Lecture 22: Lewis Dot Structures Reading: Zumdahl 13.9-13.12 Outline –Lewis Dot Structure Basics –Resonance –Those annoying exceptions.

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Presentation transcript:

Lecture 22: Lewis Dot Structures Reading: Zumdahl 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 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.

Covalent Compounds (cont.) Since we broke 4 C-H bonds with 1652 kJ in, the bond energy for a C-H bond is: 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 CH 3 Cl into its atomic constituents. What is the energy of the C-Cl bond? CH 3 Cl: 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  H rxn :  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).

Covalent Compounds (cont.) Example: Calculate  H for the following reaction using the bond enthalpy method. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) Go to Table 13.6: C-H413 O=O495 O-H467 C=O745 4 x 2 x 4 x 2 x

Covalent Compounds (cont.) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) = 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.) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) As a check: =  H° f (CO 2 (g)) + 2  H° f (H 2 O(g)) -  H° f (CH 4 (g)) - 2  H° f (O 2 (g)) 0 = kJ/mol + 2(-242 kJ/mol) - - (-75 kJ/mol) = kJ/mol

Localized Bond Models Consider our energy diagram for H 2 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]3s 1 Cl: [Ne]3s 2 3p 5 Na + : [Ne]Cl - : [Ne]3s 2 3p 6 = [Ar]

LDS Mechanics Atoms are represented by atomic symbols surrounded by valence electrons. Electron pairs between atoms indicate bond formation. Bonding Pair Lone Pair (6 x)

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: Cl 2 O 20 e - 16 e - left

LDS Mechanics (cont.) An example: CH 4 8 e - 0 e - left Done!

LDS Mechanics (cont.) An example: CO 2 16 e - 12 e - left 0 e - left Octet Violation CO double bond

LDS Mechanics (cont.) An example: NO + 10 e e - left +

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: O 3. Both structures are correct!

Resonance Structures (cont.) In this example, O 3 has two resonance structures: 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: 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”. #e Z Formal C Structure with less F. C. is more correct.

Formal Charge Example: CO 2 e Z FC 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. 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: ClF 3 (a 28 e - system) 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. 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!