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Molecular Geometry (Shapes of Molecules) VSEPR Theory Honors Chemistry http://www.scl.ameslab.gov/MacMolPlt/Surface.JPG 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 bond angles in the molecule Electron-group repulsions and the five basic molecular shapes. linear trigonal bipyramidal tetrahedral trigonal planar octahedral Electrons 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. Two electron pairs on central atom Examples: CS2, HCN, BeF2 3 electron pairs on central atom • All are in bonds • Trigonal Planar Shape Examples: SO3, BF3, NO3-, CO32- 3 Electron Pairs • 2 Bonded • 1 Un-Bonded Bent Shape • Un-bonded electron pair takes up more space and “repels” more. • The bond angle will change to less than the original 120o to about 117o-115o Other Examples: About 117o • SO2, O3, PbCl2, SnBr2 Four electron pairs on central atom • All 4 in bonds Examples: CH4, SiCl4, SO42-, ClO4- 4 Electron Pairs • 3 bonded • 1 un-bonded (lone pair) Trigonal Pyramidal Shape Examples: NH3, PF3, ClO3. H3O+ Bond angles are reduced from 109.5o to 107o due to extra repulsion by lone pair 4 Electron Pairs 2 Bonded 2 Un-bonded (lone pairs) BENT SHAPE Examples: H2O, OF2, SCl2 Bond angles are reduced a little more due to repulsion To 104.5o Five electron pairs on central atom • All Bonded = Trigonal Bipyramidal 5 Electron Pairs • 4 Bonded and 1 Un-bonded (lone pair) Also called an Irregular Tetrahedron 5 Electron Pairs • 3 Bonded • 2 Un-Bonded T-Shaped Molecule 5 Electron Pairs • 2 Bonded • 3 Un-Bonded = LINEAR Molecule!!! Bond Angles = 180o Six electron pairs on central atom • All Bonded = Octahedral Shape 6 Electron pairs • 5 Bonded • 1 Un-Bonded Lone Pair Square Pyramidal shape 6 Electron pairs • 4 Bonded • 2 Un-Bonded Lone Pair Square Planar shape The steps in determining a molecular shape Molecular 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 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 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 SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair The shape is based upon the tetrahedral arrangement. F P F The F-P-F bond angles should be <109.50 due F to the repulsion of the nonbonding electron P F pair. F F The final shape is trigonal pyramidal. 0 < 109.5 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 Sb F F F Sb 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 H tetrahedral trigonal planar O >1200 H C H H C C H HH <1200