3502_26_nove19 - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 26, November 19, 2010 (Some material in this lecture has been adapted from Cramer, C. J. Essentials of Computational Chemistry , Wiley, Chichester: 2002; pp. 111-114.) Solved Homework In the allyl system, we had H 11 = H 22 = H 33 = α , H 12 = H 21 = H 23 = H 32 = β , H 13 = H 31 = 0, S 11 = S 22 = S 33 = 1, and all other S values were 0. The difference in cyclopropenyl is that carbon atoms 1 and 3 are now neighbors, so that instead of H 13 = H 31 = 0, we have H 13 = H 31 = β . As a result, the secular equation becomes α E β β β α E β β β α E = 0 which expands as ( α E ) 3 + β 3 + β 3 – [ β • ( α E ) • β ] – β 2 ( α E ) – ( α E ) β 2 = 0 or ( α E ) 3 – 3 β 2 ( α E ) + 2 β 3 = 0 To solve the above cubic equation in E , it is convenient to set E = α + c β , in which case the equation transforms to c 3 + 3 c + 2 = 0 The roots of this equation are 1 and 2 (it turns out that 1 is a critical point where the cubic curve just touches the x axis, so there are only two roots and it must be the case that one of these roots represents a two-fold degenerate energy level). So, the allowed energy levels are E = α + 2 β (lowest energy, since α and β are negative numbers) and α β (higher energy). If we solve for the lowest energy MO we have the linear equations a 1 α α + 2 β ( ) •1 [ ] + a 2 β α + 2 β ( ) • 0 [ ] + a 3 β α + 2 β ( ) • 0 [ ] = 0 a 1 β α + 2 β ( ) • 0 [ ] + a 2 α α + 2 β ( ) • 1 [ ] + a 3 β α + 2 β ( ) • 0 [ ] = 0 a 1 β α + 2 β ( ) • 0 [ ] + a 2 β α + 2 β ( ) • 0 [ ] + a 3 α α + 2 β ( ) •1 [ ] = 0
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26-2 or 2 a 1 + a 2 + a 3 = 0 a 1 2 a 2 + a 3 = 0 a 1 + a 2 2 a 3 = 0 Casual inspection should indicate that this system of equations is solved by the relationship a 1 = a 2 = a 3 , and the normalization condition can then be solved to set a specific value of a 1 = a 2 = a 3 = 3 1/2 . So the lowest-energy MO has equal contributions from all 3 p orbitals and all 3 orbitals are in phase. This is illustrated below. What about the next energy level, α β ? That gives linear equations a 1 α α − β ( ) •1 [ ] + a 2 β α − β ( ) • 0 [ ] + a 3 β α − β ( ) • 0 [ ] = 0 a 1 β α − β ( ) • 0 [ ] + a 2 α α − β ( ) •1 [ ] + a 3 β α − β ( ) • 0 [ ] = 0 a 1 β α − β ( ) • 0 [ ] + a 2 β α − β ( ) • 0 [ ] + a 3 α α − β ( ) •1 [ ] = 0 In this case, all 3 equations reduce to a 1 + a 2 + a 3 = 0 When this equation is combined with the normalization condition, there are still only two equations in 3 unknowns, so we have remaining flexibility. This implies that we have a two-fold degenerate energy. We can arbitrarily select one set of coefficients that satisfies the linear equation and normalization. Then, since we want the next orbital of that energy to be orthogonal to the first, we will have 3 equations (the linear equation, the normalization condition, and the orthogonality condition) to determine the coefficients, and we will then be done.
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