Approximate Molecular Orbital Calculations for H 2. George M. Shalhoub

Transcription

1 Approximate Molecular Orbital Calculations for H + LA SALLE UNIVESITY 9 West Olney Ave. Philaelphia, PA 94 shalhoub@lasalle.eu Copyright. All rights reserve. You are welcome to use this ocument in your own classes but commercial use is not allowe without the permission of the author. A note about this ocument: Many of the algebraic manipulations that are shown are cumbersome for Mathca to o, an they are more reaily one by han. Any equations that are manipulate in this way are highlighte in yellow. Mathca equations that are isable for evaluation have a black square to the upper right as in r := Z Introuction The H + is the prototype molecule an is the basis for our unerstaning of molecular orbital theory. Although H + can be solve exactly in elliptical coorinates, the exact solution cannot be applie to other iatomic molecules an is therefore limite. It is more important for the unerstaning of molecular orbital concepts to approximate the solution to H +. Such approximations can be compare to the exact theoretical values an the effectiveness of the approximations can be evaluate. These approximations can then be applie to more complicates molecular systems. The H + is iagramme in the figure. There is one electron, labele e, that interacts with two nuclei, labele A an B. e r A r B A AB B Create: November H+calck.mc page of5

2 Z Z The Hamiltonian for this system is given by: H := + +. x y z r A r B All quantities are in atomic units, a set of units where h bar, a, an m e all equal ). In atomic units, the basic unit of energy is the hartree, E h. hartree = 65 kj/mol. Z represents the nuclear charge, which woul be for hyrogen. The exact solution to the Schröinger equation for H + prouces a minimum energy of -.66 E h at AB = bohrs (6 pm). This energy correspons to a issociation energy of.79 ev or 69 kj/mol. Create: November H+calck.mc page of5

3 Trial Wavefunction an the Variation Metho How might we approximate H +? We will use a trial function in the variation expression. The variation theorem states that the energy obtaine from a trial function must be greater than or equal to the true energy of the system. It seems reasonable to assume that the trial function for H + can be approximate by a combination of s atomic orbitals on nuclei A an B. The trial function is not require to have this form, but this approach seems both simple an physically reasonable. SInce two hyrogen atoms, each with a s orbital combine to form H +, it is reasonable to assume that the orbitals of H + are combinations of these s atomic orbitals. Therefore, we state that Ψ = c A φ A + c B φ B. The coefficients, c A an c B give the relative weights of the functions S A an S B in the trial function. The atomic orbitals are stanar s orbitals that can be looke up in any Physical Chemistry textbook. φ A e r A r B := φ B = The variation theorem is: Ψ H Ψ τ ε = Equation ΨΨ τ Substituting the trial function into equation prouces: ε = Expaning terms, we get: ( c A φ A + c B φ B ) H ( c A φ A + c B φ B ) ( c A φ A + c B φ B ) c A φ A + c B φ B τ τ c A φa H φ A + c A c B φ A H φ B + c B φb H φ B τ ε = Equation c A φa + c A φ A c B φ B + c B φb τ Create: November H+calck.mc page of5

4 We efine a shorthan notation for the integrals: H AA = φ A H φ A τ H AB = φ A H φ B τ S AB = φ A φ B τ With this shorthan, the expression for the variation energy, equation, becomes: ε( c A, c B ) = c A HAA c A SAA + c B HBB + c A + c B SBB + c A c B H AB c B S AB Equation Energy of the Trial Function The energy is a function of the coefficients, c A an c B. To fin the lowest energy an the best values of c A an c B, we take the erivative of the energy expression with respect to c A an c B an set the resulting expressions equal to zero. Before we take erivatives, cross-multiply the energy expression, equation, by the enominator. Take the erivative with respect to c A : Take the erivative of this expression with respect to c A ): c A SAA + c B SBB + c A c B S AB ε c A, c B := c A HAA + c B HBB + c A c B H AB c A c A SAA + c B SBB + c A c B S AB ε c A, c B = c A HAA + c B HBB + c A c B H AB c A Using the prouct rule for the left-han-sie yiels: ( c A S AA + c B S AB ) ε c A, c B + c A SAA + c B SBB + c A c B S AB ε c A, c B = c A H AA + c B H AB ca Take the erivative of the energy expression with respect to c B : c B c A SAA + c B SBB + c A c B S AB ε c A, c B = c A HAA + c B HBB + c A c B H AB c B yiels ( c B S BB + c A S AB ) ε c A, c B + c A SAA + c B SBB + c A c B S AB ε c A, c B = c B H BB + c A H AB cb Create: November H+calck.mc page 4 of5

