Lewis Structures and Bonding 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 structures1 (originally called “electronic structures”) are the map.

Lewis Dot Symbols

Each dot = one valence electron.

HHeLiBNOFNeBe

Why do bonds form?

A shared electron is attracted to both nuclei simultaneously.

This additional attraction lowers the system’s energy, forming a bond.

Potential energy of H2

Sharing electrons

A covalent bond forms when two atoms share 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

Covalent to Ionic: most bonds are somewhere in between.


Each atom has an electronegativity, a measure of how strongly it pulls on shared electrons.

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

  • Small Δχ → electrons shared equally (nonpolar covalent)
  • Moderate Δχ → electrons shared unequally (polar covalent)
  • Large Δχ → electrons transferred (ionic)

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
Placed ve:
0
12
Remaining ve:
C
Placed ve:
0
12
Total ve:
12
Remaining ve:
Step 2: Place least electronegative atom
Step 2: Place least electronegative atom (not H)
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.

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.

Your turn: CO2

Step 1: Total the number of valence electrons
Step 1: Total the number of valence electrons
CO2
2 O:
C:
Total ve:
4
12
16
Total ve:
16
Step 1: Total the number of valence electrons
2 O:
C:
4
12
CO2
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:
0
16
Remaining ve:
C
Total ve:
16
Placed ve:
0
16
Remaining ve:
Step 2: Place least electronegative atom (not H)
Step 3: Place remaining atoms around central atom
COO
Step 3: Place remaining atoms around central atom
Total ve:
16
Placed ve:
0
16
Remaining ve:
COO
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.
COO
Step 4: Draw a single bond between each terminal atom and the central atom.
Total ve:
16
Placed ve:
Remaining ve:
0
16
2 e–
4
12
2 e–
COO
Total ve:
16
Placed ve:
4
12
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.
COO
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
Total ve:
16
Placed ve:
Remaining ve:
4
12
16
0
6 e–
6 e–
COO
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
Total ve:
16
Placed ve:
16
0
Remaining ve:
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.
COOCOO
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 left oxygen
COOCOO
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 left oxygen
COOCOOCOO
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 left oxygen
Option 2: One lone pair from each oxygen
COOCOOCOO
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 pair from each oxygen
COOCOOCOOCOO
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 pair from each oxygen
Option 3: Two lone pairs from right oxygen
COOCOOCOOCOO
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 right oxygen
COOCOOCOOCOO
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 right oxygen
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
COOCOOCOO
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
+1
0
–1
0
0
0
–1
0
+1
COOCOOCOO
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
+1
+1
0
0
–1
–1
0
0
+1
+1
0
0
0
–1
–2
The best structure minimizes the overall formal charges.
COOCOOCOO
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
The best structure minimizes the overall formal charges.
0
0
0
0
0
+1
–1
–2
+1
0
0

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.

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).
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

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.
CO
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.
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.
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.
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
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
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. 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.
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.

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−)


COOOCOOOCOOOCOOOCOOOCOOOCOOO2−2−2−

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)

CCCCCCHHHHHH
Draw Lewis dot symbols
Match up lone pairs
CCCCCCHHHHHH
Match up lone pairs
CCCCCCHHHHHH
Match up lone pairs
CCCCCCHHHHHH
Match up lone pairs
Form single bonds
CCCCCCHHHHHHCCCCCCHHHHHHCCCCCCHHHHHH
Form single bonds
Remaining electrons can form double bonds.
CCCCCCHHHHHHCCCCCCHHHHHH
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.
CCCCCCHHHHHH
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

  1. Fewer than an octet
  2. More than an octet (termed ‘Expanded Octet’: a matter of convenience, not science)
  3. Odd-electron species

Fewer than an octet: BF3

BFFFBFFF
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

PFFFFF
SFFFFFF

A matter of convenience

IUPAC Graphical Representation Standards (GR-8):2

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.”


SF6: convenience vs. correctness

“Expanded octet”: 1 structure

SFFFFFF

Violates octet. Readable.

Octet-compliant: 15 resonance structures

S2+FFFFF−F−F−F−S2+FFFFS2+FFFFF−F−S2+FFFFF−F−S2+FFFFF−F−F−S2+FFFFF−F−F−S2+FFFFS2+FFFFF−F−S2+FFFFF−F−S2+FFFFF−F−S2+FFFFF−F−F−S2+FFFFS2+FFFFS2+FFFFF−F−F−F−F−F−S2+FFFFF−

Obeys octet. Good luck.

XeF4

36 valence electrons. Four bonds + two lone pairs on xenon.

XeFFFF
F−F−Xe2+FF

Bartlett’s synthesis of XePtF63 overturned the belief that noble gases were inert. XeF2 followed months later.4

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)5 showed computationally that d-orbital participation in bonding is negligible; they do not contribute to ‘expanded octets’.

Odd-electron species

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

NO (11 e−)

NO

NO2 (17 e−)

NOO

Free radicals are typically very reactive.

References

(1)
Lewis, G. N. The Atom and the Molecule. J. Am. Chem. Soc. 1916, 38 (4), 762–785. https://doi.org/10.1021/ja02261a002.
(2)
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)
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