3502_26_nove19

# 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
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## This note was uploaded on 12/18/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

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

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