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Unformatted text preview: Chem 120A Hybrid Orbitals/Spin 03/20/06 Spring 2006 Lecture 25 READING: Engel Ch 14.114.4, 10.210.4 1 Hybrid Orbitals You may have been introduced to hybrid orbitals in a general chemistry or organic chemistry course. Using our knowledge of atomic orbitals (wavefunctions) from the last few lectures, we can present a more rigorous treatment of hybrid orbitals. Hybrid orbitals are linear superpositions of atomic orbitals. Remember that if a set of wavefunctions are eigenstates of the Hamiltonian then all linear combinations of these wavefunctions are eigenstates as well. Thus we can take our atomic wavefunctions and use them to form hybrid orbitals. Hybrid orbitals are useful models for decribing the bonding geometries of molecules. For example, you have seen the various ways the 2 s and 2 p orbitals can hybridize in carbon to form organic molecules (See Figure 1) Figure 1: a summary of the hybrid orbitals which may be formed from s and p orbitals. Image taken from: http://www.cms.k12.nc.us/allschools/providence/keenan/chem/images/hybrid orbitals.gif It is very important to remember that the total number of hybrid orbitals will be the same as the orginal number of atomic orbitals we started with. So if we take one s orbital and one p orbital we will end up with Chem 120A, Spring 2006, Lecture 25 1 two sp hybrid orbitals. The sp hybrid orbitals are formed by taking the following linear combinations: a = 1 2 ( s + p z ) b = 1 2 ( s p z ) (1) where the factor of 1 2 normalizes the hybrid orbitals. The sp hybrid orbitals are shown below. Notice that their geometry is linear Figure 2: Combining one s orbital and one p orbital to form two sp hybrids (only the angular part of the wavefunction for each of the orbitals is shown). Image taken from: http://www.chemistry.ohiostate.edu/ grandinetti/teaching/Chem121/lectures/hybridization/hybrid.html We can form three sp 2 are formed by taking three different cominations of the s , p x and p y orbitals. a = c 1 p x + c 2 s + c 3 p y b = c 4 p x + c 5 s + c 6 p y c = c 7 p x + c 8 s + c 9 p y (2) We can determine the constants, c 1 through c 9 by the fact that the three sp 2 orbitals must be orthonormal. A few aspects of the chosen geometry simplify the task of determining the coefficients. For example, because the s orbital is spherically symetric, it will contribute equally to each of the hybrid orbitals, so c 2 = c 5 = c 8 . These coefficients describe the amount of s character in each of the sp 2 orbitals. As the total amount of scharacter is 1, : i ( c i ) 2 = 1. Therefore c 2 = c 5 = c 8 = 1 3 . The sp 2 orbitals are shown below, notice that they are all in the xy plane....
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This note was uploaded on 09/29/2009 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at University of California, Berkeley.
 Spring '07
 Whaley
 Physical chemistry, Organic chemistry, Atom, pH

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