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Visualizing Molecules - VSEPR, Valence Bond Theory, and Molecular Orbital Theory

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Visualizing Molecules - VSEPR, Valence Bond Theory, and Molecular Orbital Theory Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Visualizing Molecules - VSEPR, Valence Bond Theory, and Molecular Orbital Theory Nestor S. Valera 1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Visualizing Molecules 1. Lewis Structures and Valence-Shell Electron-Pair Repulsion Theory 2. Valence Bond Theory and Hybridization 3. Molecular Orbital Theory 2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Shapes of Molecules 10.1 Depicting Molecules and Ions with Lewis Structures 10.2 Valence-Shell Electron-Pair Repulsion (VSEPR) Theory and Molecular Shape 10.3 Molecular Shape and Molecular Polarity 3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 10.1 The steps in converting a molecular formula into a Lewis structure. Place atom Molecular Step 1 with lowest formula EN in center Atom Add A-group Step 2 placement numbers Sum of Draw single bonds. Step 3 valence e- Subtract 2e- for each bond. Give each Remaining Step 4 atom 8e- valence e- (2e- for H) Lewis structure 4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Molecular formula For NF3 Atom placement : : N 5e- : F: : F: : Sum of N F 7e- X 3 = 21e- valence e- - : F: Total 26e Remaining : valence e- Lewis structure 5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 10.1 Writing Lewis Structures for Molecules with One Central Atom PROBLEM: Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone. PLAN: Follow the steps outlined in Figure 10.1 . SOLUTION: Cl Step 1: Carbon has the lowest EN and is the central atom. Cl C F The other atoms are placed around it. F Steps 2-4: C has 4 valence e-, Cl and F each have 7. The : : Cl : sum is 4 + 4(7) = 32 valence e-. : : :Cl C F : Make bonds and fill in remaining valence : : electrons placing 8e- around each atom. : F : : 6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with More than One Central Atom PROBLEM: Write the Lewis structure for methanol (molecular formula CH4O), an important industrial alcohol that is being used as a gasoline alternative in car engines. SOLUTION: Hydrogen can have only one bond so C and O must be next to each other with H filling in the bonds. There are 4(1) + 4 + 6 = 14 valence e-. C has 4 bonds and O has 2. O has 2 pair of nonbonding e-. H : H C O H : H 7 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. SAMPLE PROBLEM 10.3 Writing Lewis Structures for Molecules with Multiple Bonds. PROBLEM: Write Lewis structures for the following: (a) Ethylene (C2H4), the most important reactant in the manufacture of polymers (b) Nitrogen (N2), the most abundant atmospheric gas PLAN: For molecules with multiple bonds, there is a Step 5 which follows the other steps in Lewis structure construction. If a central atom does not have 8e-, an octet, then two e- (either single or nonbonded pair)can be moved in to form a multiple bond. SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e-. H can have only one bond per atom. H : H H H C C C C H H H H - (b) N2 has 2(5) = 10 valence e . Therefore a triple bond is required to make the octet around each N. : : : : : : N N N N N N . 8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Here are some helpful generalizations used in constructing Lewis structures • H atoms form one bond • Carbon atoms form four bonds • Nitrogen atoms form three bonds • Oxygen atoms form two bonds • Halogen atoms form one bond when they are “surrounding” atoms; but more than one bond when they are “central” atoms. Fluorine is always a “surrounding” atom. • The “central” atom is usually large and of low EN • The “octet” rule may be violated by almost all atoms except those that belong to the second period. 9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Writing Lewis Structures for Molecules with One Central Atom PROBLEM: Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone. SOLUTION: Cl Step 1: Carbon has the lowest EN and is the central atom. Cl C F The other atoms are placed around it. F Steps 2-4: C has 4 valence e-, Cl and F each have 7. The : : Cl : sum is 4 + 4(7) = 32 valence e-. : : :Cl C F : Make bonds and fill in remaining valence : : electrons placing 8e- around each atom. F : : : 10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Writing Lewis Structure for Molecules with More than One Central Atom PROBLEM: Write the Lewis structure for methanol (molecular formula CH4O), an important industrial alcohol that is being used as a gasoline alternative in car engines. SOLUTION: Hydrogen can have only one bond so C and O must be next to each other with H filling in the bonds. There are 4(1) + 4 + 6 = 14 valence e-. C has 4 bonds and O has 2. O has 2 pair of nonbonding e-. H : H C O H : H 11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Writing Lewis Structures for Molecules with Multiple Bonds. Write Lewis structures for the following: (a) Ethylene (C2H4), the most important reactant in the manufacture of polymers (b) Nitrogen (N2), the most abundant atmospheric gas - SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e . H can have only one bond per atom. H H H H C : C C C H H H H - (b) N2 has 2(5) = 10 valence e . Therefore a triple bond is required to make the octet around each N. : : : : : : N N N N N N . 12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. It is possible to have more than one satisfactory Lewis structure. O can be drawn in 2 ways - 3 O O O O O O Neither structure is actually correct but can be drawn to represent a structure which is a hybrid of the two - a resonance structure. Resonance structures are NOT capable of independent existence. B B O O O O O O O O O A C A C Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs. is used to indicate that resonance occurs. 13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Can you see the “resonance” or “canonical” figures? 14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. - PROBLEM: Write resonance structures for the nitrate ion, NO3 . SOLUTION: Nitrate has 1(5) + 3(6) + 1 = 24 valence e- O O O N does not have an octet; a pair of e- N N N will move in to form O O O O O O a double bond. O O O N N N O O O O O O 15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Some resonance structures are more “important” or dominant than others. 16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The formal charge helps us selecting the best resonance structure. Formal charge is the charge an atom would have if the bonding electrons were shared equally. An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons) B For OC For OA O # valence e- = 6 # valence e- = 6 O O # nonbonding e- = 6 # nonbonding e- = 4 A C For OB # bonding e- = 2 X 1/2 = 1 # bonding e- = 4 X 1/2 = 2 - # valence e = 6 Formal charge = -1 Formal charge = 0 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1 17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Resonance (continued) Three criteria for choosing the more important resonance structure: Smaller formal charges (either positive or negative) are preferable to larger charges. Avoid like charges (+ + or - - ) on adjacent atoms. A more negative formal charge should exist on an atom with a larger EN value. 18 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Resonance (continued) EXAMPLE: NCO- has 3 possible resonance forms - N C O N C O N C O A B C formal charges -2 0 +1 -1 0 0 0 0 -1 N C O N C O N C O Forms B and C have negative formal charges on N and O; this makes them more preferred than form A. Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid. 19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The “octet” rule is often violated especially for atoms not belonging to the second period. PROBLEM: Write Lewis structures for (a) H3PO4 (pick the most likely structure); (b) BFCl2. Draw the Lewis structures for the molecule and determine STRATEGY: if there is an element which can be an exception to the octet rule. Note that (a) contains P which is a Period-3 element and can have an expanded valence shell.
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