Lecture 22: Lewis Dot Structures
• Reading: Zumdahl 13.9-13.12
• Outline
– Lewis Dot Structure Basics
– Resonance
– Those annoying exceptions
Partial Ionic Compounds (cont.)
• We can define the ionic character of bonds as follows:
% Ionic Character =
(dipole moment X-Y)experimental
(dipole moment X+Y-)calculated
x 100%
Partial Ionic Compounds (cont.)
Covalent
Ionic
Increased Ionic Character
Polar Covalent
Covalent Compounds (cont.)
• The same concept can be envisioned for other covalent
compounds:
Think of the covalent bond as the
electron density existing
between the C and H atoms.
Covalent Compounds (cont.)
• We can quantify the degree of stabilization by seeing how
much energy it takes to separate a covalent compound into
its atomic constituents.
C(g) + 4H(g)
CH4(g)
q
Covalent Compounds (cont.)
• Since we broke 4 C-H bonds with 1652 kJ in, the bond
energy for a C-H bond is:
1652 kJ mol
4
 413 kJ mol
• We can continue this process for a variety of compounds to
 a table of bond strengths.
develop
Covalent Compounds (cont.)
• Example: It takes 1578 kJ/mol to decompose CH3Cl into
its atomic constituents. What is the energy of the C-Cl
bond?
CH3Cl: 3 C-H bonds and 1 C-C bond.
3 (C-H bond energy) + C-Cl bond energy = 1578 kJ/mol
413 kJ/mol
1239 kJ/mol + C-Cl bond energy = 1578 kJ/mol
C-Cl bond energy = 339 kJ/mol
Covalent Compounds (cont.)
• We can use these bond energies to determine DHrxn:
DH = sum of energy required to break bonds (positive….heat
into system) plus the sum of energy released when the
new bonds are formed (negative….heat out from system).
DHrxn   Dbonds broken   Dbonds formed
Covalent Compounds (cont.)
• Example: Calculate DH for the following reaction using
the bond enthalpy method.
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
Go to Table 13.6:
4 x C-H 413
2 x O=O 495
4 x O-H
2 x C=O
467
745
Covalent Compounds (cont.)
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
DHrxn   Dbonds broken   Dbonds formed
= 4D(C-H) + 2D(O=O) - 4D(O-H) - 2D(C=O)

= 4(413) + 2(495) - 4(467) - 2(745)
= -716 kJ/mol
• Exothermic, as expected.
Covalent Compounds (cont.)
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
• As a check:
o
DHrxn
 DH of (prod.)  DH of (react.)

0
= DH°f(CO2(g)) + 2DH°f(H2O(g))
- DH°f(CH4(g)) - 2 DH°f(O2(g))
= -393.5 kJ/mol + 2(-242 kJ/mol) - (-75 kJ/mol)
= -802.5 kJ/mol
Localized Bond Models
• Consider our energy diagram for H2 bonding:
Localized Model Limitations
• It is important to keep in mind that the models we
are discussing are just that…..models.
• We are operating under the assumption that when
forming bonds, atoms “share” electrons using
atomic orbitals.
• Electrons involved in bonding: “bonding pairs”.
Electrons not involved in bonding: “lone pairs”.
Lewis Dot Structures (cont.)
• Developed by G. N. Lewis to serve as a
way to describe bonding in polyatomic
systems.
• Central idea: the most stable arrangement of
electrons is one in which all atoms have a “noble”
gas configuration.
• Example: NaCl versus Na+ClNa: [Ne]3s1
Cl: [Ne]3s23p5
Na+: [Ne]
Cl-: [Ne]3s23p6 = [Ar]
LDS Mechanics
F
C
Lone Pair (6 x)
F
F
Bonding Pair
• Atoms are represented
by atomic symbols
surrounded by valence
electrons.
• Electron pairs between
atoms indicate bond
formation.
LDS Mechanics (cont.)
• Three steps for “basic” Lewis structures:
1. Sum the valence electrons for all atoms to determine
total number of electrons.
2. Use pairs of electrons to form a bond between each
pair of atoms (bonding pairs).
3. Arrange remaining electrons around atoms (lone pairs)
to satisfy the “octet rule” (“duet” rule for hydrogen).
LDS Mechanics (cont.)
• An example: Cl2O
O
Cl
Cl
20 e-
16 e- left
Cl O Cl
Cl O Cl
LDS Mechanics (cont.)
• An example: CH4
H H
H H
C
8 e-
H
H C H
H
Done!
0 e- left
LDS Mechanics (cont.)
• An example: CO2
O
O
C
O C O
Octet Violation
CO double bond
16 e12 e- left
O C O
O
C O
0 e- left
LDS Mechanics (cont.)
• An example: NO+
+
N
+
N
O
10 e-
8 e- left
N O
+
N
O
O
Resonance Structures
• We have assumed up to this point that there is one
correct Lewis structure.
• There are systems for which more than one Lewis
structure is possible:
– Different atomic linkages: Structural Isomers
– Same atomic linkages, different bonding: Resonance
Resonance Structures (cont.)
• The classic example: O3.
O
O
O O
O
O
O
O
O
Both structures are correct!
Resonance Structures (cont.)
• In this example, O3 has two resonance structures:
O
O
O
• Conceptually, we think of the bonding being an
average of these two structures.
• Electrons are delocalized between the oxygens
such that on average the bond strength is
equivalent to 1.5 O-O bonds.
Structural Isomers
• What if different sets of atomic linkages can be
used to construct correct LDSs:
Cl O Cl
Cl O
Cl
Cl Cl O
Cl Cl O
• Both are correct, but which is “more” correct?
Formal Charge
• Formal Charge: Compare the nuclear charge (+Z)
to the number of electrons (dividing bonding
electron pairs by 2). Difference is known as the
“formal charge”.
Cl O Cl
Cl Cl O
#eZ+
7
7
6
6
7
7
7
7
6
7
Formal C.
0
0
0
0
+1 -1
7
6
• Structure with less F. C. is more correct.
Formal Charge
• Example: CO2
O
C
e-
6
4
6
6
4
6
7
4
5
Z+
6
4
6
6
6
4
6
6
4
FC
0
0
0
0
+2
-2
-1
+2 -1
O
More Correct
O
O C
O
O C
Beyond the Octet Rule
• There are numerous exceptions to the octet rule.
• We’ll deal with three classes of violation here:
– Sub-octet systems
– Valence shell expansion
– Odd-electron systems
Beyond the Octet Rule (cont.)
• Some atoms (Be and B in particular) undergo
bonding, but will form stable molecules that do
not fulfill the octet rule.
F
F
B
F
F
F
B
F
• Experiments demonstrate that the B-F bond
strength is consistent with single bonds only.
Beyond the Octet Rule (cont.)
• For third-row elements (“Period 3”), the energetic
proximity of the d orbitals allows for the
participation of these orbitals in bonding.
• When this occurs, more than 8 electrons can
surround a third-row element.
• Example: ClF3 (a 28 e- system)
F obey octet rule
F
F
Cl
F
Cl has 10e-
Beyond the Octet Rule (cont.)
• Finally, one can encounter odd electron systems
where full pairs will not exist.
• Example: Chlorine Dioxide.
O Cl O
Unpaired electron
Summary
• Remember the following:
– C, N, O, and F almost always obey the octet rule.
– B and Be are often sub-octet
– Second row (Period 2) elements never exceed the octet
rule
– Third Row elements and beyond can use valence shell
expansion to exceed the octet rule.
• In the end, you have to practice…..a lot!