orbital is the LUMO, an antibonding orbital. The four Hückel molecular orbitals for 1,3-butadiene are* = 0.3720 +0.602d2 +0.60203 + 0.3720 2 = 0.6026 + 0.372, – 0.372d3 - 0.60264 3 = 0.602d - 0.37262 - 0.37203 + 0.60264 a = 0.3726 - 0.60262 + 0.60203 – 0.37264 (11.75) These four molecular orbitals are indicated in Fig. 11.21. Notice that the r orbitals extend the entire length of the molecule. (For the calculation of these coefficients, see Computer Problem 11.C.) The Hückel secular determinant for benzene is a - E B B B E 0. B = 0 (11.76) a – E B a - E B B a - E B The six roots are E = a + 2B E2 = E3 = a +B E4 = Es = a -B E6 = a - 2B (11.77) Since benzene has six 7 electrons, pairs of electrons go in the three lowest energy orbitals (those with plus signs in E;). Thus the n electronic energy in benzene is E, = 2(a + 2B) + 4(a + B) = 6a + 8B (11.78) The equations for the six Hückel molecular orbitals for benzene are not given here, but the corresponding electron densities are shown in Fig. 11.22. Note that the electronic energy in C,H6 is more negative than three times the value in C,H4, indicating that C,H does not contain three double bonds. The difference is called the delocalization energy. The Hückel theory is an example of a semiempirical molecular orbital method. We have used a simple Hamiltonian (neglecting many terms) to find the orbitals and their energies. We now can use experimental quantities to fit a and B for ethylene. We then use the values to make predictions for butadiene, benzene, and so on. This method does not give quantitative results, but it does provide us with qualitative insights about larger systems for which the more com- putationally intensive methods are too costly or time-consuming, and it gives us insight into the excited electronic states of conjugated 7 electron molecules.
orbital is the LUMO, an antibonding orbital. The four Hückel molecular orbitals for 1,3-butadiene are* = 0.3720 +0.602d2 +0.60203 + 0.3720 2 = 0.6026 + 0.372, – 0.372d3 - 0.60264 3 = 0.602d - 0.37262 - 0.37203 + 0.60264 a = 0.3726 - 0.60262 + 0.60203 – 0.37264 (11.75) These four molecular orbitals are indicated in Fig. 11.21. Notice that the r orbitals extend the entire length of the molecule. (For the calculation of these coefficients, see Computer Problem 11.C.) The Hückel secular determinant for benzene is a - E B B B E 0. B = 0 (11.76) a – E B a - E B B a - E B The six roots are E = a + 2B E2 = E3 = a +B E4 = Es = a -B E6 = a - 2B (11.77) Since benzene has six 7 electrons, pairs of electrons go in the three lowest energy orbitals (those with plus signs in E;). Thus the n electronic energy in benzene is E, = 2(a + 2B) + 4(a + B) = 6a + 8B (11.78) The equations for the six Hückel molecular orbitals for benzene are not given here, but the corresponding electron densities are shown in Fig. 11.22. Note that the electronic energy in C,H6 is more negative than three times the value in C,H4, indicating that C,H does not contain three double bonds. The difference is called the delocalization energy. The Hückel theory is an example of a semiempirical molecular orbital method. We have used a simple Hamiltonian (neglecting many terms) to find the orbitals and their energies. We now can use experimental quantities to fit a and B for ethylene. We then use the values to make predictions for butadiene, benzene, and so on. This method does not give quantitative results, but it does provide us with qualitative insights about larger systems for which the more com- putationally intensive methods are too costly or time-consuming, and it gives us insight into the excited electronic states of conjugated 7 electron molecules.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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What would be the lowest possible excitation energy for benzene in this model? Give your answer in units of β.
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