1. Molecular Geometry and Polarity http://www.scl.ameslab.gov/MacMolPlt/Surface.JPG

2. Bond Angles in Carbon Compounds electron configuration = 1s 2 2s 2 2p 2 If they can, the bond angles should be 90 o. But…the bond angles are 109.5 o ! 2p orbitals with one electron in each. Orbitals with one electron in each will overlap to form single bonds. Can p orbitals with one electron in each find the place where the 3 rd p orbital should be?

3. It’s All in the Shape… So what’s going on? Think back to the lab… What is the primary reason molecules form the geometry we find? Electron Pair Repulsion

4. 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 minimize 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. AX m E n A - central atomX -surrounding atomE -nonbonding valence electron-group integers Silberberg, Principles of Chemistry

5. Figure 10.3 Electron-group repulsions and the five basic molecular shapes. linear trigonal planar tetrahedral trigonal bipyramidal octahedral Silberberg, Principles of Chemistry

6. Figure 10.4 The single molecular shape of the linear electron-group arrangement. Examples: CS 2, HCN, BeF 2 Silberberg, Principles of Chemistry

7. Figure 10.5 The two molecular shapes of the trigonal planar electron- group arrangement. Class Shape Examples: SO 3, BF 3, NO 3 -, CO 3 2- Examples: SO 2, O 3, PbCl 2, SnBr 2 Silberberg, Principles of Chemistry

8. Figure 10.6 The three molecular shapes of the tetrahedral electron-group arrangement. Examples: CH 4, SiCl 4, SO 4 2-, ClO 4 - NH 3 PF 3 ClO 3 H 3 O + H 2 O OF 2 SCl 2 Silberberg, Principles of Chemistry

9. Figure 10.8 The four molecular shapes of the trigonal bipyramidal electron- group arrangement. SF 4 XeO 2 F 2 I F 4 + I O 2 F 2 - ClF 3 BrF 3 XeF 2 I 3 - I F 2 - PF 5 AsF 5 SOF 4 Silberberg, Principles of Chemistry

10. Figure 10.9 The three molecular shapes of the octahedral electron-group arrangement. SF 6 I OF 5 BrF 5 TeF 5 - XeOF 4 XeF 4 I Cl 4 - Silberberg, Principles of Chemistry

11. 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. ideal 120 0 larger EN greater electron density 122 0 116 0 real Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. 95 0 Effect of Double Bonds Effect of Nonbonding(Lone) Pairs Silberberg, Principles of Chemistry

12. Figure 10.10 The steps in determining a molecular shape. Molecular formula Lewis structure Electron-group arrangement Bond angles Molecular shape (AX m E n ) Count all e - groups around central atom (A) Note lone pairs and double bonds Count bonding and nonbonding e - groups separately. Step 1 Step 2 Step 3 Step 4 Silberberg, Principles of Chemistry

13. SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups PROBLEM:Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF 3 and (b) COCl 2. SOLUTION:(a) For PF 3 - there are 26 valence electrons, 1 nonbonding pair The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be <109.5 0 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. <109.5 0 The type of shape is AX 3 E Silberberg, Principles of Chemistry

14. SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups continued (b) For COCl 2, 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. 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 120 0 due to the electron density of the C=O. 124.5 0 111 0 Type AX 3 Silberberg, Principles of Chemistry

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

16. SAMPLE PROBLEM 10.8 Predicting Molecular Shapes with More Than One Central Atom SOLUTION: PROBLEM:Determine the shape around each of the central atoms in acetone, (CH 3 ) 2 C=O. PLAN:Find the shape of one atom at a time after writing the Lewis structure. tetrahedral trigonal planar >120 0 <120 0 Silberberg, Principles of Chemistry

17. Molecular Polarity Just like bonds can be polar because of even electron distribution, molecules can be polar because of net electrical imbalances. These imbalances are not the same as ion formation. How do we know when a molecule is polar?

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

19. SAMPLE PROBLEM 10.9Predicting the Polarity of Molecules (a) Ammonia, NH 3 (b) Boron trifluoride, BF 3 (c) Carbonyl sulfide, COS (atom sequence SCO) 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) NH 3 EN N = 3.0 EN H = 2.1 bond dipoles molecular dipole The dipoles reinforce each other, so the overall molecule is definitely polar.

20. SAMPLE PROBLEM 10.9Predicting the Polarity of Molecules continued (b) BF 3 has 24 valence e - and all electrons around the B will be involved in bonds. The shape is AX 3, 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. 120 0 (c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(  EN) so the molecule is polar overall. Silberberg, Principles of Chemistry

21. More Molecular Polarity… http://academic.pgcc.edu/~ssinex/polar ity/polarity.htm http://academic.pgcc.edu/~ssinex/polar ity/polarity.htm Work through the site listed above.