Hybridization - Southwestern Oklahoma State University
I. VSEPR = Valence Shell Electron-Pair Repulsion A. Electron pairs repel each other 1. This applies to bonding pairs and lone pairs alike 2. Steps in Applying VSEPR a. Draw the Lewis Structure b. Count atoms and lone pairs and arrange them as far apart as possible
c. Determine the name of the geometry based only on where atoms are B. Complications to VSEPR 1. Lone pairs count for arranging electrons, but not for naming geometry 2. Example: NH3 (ammonia) .. H N H Tetrahedral
Trigonal pyramidal H 3. Lone pairs are larger than bonding pairs, resulting in adjusted geometries a. Bond angles are squeezed to accommodate lone pairs b. Lone pairs must be as far from each other as possible
4. Double and triple bonds are treated as only 1 pair of electrons 5. Names for and examples of complicated VSEPR Geometries linear trigonal planar bent tetrahedral trigonal pyramidal bent
trigonal bipyramidal. octahedral see-saw T-shaped linear square pyramidal
square planar II. Hybridization A. Localized Electron Model 1. Molecules = atoms bound together by sharing e- pairs between atomic orbitals 2. Lewis structures show the arrangement of electron pairs 3. VSEPR Theory predicts the geometry based on e- pair repulsion 4. Hybridization describes formation and properties of the orbitals involved in bonding more specifically
B. sp3 Hybridization 1. Methane, CH4, will be our example 2. Bonding only involves valence ea. H = 1s b. C = 2s, 2p 3. If we use atomic orbitals directly, the predicted shape doesnt match what we predict by VSEPR and what we observe experimentally a. One C2s--H1s bond would be shorter than others b. Three C2p--H1s bonds would be at right
angles to each other a. We know CH4 is tetrahedral, symmetric 4. Free elements (C) use pure atomic orbitals 5. Elements involved in bonding will modify their orbitals to reach the minimum energy configuration 6. When C bonds to four other atoms, it hybridized its 2s and 2p atomic orbitals to form 4 new sp3 hybrid orbitals a. The sp3 orbital shape is between 2s/2p; one large lobe dominates
b. When you hybridize atomic orbitals, you always get an equal number of new orbitals 7. The new sp3 orbitals are degenerate due to the mixing 8. The H atoms of methane can only use their 1s orbitals for bonding. The shared electron pair can be found in the overlap area of H1sCsp 3 9. NH3 and H2O also use sp3 hybridization, even though lone pairs occupy some of the hybrid orbitals.
C. sp2 Hybridization 1. C2H4, ethylene is our example H H C H
C H 2. Lewis and VSEPR structures tell us what to expect 3. H atoms still can only use 1s orbitals 4. C atom hybridizes 2s and two 2p orbitals into 3 sp 2 hybrid orbitals 5. The new sp2 orbitals are degenerate and in the same plane 6. One 2p orbital is left unchanged and is perpendicular to that plane
7. One CC bond of ethylene forms by overlap of an sp 2 orbital from each of the two sp2 hybridized C atoms. This is a sigma () bond because the overlap is along the internuclear axis. 8. The second CC bond forms by overlap of the remaining single 2p orbital on each of the carbon atoms. This is a pi () bond because the overlap is perpendicular to the internuclear axis. H H
C C H H 9. Pictorial views of the orbitals in ethylene
D. sp Hybridization 1. CO2, carbon dioxide is our example O C O 2. Lewis and VSEPR predict linear structure
3. C atom uses 2s and one 2p orbital to make two sp hybrid orbitals that are 180 degrees apart 4. We get 2 degenerate sp orbitals and two unaltered 2p orbitals 5. Oxygen uses sp2 orbitals to overlap and form sigma bonds with C 6. Free p orbitals on the O and C atoms form pi bonds to complete bonding E. d-orbitals can also be involved in hybridization 1. dsp3 hybridization in PCl5
2. d2sp3 hybridization in SF6 F. The Localized Electron Model 1. Draw the Lewis structure(s) 2. Determine the arrangement of electron pairs (VSEPR model). 3. Specify the necessary hybrid
orbitals. III.Polar Bonds A. Any bond between atoms of different electronegativities is polar Electrons concentrate on one side of the bond One end of the molecule is (+) and one end is (-) B. Complex molecules: vector addition of all bond dipoles C. Examples:
D. Polarity greatly effects molecular properties Oppositely charged ends of polar molecules Attract each other like opposite poles of magnets
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