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Chapter 10

Chapter 10. The Shapes of Molecules. 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. Figure 10.1.

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Chapter 10

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  1. Chapter 10 The Shapes of Molecules

  2. 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. Figure 10.1 The steps in converting a molecular formula into a Lewis structure. Place atom with lowest EN in center Molecular formula Step 1 Atom placement Add A-group numbers Step 2 Sum of valence e- Draw single bonds. Subtract 2e- for each bond. Step 3 Give each atom 8e- (2e- for H) Remaining valence e- Step 4 Lewis structure

  4. Molecular formula For NF3 Atom placement : : N 5e- : F : : F : : Sum of valence e- F 7e- X 3 = 21e- N Total 26e- : F : : Remaining valence e- Lewis structure

  5. 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 . : : : Cl : : : Cl : C F : : : : F : SAMPLE PROBLEM 10.1 Writing Lewis Structures for Molecules with One Central Atom SOLUTION: Cl Step 1: Carbon has the lowest EN and is the central atom. The other atoms are placed around it. Cl C F F Steps 2-4: C has 4 valence e-, Cl and F each have 7. The sum is 4 + 4(7) = 32 valence e-. Make bonds and fill in remaining valence electrons placing 8e- around each atom.

  6. 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. SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with More than One Central Atom 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. H H H H C C C C H H H H N : N : . . N : N : N : N : . . . . 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 5which 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. : (b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make the octet around each N.

  8. is used to indicate that resonance occurs. Resonance: Delocalized Electron-Pair Bonding O3 can be drawn in 2 ways - 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 have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs.

  9. SAMPLE PROBLEM 10.4 Writing Resonance Structures PROBLEM: Write resonance structures for the nitrate ion, NO3-. PLAN: After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms. SOLUTION: Nitrate has 1(5) + 3(6) + 1 = 24 valence e- N does not have an octet; a pair of e- will move in to form a double bond.

  10. For OC For OA # valence e- = 6 # valence e- = 6 # nonbonding e- = 6 # nonbonding e- = 4 # bonding e- = 2 X 1/2 = 1 # bonding e- = 4 X 1/2 = 2 For OB Formal charge = -1 Formal charge = 0 # valence e- = 6 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1 Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge is the charge an atom would have if the bonding electrons were shared equally. Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons)

  11. Three criteria for choosing the more important resonance structure: Resonance (continued) 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.

  12. A B C Resonance (continued) EXAMPLE: NCO- has 3 possible resonance forms - formal charges -2 0 +1 -1 0 0 0 0 -1 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.

  13. PROBLEM: Write Lewis structures for (a) H3PO4 (pick the most likely structure); (b) BFCl2. SAMPLE PROBLEM 10.5 Writing Lewis Structures for Octet Rule Exceptions PLAN: Draw the Lewis structures for the molecule and determine 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. SOLUTION: (a) H3PO4 has two resonance forms and formal charges indicate the more important form. -1 0 +1 0 (b) BFCl2 will have only 1 Lewis structure. 0 0 0 0 0 0 0 0 0 0 more stable 0 0 lower formal charges

  14. A - central atom X -surrounding atom E -nonbonding valence electron-group integers VSEPR - Valence Shell Electron Pair Repulsion Theory Each group of valence electrons around a central atom is located as far away as possible from the others in order to maximize repulsions. These repulsions maximize the space that each object attached to the central atom occupies. The result is five electron-group arrangements of minimum energy seen in a large majority of molecules and polyatomic ions. The electron-groups are defining the object arrangement,but the molecular shape is defined by the relative positions of the atomic nuclei. Because valence electrons can be bonding or nonbonding, the same electron-group arrangement can give rise to different molecular shapes. AXmEn

  15. linear tetrahedral trigonal planar trigonal bipyramidal octahedral Figure 10.2 Electron-group repulsions and the five basic molecular shapes.

  16. Figure 10.3 The single molecular shape of the linear electron-group arrangement. Examples: CS2, HCN, BeF2

  17. Class Shape Figure 10.4 The two molecular shapes of the trigonal planar electron-group arrangement. Examples: SO2, O3, PbCl2, SnBr2 Examples: SO3, BF3, NO3-, CO32-

  18. 1220 Effect of Double Bonds 1160 real Effect of Nonbonding(Lone) Pairs Factors Affecting Actual Bond Angles Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. 1200 larger EN 1200 ideal greater electron density Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. 950

  19. The three molecular shapes of the tetrahedral electron-group arrangement. Figure 10.5 Examples: CH4, SiCl4, SO42-, ClO4- NH3 PF3 ClO3 H3O+ H2O OF2 SCl2

  20. Lewis structures and molecular shapes. Figure 10.6

  21. Figure 10.7 The four molecular shapes of the trigonal bipyramidal electron-group arrangement. PF5 AsF5 SOF4 SF4 XeO2F2 IF4+ IO2F2- XeF2 I3- IF2- ClF3 BrF3

  22. Figure 10.8 The three molecular shapes of the octahedral electron-group arrangement. SF6 IOF5 BrF5 TeF5- XeOF4 XeF4 ICl4-

  23. A summary of common molecular shapes with two to six electron groups. Figure 10.9

  24. See Figure 10.1 The steps in determining a molecular shape. Figure 10.10 Molecular formula Step 1 Lewis structure Count all e- groups around central atom (A) Step 2 Electron-group arrangement Note lone pairs and double bonds Step 3 Count bonding and nonbonding e- groups separately. Bond angles Step 4 Molecular shape (AXmEn)

  25. PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b)COCl2. SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. <109.50 The type of shape is AX3E

  26. 124.50 1110 SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups continued (b) For COCl2, C has the lowest EN and will be the center atom. There are 24 valence e-, 3 atoms attached to the center atom. C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond. Type AX3 The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar. The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

  27. PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5. SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal. SAMPLE PROBLEM 10.7 Predicting Molecular Shapes with Five or Six Electron Groups (b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

  28. PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O. PLAN: Find the shape of one atom at a time after writing the Lewis structure. tetrahedral tetrahedral trigonal planar >1200 <1200 SAMPLE PROBLEM 10.8 Predicting Molecular Shapes with More Than One Central Atom SOLUTION:

  29. ethanol ethane CH3CH2OH CH3CH3 The tetrahedral centers of ethane and ethanol. Figure 10.11

  30. Electric field OFF Electric field ON The orientation of polar molecules in an electric field. Figure 10.12

  31. PROBLEM: From electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity. SOLUTION: (a) NH3 SAMPLE PROBLEM 10.9 Predicting the Polarity of Molecules (a) Ammonia, NH3 (b) Boron trifluoride, BF3 (c) Carbonyl sulfide, COS (atom sequence SCO) The dipoles reinforce each other, so the overall molecule is definitely polar. ENN = 3.0 ENH = 2.1 molecular dipole bond dipoles

  32. SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules continued (b) BF3 has 24 valence e- and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar. F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar. 1200 (c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(DEN) so the molecule is polar overall.

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