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.
The sum of the formal charges on a Lewis structure equals the charge on the structure.
Two ways to count
Same bond, different bookkeeping.
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).
HFF−HHFF
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
CO
Step 1: Total the number of valence electrons
Step 1: Total the number of valence electrons
CO
O:
C:
Total ve:
4
6
10
Total ve:
10
Step 1: Total the number of valence electrons
O:
C:
4
6
CO
Step 2: Place least electronegative atom (not H)
Placed ve:
0
10
Remaining ve:
C
Step 2: Place least electronegative atom (not H)
Total ve:
10
Placed ve:
0
10
Remaining ve:
C
Total ve:
10
Placed ve:
0
10
Remaining ve:
Step 2: Place least electronegative atom (not H)
Step 3: Place remaining atoms around central atom
CO
Step 3: Place remaining atoms around central atom
Total ve:
10
Placed ve:
0
10
Remaining ve:
CO
Total ve:
10
Placed ve:
0
10
Remaining ve:
Step 3: Place remaining atoms around central atom
Step 4: Draw a single bond between each
terminal atom and the central atom.
CO
Step 4: Draw a single bond between each
terminal atom and the central atom.
Total ve:
10
Placed ve:
Remaining ve:
0
10
2 e–
2
8
CO
Total ve:
10
Placed ve:
2
8
Remaining ve:
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.
CCOO
Step 5: Distribute remaining electrons as lone pairs on terminal atoms.
If any remain, place on central atom.
Total ve:
10
Placed ve:
Remaining ve:
2
8
10
0
2 e–
6 e–
CO
Step 5: Distribute remaining electrons as lone pairs on terminal atoms.
If any remain, place on central atom.
Total ve:
10
Placed ve:
10
0
Remaining ve:
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.
COO
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.
CO
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.
Octets satisfied.
CO
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.
Octets satisfied.
Formal charge analysis
CO
Formal charge analysis
The 50/50 assumption breaks down here
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.
COONN
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
CONCOON
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
CONCONCOONN
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
CONCONCONCONN
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
0
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
0
0
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
0
+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
0
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.
−
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
OO−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−O−CCNN
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−ONNCC
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.
Matches lines-first approach.
Sulfur dioxide (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.
SSOOOO
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.
SOOOSOOOSOO
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.
What is resonance?
Sometimes a double bond could be assigned to 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 (or blend) of all contributors.
Spotting resonance: If you can move a double bond to a new position without moving atoms, resonance structures exist.
Ozone (O3)
OOOOOO
Both O–O bonds are identical: 1.28 Å, bond order 1.5
(between single at 1.48 Å and double at 1.21 Å)
Carbonate ion (CO32−)
COOOCOOOCOOOCOOOCOOOOCOOOOCOOOO2−2−2−
All three C–O bonds are identical: 1.28 Å, bond order 4/3 (or ~1.33)
(between single at 1.43 Å and double at 1.23 Å)
Benzene (C6H6)
CCCCCCCCCCCCHHHHHHHHHHH
Draw Lewis dot symbols
Match up lone pairs
CCCCCCCCCCCCHHHHHHHHHHHH
Match up lone pairs
CCCCCCCCHHHHHH
Match up lone pairs
CCCCCCCCCCCCHHHHHHHHHHHH
Match up lone pairs
Form single bonds
CCCCCCHHHHHHCCCCCCHHHHHHCCCCCCHHHHHH
Form single bonds
Remaining electrons can form double bonds.
CCCCCCCCCCCCHHHHHHCCCCCCCCCCCCHHHHHH
Remaining electrons can form double bonds.
CCCCCCCCCCCCHHHHHHHHHHHH
There are two Kekulé resonance structures.
The six electrons that form the three double bonds are delocalized.
CCCCCCHHHHHH
There are two Kekulé resonance structures.
The six electrons that form the three double bonds are delocalized.
CCCCCCCCCCCCHHHHHH
The circle notation replaces the two Kekulé structures with a single drawing: the circle represents six delocalized electrons shared equally across the ring.
CCCCCCHHHHHH
The circle notation replaces the two Kekulé structures with a single drawing: the circle represents six delocalized electrons shared equally across the ring.
Bond order 1.5 throughout.
Three categories
Fewer than an octet
More than an octet (termed ‘Expanded Octet’: a matter of convenience, not science)
Odd-electron species
Fewer than an octet: BF3
BFFFBBFFFFFF
Electron deficient center.
Only six electrons.
Unoccupied p orbital on B.
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
Functional groups containing sulfur… are preferably depicted with normal single and double bonds and without the addition of extra formal charges, even though such representations will violate the “octet rule.” “…this depiction style should be preserved even for anions. The four oxygen atoms are chemically equivalent, but that equivalence is commonly understood through resonance.”
Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure Appl. Chem.2008, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277.
(3)
Bartlett, N. Xenon Hexafluoroplatinate(V) Xe\(^+\)[PtF\(_6\)]\(^-\). Proc. Chem. Soc.1962, 218.
(4)
Chernick, C. L.; Claassen, H. H.; Fields, P. R.; Hyman, H. H.; Malm, J. G.; Manning, W. M.; Matheson, M. S.; Quarterman, L. A.; Schreiner, F.; Selig, H. H.; Sheft, I.; Siegel, S.; Sloth, E. N.; Stein, L.; Studier, M. H.; Weeks, J. L.; Zirin, M. H. Fluorine Compounds of Xenon and Radon. Science1962, 138 (3537), 136–138. https://doi.org/10.1126/science.138.3537.136.
(5)
Magnusson, E. Hypercoordinate Molecules of Second-Row Elements: D Functions or d Orbitals? J. Am. Chem. Soc.1990, 112 (22), 7940–7951. https://doi.org/10.1021/ja00178a014.