Molecular Geometry and Polarity http://www.scl.ameslab.gov/MacMolPlt/Surface.JPG Bond Angles in Carbon Compounds electron configuration = 1s22s22p2 2p orbitals with one electron in each. Can p orbitals with one electron in each find the place where the 3rd p orbital should be? If they can, the bond angles
should be 90o. Orbitals with one electron in each will overlap to form single bonds. Butthe bond angles are 109.5o! Its All in the Shape So whats going on? Think back to the lab What is the primary reason molecules form the geometry we find?
Electron Pair Repulsion VSEPR Theory Electron groups around the central atom will be most stable when they are as far apart as possible we call this valence shell electron pair repulsion theory because electrons are negatively charged, they should be most stable when they are separated as much as possible The resulting geometric arrangement will allow us to predict the shapes and
Electron-group repulsions and the five basic molecular shapes. linear trigonal bipyramidal tetrahedral trigonal planar octahedral Two electron pairs on central atom
Examples: CS2, HCN, BeF2 Electron vs Molecular Geometry The geometry of electron pairs around a central atom is called the electron geometry. The arrangement of bonded nuclei around a central atom forms the molecular geometry. Lone pair electrons on a central atom will repel other pairs but will not be visible in the molecular geometry (no nuclei) If there are lone pairs on the central
atom the electron geometry and the molecular geometry will differ. Three electron pairs on central atom Examples: SO3, BF3, NO3-, CO32- Examples: SO2, O3, PbCl2, Four electron pairs on central atom Examples: CH4, SiCl4,
SO42-, ClO4- Examples: NH3, PF3, ClO3. H3O+ Examples: H2O, OF2, SCl2 Five electron pairs on central atom Six electron pairs on central atom
Representing 3-Dimensional Shapes on a 2-Dimensional Surface One of the problems with drawing molecules is trying to show their dimensionality By convention, the central atom is put in the plane of the paper Put as many other atoms as possible in the same plane and indicate with a straight line For atoms in front of the plane, use a solid wedge For atoms behind the plane, use a hashed wedge
The steps in determining a molecular shape Molecula r formula Step 1 Lewis structure Step 2 Electron-group
arrangement (electron geometry) Count all e- pairs around central atom Step 3 Bond angles Note lone pairs and double bonds Step 4
Consider bonding e- pairs only Molecular geometry 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. H Effect of Double Bonds 120 ideal
0 1200 C O H larger EN 0 greater electro
n Effect of Nonbonding(Lone) Pairsdensity Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. C 116 H 1220 H real
Sn Cl Cl 95 0 O 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) PF3 and (b) COCl2. SOLUTION: (a) For PF3 - there are 26 valence electrons, 1
nonbonding pair The shape is based upon the tetrahedral F P F arrangement. 0 The F-P-F bond angles should be <109.5 F due to the repulsion of the nonbonding P F electron pair.
F F The final shape is trigonal pyramidal. <109.5 0 Predicting Molecular Shapes with Two, Three, or Four Electron Groups (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. Cl C O
Cl O C Cl Cl 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. O The Cl-C-Cl bond angle will be less than 1200
due to the electron density of the C=O. 124.50 C Cl 111 0 Cl 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) SbF5 and (b) BrF5. SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is trigonal bipyramidal. F F F F F Sb F Sb F F F F
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is square pyramidal. F F F Br F F Predicting Molecular Shapes with More Than One Central Atom PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O. Find the shape of one atom at a time after writing the
Lewis structure. SOLUTION: tetrahedral H H C H O C H C H tetrahedral
H trigonal planar O H C H C C H HH
>120 H 0 <120 0 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? The orientation of polar molecules in an electric field. Electric field OFF Electric field ON Polarity of Molecules For a molecule to be polar it must 1. have polar bonds
2. electronegativity difference - theory bond dipole moments - measured have an unsymmetrical shape vector addition Nonbonding pairs affect molecular polarity, strong pull in their direction
Molecule Polarity The HCl bond is polar. The bonding electrons are pulled toward the Cl end of the molecule. The net result is a polar molecule. 27 Molecule Polarity The OC bond is polar. The bonding electrons are pulled equally toward both O ends of the molecule. The net result is a nonpolar molecule. 28
Molecule Polarity The HO bond is polar. Both sets of bonding electrons are pulled toward the O end of the molecule. The net result is a polar molecule. 29 Predicting the Polarity of Molecules 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: (a) Ammonia, NH3
(b) Boron trifluoride, BF3 (c) Carbonyl sulfide, COS (atom sequence SCO) Draw the shape, find the EN values and combine the concepts to determine the polarity. SOLUTION: (a) NH3 The dipoles reinforce each other, so the overall molecule is definitely polar. ENN = 3.0 H ENH = 2.1
N H H H N H bond dipoles H H
N H H molecular dipole Predicting the Polarity of Molecules (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 F than B (EN 2.0) and all of the
dipoles will be directed from B to B F F 120 F. Because all are at the same angle and of the same 0 magnitude, the molecule is (c) COS is linear. nonpolar. C and S have the same EN (2.0) but the C=O bond is quite polar(EN) so the molecule is polar overall. S C O
More Molecular Polarity http://academic.pgcc.edu/~ssinex/po larity/polarity.htm
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