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 = x 100%(dipole moment X-Y)experimental

(dipole moment X+Y-)calculated

Partial Ionic Compounds (cont.)

Covalent

Polar Covalent

Ionic

Increased Ionic Character

Covalent Compounds (cont.)

• The same concept can be envisioned for other covalent compounds:

Think of the covalent bond as theelectron density existingbetween 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.

q

CH4(g)

C(g) + 4H(g)

Covalent Compounds (cont.)• Since we broke 4 C-H bonds with 1652 kJ in, the bond

energy for a C-H bond is:

1652kJ mol4

413kJ mol

• We can continue this process for a variety of compounds to

develop a table of bond strengths.

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 Hrxn:

H = 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).

Hrxn Dbonds broken Dbonds formed

Covalent Compounds (cont.)• Example: Calculate H for the following reaction using

the bond enthalpy method.

CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g)

Go to Table 13.6:

C-H 413O=O 495

O-H 467C=O 745

4 x2 x

4 x2 x

Covalent Compounds (cont.)CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g)

Hrxn 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:

Hrxno H f

o (prod.) H fo (react .)

= H°f(CO2(g)) + 2H°f(H2O(g)) - H°f(CH4(g)) - 2 H°f(O2(g))

0

= -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+Cl-

Na: [Ne]3s1 Cl: [Ne]3s23p5

Na+: [Ne] Cl-: [Ne]3s23p6 = [Ar]

LDS Mechanics

• Atoms are represented by atomic symbols surrounded by valence electrons.

F C

• Electron pairs between atoms indicate bond formation.

F F

Bonding Pair

Lone Pair (6 x)

LDS Mechanics (cont.)• Three steps for “basic” Lewis structures:

1. Sum the valence electrons for all atoms to determinetotal 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

20 e-O Cl Cl

OCl Cl 16 e- left

OCl Cl

LDS Mechanics (cont.)• An example: CH4

8 e-

0 e- left

H CH HH

H C

H

H

H

Done!

LDS Mechanics (cont.)• An example: CO2

16 e-

12 e- left

O CO

OCO

OCO 0 e- leftOctet Violation

OCOCO double bond

LDS Mechanics (cont.)• An example: NO+

10 e-ON+

ON+

ON 8 e- left

ON+

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:

OCl Cl ClCl O

OCl Cl ClCl 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”.

OCl Cl ClCl O

#e- 7 6 7 7 6 7Z+ 7 6 7 7 7 6

Formal C. 0 0 0 0 +1 -1

• Structure with less F. C. is more correct.

Formal Charge• Example: CO2

OCO O O C O 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

More Correct

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.

FBF

F

FBF

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)

FClF

F F obey octet rule

Cl has 10e-

Beyond the Octet Rule (cont.)• Finally, one can encounter odd electron systems where full pairs will not

exist.

ClO O

• Example: Chlorine Dioxide.

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!