond between each terminal atom and the central atom.
0
16
4
12
2 e–
2 e–
CON
Total ve:
16
Placed ve:
4
12
Remaining ve:
2 e–
2 e–
Step 4: Draw a single bond between each terminal atom and the central atom.
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
CON
Total ve:
16
Placed ve:
Remaining ve:
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
4
12
16
0
6 e–
6 e–
CONCON
Total ve:
16
Placed ve:
16
0
Remaining ve:
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
6 e–
6 e–
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: Two lone pairs from oxygen.
CONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: Two lone pairs from oxygen.
CONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: Two lone pairs from oxygen.
Option 2: One lone pairs from oxygen and one from nitrogen.
CONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 2: One lone pairs from oxygen and one from nitrogen.
Option 2: One lone pairs from oxygen and one from nitrogen.
CONCONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 2: One lone pairs from oxygen and one from nitrogen.
Option 3: Two lone pairs from nitrogen.
CONCONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 3: Two lone pairs from nitrogen.
CONCONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 3: Two lone pairs from nitrogen.
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
CONCONCON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
+1
0
–2
0
0
–1
–1
0
+1
CONCONCON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
+1
+1
0
0
–2
–2
0
0
0
0
–1
–1
–1
–1
0
0
+1
+1
The best structure has the negative formal charge on the more electronegative atom and the positive formal charge on the less electronegative atom.
OCNCONCON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
The best structure has the negative formal charge on the more electronegative atom and the positive formal charge on the less electronegative atom.
–1
0
+1
+1
0
–2
0
0
–1
CON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
–1
0
+1
The best structure has the negative formal charge on the more electronegative atom and the positive formal charge on the less electronegative atom.
Step 8: Because the structure is ionic, encapsulate in square brackets and add the charge.
CON
Step 8: Because the structure is ionic, encapsulate in square brackets and add the charge.
−

Formal Charge

Bookkeeping, not real charge.

Split bonding electrons 50/50, then ask:

Does each atom own what it started with?

\[\mathrm{FC} = \underset{\small free~atom}{\mathrm{valence~e^-}} ~-~ \underset{\small owned}{\mathrm{dots}} ~-~ \underset{\small half~shared}{\mathrm{lines}}\]

FC is not actual charge

The formula assumes equal sharing. Real bonds are almost never 50/50.

Formal charge compares Lewis structures to each other.

It does not measure charge distribution.

Two ways to count

Same bond, different bookkeeping. Step through to compare.

HFHF
Oxidation State
Oxidation State
Formal Charge
Formal Charge
Assume all bonding electrons go to the more electronegative atom (heterolytic cleavage).
Assume all bonding electrons go to the more electronegative atom (heterolytic cleavage).
Split bonding electrons equally between atoms (homolytic cleavage).
Split bonding electrons equally between atoms (homolytic cleavage).
HF−HF
Oxidation State
Formal Charge
Assume all bonding electrons go to the more electronegative atom (heterolytic cleavage).
Split bonding electrons equally between atoms (homolytic cleavage).
This gives each atom a hypothetical charge called its oxidation state.
Count the electrons each atom owns, then compare to its free-atom count: more → negative; fewer → positive.
–1
+1
0
0

Carbon monoxide (CO)

10 valence electrons. The negative charge lands on carbon.

One of the few molecules where FC points in the “wrong” direction. The triple bond is still correct.

Dots-First Approach

OCN
Step 1: Draw Lewis dot symbols for each atom
OCN–
OCN
Step 1: Draw Lewis dot symbols for each atom
Step 2: Place extra electrons on the most electronegative atom(s); remove electrons from the least electronegative
O−CN
Step 2: Place extra electrons on the most electronegative atom(s); remove electrons from the least electronegative
O−CN
Step 2: Place extra electrons on the most electronegative atom(s); remove electrons from the least electronegative
Step 3: Pair all unpaired electrons between adjacent atoms to form bonds
O−−CN
Step 3: Pair all unpaired electrons between adjacent atoms to form bonds
O−NC
Step 3: Pair all unpaired electrons between adjacent atoms to form bonds
O−NC
Step 3: Pair all unpaired electrons between adjacent atoms to form bonds
−CON−−
Step 3: Pair all unpaired electrons between adjacent atoms to form bonds
Step 4: Check octets. If any atom is short, promote a lone pair from a neighbor into a bonding pair.
Octets are satisfied. No promotion necessary.

What is resonance?

Sometimes a double bond could go in more than one place.

The molecule doesn’t pick one. The electrons are delocalized.

Resonance structures are not different molecules. The real molecule is a hybrid, a blend of all contributors.

