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أستاذ المادة عباس عبد علي دريع الصالحي18/12/2017 16:59:36

University of Babylon Undergraduate Studies College of Science Department of Chemistry Course No. Chsc 424 Physical chemistry Fourth year - Semester 1 Credit Hour: 3 hrs.

Lectures of Quantum mechanics Second Semester, Scholar year 2017-2018 Prof. Dr. Abbas A-Ali Drea ----------------------------------------------------------------------------- Lecture No. Fourteen: Molecular Orbital theory 1-Introudction. 2-Forming Molecular Orbitals. 3-Born Oppenheimer approximation. 4- Linear combination of atomic orbital (LCAO). 5- Application of Molecular Orbital theory.

1-Introudction: Molecular orbital (MOT) theory is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. The spatial and energetic properties of electrons within atoms are fixed by quantum mechanics to form orbitals that contain these electrons. While atomic orbitals contain electrons ascribed to a single atom, molecular orbitals, which surround a number of atoms in a molecule, contain valence electrons between atoms. Molecular orbital theory was proposed in the early twentieth century, revolutionized the study of bonding by approximating the positions of bonded electrons, the molecular orbitals as linear combinations of atomic orbitals (LCAO). These approximations are now made by applying the density functional theory (DFT) or Hartree–Fock (HF) models to the Schr?dinger equation. 2-Forming Molecular Orbitals. Molecular orbitals are obtained by combining the atomic orbitals on the atoms in the molecule. Consider the H2 molecule, for example. One of the molecular orbitals in this molecule is constructed by adding the mathematical functions for the two 1s atomic orbitals that come together to form this molecule. Another orbital is formed by subtracting one of these functions from the other, as shown in the figure below.

Forming of Molecular orbitals. One of these orbitals is called a bonding molecular orbital because electrons in this orbital spend most of their time in the region directly between the two nuclei. It is called a sigma (?) molecular orbital because it looks like an s orbital when viewed along the H-H bond. Electrons placed in the other orbital spend most of their time away from the region between the two nuclei. This orbital is therefore an antibonding, or sigma star (?*), molecular orbital.