Ozone (O3)

18 valence electrons, two equivalent structures

Both O-O bonds are identical: 128 pm, bond order 1.5 (between single at 148 and double at 121)

Nitrate ion (SO2⁻)

Step 1: Total the number of valence electrons
Step 1: Total the number of valence electrons
SO2
2 O:
S:
Total ve:
6
12
18
Total ve:
18
Step 1: Total the number of valence electrons
2 O:
S:
6
12
SO2
Step 2: Place least electronegative atom (not H)
Placed ve:
0
18
Remaining ve:
S
Step 2: Place least electronegative atom (not H)
Total ve:
18
Placed ve:
0
18
Remaining ve:
S
Total ve:
18
Placed ve:
0
18
Remaining ve:
Step 2: Place least electronegative atom (not H)
Step 3: Place remaining atoms around central atom
SOO
Step 3: Place remaining atoms around central atom
Total ve:
18
Placed ve:
0
18
Remaining ve:
SOO
Total ve:
18
Placed ve:
0
18
Remaining ve:
Step 3: Place remaining atoms around central atom
Step 4: Draw a single bond between each terminal atom and the central atom.
SOO
Step 4: Draw a single bond between each terminal atom and the central atom.
Total ve:
18
Placed ve:
Remaining ve:
0
18
4
14
2 e–
2 e–
SOO
Total ve:
18
Placed ve:
Placed ve:
4
4
14
14
Remaining ve:
Remaining ve:
Step 4: Draw a single bond between each terminal atom and the central atom.
2 e–
2 e–
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
SOO
Total ve:
Placed ve:
Remaining ve:
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
18
4
14
16
16
0
6 e–
2 e–
6 e–
SOOSOOSOO
Total ve:
16
Placed ve:
16
0
Remaining ve:
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
6 e–
2 e–
6 e–
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: One lone pair from left oxygen.
Option 2: One lone pair from right oxygen.
SOOSOOSOO
Option 1: One lone pair from left oxygen.
Option 2: One lone pair from right oxygen.
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
SOOSOOSOO
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: One lone pair from left oxygen.
Option 2: One lone pair from right oxygen.
Octets satisfied. Two resonance structures connected by double-headed arrow.

Carbonate ion (CO3²⁻)

24 valence electrons, three equivalent structures

Benzene (C6H6)

Two Kekulé structures; six equivalent C-C bonds

Bond order 1.5 throughout. Unusual stability from delocalization.

Spotting resonance

If you can move a double bond to a new position without moving atoms, resonance structures exist.

Common patterns:

  • Multiple bond adjacent to an atom with a lone pair
  • Equivalent atoms around a central atom (NO3⁻, CO3²⁻, O3)

Three categories

  1. Fewer than an octet
  2. More than an octet
  3. Odd-electron species

Fewer than an octet: BF3

24 valence electrons. Boron has only 6.

Electron-deficient molecules are Lewis acids: they accept electron pairs from other molecules.

More than an octet

Period 3+ central atoms can accommodate more neighbors

XeF2

22 valence electrons. Two bonds + three lone pairs on xenon.

One of the first noble gas compounds, overturning the belief that noble gases were completely inert.

What is actually happening?

These bonds are highly polar. Electron density sits on the terminal atoms.

The central atom maintains roughly an octet. “Expanded octet” is a notation artifact, not a physical reality.

Magnusson (1990) showed computationally that d-orbital participation is negligible. The old sp³d/sp³d² explanation has been abandoned.

Odd-electron species

Odd total electrons = at least one unpaired electron (free radical)

NO (11 e⁻)

NO2 (17 e⁻)

Free radicals are typically very reactive. Both play key roles in atmospheric chemistry.

Electronegativity and bond character

The EN difference between two atoms determines how electrons are shared.

Small ΔEN → electrons shared equally (nonpolar covalent)

Large ΔEN → electrons transferred (ionic)

Back to formal charge

FC assumes electrons are shared 50/50.

That works for nonpolar bonds (C-C, C-H).

It fails for very polar bonds (S-F, S-O, P-O).

For polar bonds, non-minimized formal charges may reflect real charge separation. This is why “minimize FC” is a guideline, not a law.

What Lewis structures cannot show

Every bond line looks the same.

H-F is drawn the same as H-H, but the electrons are not shared equally.

Lewis structures also cannot explain why O₂ is paramagnetic (attracted to magnets). That requires molecular orbital theory.

One trend, three consequences

More shared electrons =

shorter bond + stronger bond + higher bond order

Bond length

Bond Order Length (pm)
C–C 1 154
C=C 2 134
C≡C 3 120
N–N 1 146
N=N 2 125
N≡N 3 110

Bond energy

Bond Order Energy (kJ/mol)
C–C 1 346
C=C 2 614
C≡C 3 839
N–N 1 160
N=N 2 418
N≡N 3 945

Bond energies are averages. The exact energy depends on molecular context.

Resonance and bond order

Resonance hybrids have fractional bond orders.

Ozone: bond order = (2 + 1) / 2 = 1.5

Bond length: 128 pm (between 148 for single and 121 for double)

Nitrate: bond order = (1 + 1 + 2) / 3 = 4/3

Summary

Lewis structures show connectivity, lone pairs, and bond order.

Combined with formal charge and resonance, they predict geometry and reactivity.

Next chapter: VSEPR theory. From flat Lewis structures to 3D molecular geometry.