Energy diagram of Molecular orbitals

3-Born Oppenheimer approximation: This principles depend on the large differences in speed moving between nuclei and their electrons (electron lighter than nuclei by 1837 times). The slow motion nuclei have stationary position relatively, therefore can be find accurate precisely wave functions of electrons at approximated constant position of nuclei. Hamiltonian operator of molecular system have nucleus µ and ? with their electrons i, and j respectively. H= -1/2 ?_???(?^2 ?)/M?-1/2 ?_i???_i^2-?_i??_(? )??Z?/?r_(i ) ? ?_( ) +?_(? The first term of eq. 1, is represent the kinetics energies of nucleus (Mµ is mass of nuclei µ). Second term is represent the kinetics energies of electrons. Third terms is represent the potential energies of attractive nucleus-electrons (rµ distance between electron i and nuclei µ). Fourth terms is represent the potential energies of repulsion of nuclei-nuclei (Rµ-? distance between nuclei µ and nuclei ?). Fifth term is represent the potential energies of repulsion electron i -- electron j (rij distance between electron i and electron j). This Hamiltonian operator is identical with Schrodinger equation. H?(r,R)=E?(r,R)------2 The term of E is the total energy of system, and ? is a function for electronic dimensions (r) and nuclear dimensions (R), this approximation give up ability to approximate ? into two sub wave functions. ?(r,R)=?_R (r).?(R)-------3 The term of ?_R (r),is wave function of electrons, at stationary position of nuclei is represent a function of electronic dimensions, and it s depend on quantized states of electrons. This function is change according to the different position of nucleus, since symbol R at the function is depend on the nuclear dimensions. Function .?(R) is nuclear wave function but it s depend on electronic energies and equilibrium to the potential energy for unclear motion. The electronic wave function is equal to following equation: H_e (r,R)?_R (r)=E_e ?(R)??_R (r)------4 The Hamiltonian operator He is electronic Hamiltonian operator that’s same operator in eq.1, but without nuclear kinetic energy terms. The Eigen value (energy) depend on nuclear dimensions, the nuclear wave function represented by: [H_n (R)+E_e (R)]?(R)=E?(R)-------5 The Hamiltonian operator Hn is qualify to first term of equation 1 of nuclear kinetic energy. Equations 3, 4, and 5 are results of Born Oppenheimer approximation. If the position of nucleus are constant, therefore the first term equal to zero and the fourth term equal to constant quantity. The equation 1, became for electronic motion only as follow: H= -1/2 ?_i^2-?_i??_???Z?/r_i? +? ?_(i At the same approximation can be separate the electronic motion than nuclear motion as in eq.3, so that can be solve the equation of electronic motion in eq.4 to find up the functions and electronic energies for molecule at last get up the nuclear function ?R, total energy and the total wave function. 4- Linear combination of atomic orbital (LCAO): Variation theory have been used to produce approximated wave functions for Schrodinger equations in the molecular states. Experimental wave function is comes out through linear combination of atomic orbitals for atoms that’s consisted a molecule. Positive Hydrogen ion molecule is consisted from two protons and only one electron. If that first nuclei effect on the lonely electron is larger than second nuclei effect, so can be supposed that the molecular orbital of electron is the same of atomic orbital that’s nuclei of large effect. Supposed that wave function of atomic orbitals ?i, so that: ?=a_1 ?_1+a_2 ?_2+a_3 ?_3+,+?+a_n ?_n= ?_i??a_i ?_i ?-----7 By using variation method can be find the best value of coefficient a_i , that’s produce lowest value of energy, so can be write the following: (?( ?_i??a_i ?_i ?)?(?_j??a_j ?_j ?)??)/(?( ?_i??a_i ?_i ?).(?_j??a_j ?_j ?)??)?E0 ----8 Since ?, is the correct Hamiltonian operator, Eo is lowest energy value (associated value or Eigen value). By substituted with symbols of Hij for ????_i^* H?_j ??? and Sij for????_i^* ?_j ???, eq. 8 can be written as follow: (?_ij??a_i a_j H_ij ?)/(?_ij???a_i a?_j S_ij ?)?E0 ----9 S_ij is over lap integrals,its very important role in velance bound theory. H_ij is matrix elemente integrals,its very important role in limitation of Hamlitonian operator. Both are i and j takes all possible values, according variation theory the left side will decreased to lowest value of eq. 9, to find the best value of coefficient a_i . E=(?_ij??a_i a_j H_ij ?)/(?_ij???a_i a?_j S_ij ?) ----10 Or ?_ij??a_i a_j ?(H?_ij ?-ES_ij)=0 ----11 By differential eq.11 accordance to the operator a_k , at the same time all other rest operators are constant. ?_i?a_i H_ki+?_i?a_i H_ik-E(?_i?a_i S_ki+?_i?a_i S_ik )-?E/?ak ?_ij??a_i a_j ? S_ij=0-12 To be the energy value in minimum value (dE/dak=0), If supposed the functions are true completely (Sik =Ski) and hermitic conditions (Hki = Hik), so that ?_i??a_i ?(H?_ki ?-ES_ki)=0 ----13 Since, k=1, 2, 3, 4… n , That’s mean they found a number of n equations likes eq.13. Every equation can be vary for the every operator, this state produce n groups of linear equations, since n is unknown. Eq.13 have up normal solutions, if the determinate of operators (unknown operatora_i ) equal to zero. That’s mean: |?(H_11-ES_11&H_12-ES_12&H_1n-ES_1n@H_21-ES_21&H_22-ES_22&H_2n-ES_2n@H_n1-ES_n1&H_n2-ES_n2&H_nn-ES_nn )|=0 ---14 Eq.14 can be summarized into: |H_ki-ES_ki |=0 -----15 This determinates are called Secular determinates or equation parent. At expansions, they will gave n degree of multi terminal or polynomial of energy and the lowest energy value of n is the best value of energy at ground state, that’s defend with eq.7 and the best state of this equation is: ?=a_1 ?_1+a_2 ?_2------16 According eq. 14, can be write the solvation of eq.16 as follow: a_1 (H_11-E_min S_11 )+a_2 (H_21-E_min S_21 )=0 ----17 a_1 (H_21-E_min S_21 )+a_2 (H_22-E_min S_22 )=0 ----18 Coefficient value (a_1), can be extracted from eq.17 to substitute into eq.18 as follow: a_1=(-a_2 (H_21-E_min S_21 ))/((H_11-E_min S_11 ) ) And (H_11-E_min S_11 )(H_22-E_min S_22 )+(H_21-E_min S_21 )^2=0-19,. They concluded, that’s found two answers from state of eq.16 agreements with eq.19, due its second order equations. If that’s ?_1 and ?_2 are normalized wave functions, that’s mean: S_22=S_11=1 Therefore (H_11-E_min )(H_22-E_min )-(H_21-E_min S_21 )^2=0----20 If that? S?_21=1, means the atomic orbitals (?1 & ?2) are completely combination, but when the value (?0 The first, is lowest energy value (Emin)1 that’s values is lowest than H11 : ((Emin)1< H11) The second, is lowest energy value (Emin)2 that’s values is largest than H22 : ((Emin)2> H22) The second lowest energy value (Emin)2 is larger than lowest energy value (Emin)1 by small difference. Proofing this case come out by making eq.20 equal to function, that’s suited to (Emin) at condition f=0. f=(H_11-E_min )(H_22-E_min )-(H_21-E_min S_21 )^2=0---21 That’s mean:

Two minimum values of wave function. Both a, and b points are values of function at zero, that’s equal to minimum energy values. The two probable answers of eq.16 are (Emin)2 and (Emin)1 ,one of them represent the ground state and second represent the excited state. Can be concluded from this results that’s combination between two atomic orbitals, produced two molecular orbitals for ground state suited to (Emin)1 and excited state suited to (Emin)2 respectively. The first molecular orbital is called Bonding orbital and the second molecular orbital is called Anti Bonding orbital. 5- Application: Q/Find out the formalism structure of positive hydrogen molecule. Solution: The Hydrogen positive ion molecule (H2+) is simplest known molecules, following figure represented nuclear distance for hydrogen ion molecule.

Nuclear distance of Hydrogen ion molecule. By using Born Oppenheimer approximation for relatively stationary nucleus, considering the system is consisted from one mobile electron and two stationary protons. Hamiltonian operator take out the following conformation: H=-1/2 ?^2-1/r_A -1/r_B -----22 Note, they didn’t including inter nuclear repulsion terms (1/R) because it s constant at this approximation according to Born Oppenheimer approximation and they be consider later. Schrodinger equation of this system is: H?_n=E_n ?_n All quantity of this equation depending on Inter nuclear distances. If that, R=0, the system became likely the ionized system of Helium atom (+He), the charge of nuclei equal to +2. If that R=?, the system became likely the system of Hydrogen atom. When they add term of inter nuclear repulsion energy (E1), so can be get the total energy (ER) as follow: E_((R))=E_1+1/R ----23 If they determine the atomic orbitals for ionic hydrogen molecule as (1sA) into first atom and (2sB) into second atom, therefore the suitable state for eq.14, as follow: |?(H_AA-E_n&H_AB-E_n S@H_BA-E_n S&H_BB-E_n )|=0 ----24 Since that: H_AA=????1S?_A.H?.1S?_A.??=H_BB=????1S?_B.H?.1S?_B.???? H_AB=????1S?_A.H?.1S?_B.??=H_AB=????1S?_B.H?.1S?_A.???? Note: the value of integral is not equal to zero, because the two orbitals depended on each other. Also: S_AA=S_BB=????1S?_A.?1S?_A.??=????1S?_B.?1S?_B.???? Eq.24, didn’t included the quantitiesS_AA, S_BB . Due that HAA= HBB; HAB = HBA, therefore can be write eq.24 as follow: |?(H_AA-E_n&H_AB-E_n S@H_BA-E_n S&H_BB-E_n )|=0

??-(H?_AA-E_n)?^2-(H_AB-E_n S)^2=0 H_AA-E_n=±H_AB-E_n S H_AA=±H_AB-E_n (1±S) E_n=(H_AA±H_AB)/((1±S)) ----25 That’s mean, they found two energetic values suited to eq.24, since system of ionic hydrogen molecule is the simplest state of eq.24. The first lowest energy value (Emin)1= E_g=(H_AA+H_AB)/((1+S)) ----26 The second lowest energy value (Emin)2= E_u=(H_AA-H_AB)/((1-S)) ----27 Subscribes g and u are come out from German words (Grade and Ungraded). They are represented types of molecular orbital according to the point of conversion in the area between two nuclei. E_gis the energy of Bonding molecular orbital and E_uis the energy of Anti Bonding molecular orbital respectively. The wave functions for this orbitals, can be extract from eq.13, as follow: ?_g=(?1S?_A+?1S?_B)/?(2+S)?^0.5 -----28 ?_u=(?1S?_A-?1S?_B)/?(2-S)?^0.5 -----29 The values at the base of two equations are constants of variation.

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