5 To have Mathca solve for ε c A, c B c A, we change the name of ε c A, c B c A variable an we can symbolically solve for variable. The result of the substitution is: to ε eriv. ε eriv is a ( c A S AA + c B S AB ) ε + c A SAA + c B SBB + c A c B S AB ε eriv = c A has solution(s) ( ε c A S AA ε c B S AB + c A H AA + c B H AB ) c A SAA + c B SBB + c A c B S AB epeat the process for the equation with ε c A, c B c B ( c B S BB + c A S AB ) ε + c A SAA + c B SBB + c A c B S AB ε eriv = c B has solution(s) ( ε c B S BB ε c A S AB + c B H BB + c A H AB ) c A SAA + c B SBB + c A c B S AB H AA + H BB + c B c A H AB H AB We now have two equations containing the energy, ε: ε c A S AA ε c B S AB + c A H AA + c B H AB = c A SAA + c B SBB + c A c B S AB Equations 4 ε c B S BB ε c A S AB + c B H BB + c A H AB = c A SAA + c B SBB + c A c B S AB We assume that the atomic orbitals are normalize an S BB equal to. Equations 4 simplify to; ε c A ε c B S AB + c A H AA + c B H AB = c A + c B + c A c B S AB ε c B ε c A S AB + c B H BB + c A H AB = c A + c B + c A c B S AB φ A φ A τ =, so we can set S AA Equations 5 Create: November H+calck.mc page 5 of5

6 Set equations 5 equal to each other: ε c A ε c B S AB + c A H AA + c B H AB ε c B ε c A S AB + c B H BB + c A H AB = = c A + c B + c A c B S AB c A + c B + c A c B S AB Simplify by han, Mathca apparently oes not see that the term c A + c B + c A c B S AB cancels out in this expression. ε c A ε c B S AB + c A H AA + c B H AB = ε c B ε c A S AB + c B H BB + c A H AB = Write as two separate equations: ε c A ε c B S AB + c A H AA + c B H AB = ε c B ε c A S AB + c B H BB + c A H AB = Have Mathca collect on c A then collect on c B. Copy the first equation to the space below. Click symbolics, evaluation style, evaluate in place. ( ε + H AA ) c A + ( ε S AB + H AB ) c B = ( ε S AB + H AB ) c A + ( ε + H BB ) c B = Equation 6 a Equation 6 b These two equations are simultaneous equations in two unknowns, c A an c B. We can write these two equations in matrix form: H AA ε H AB ε S AB H AB ε S AB H BB ε c A c B = For H +, H AA = H BB, since these are both integrals of s orbitals. Get the eterminant of the matrix (Symbolics, Matrix, Determinant) an solve for the energy H AA ε H AB ε S AB H AB ε S AB H AA ε has eterminant ε H AA ε H AA ε + S AB + ε S AB H AB H AB Create: November H+calck.mc page 6 of5

7 has solution(s) ( H AA + H AB ) ( + S AB ) ( H AA H AB ) ( S AB ) ewrite the solutions in a more convenient form. These are the electronic energies of H + ; they o not inclue the nuclear repulsion. H AA + H AB H AA H AB Equation 7 ε := ε := ( + S AB ) ( S AB ) These are the electronic energies of H +. we will evaluate them shortly. Form of the Wavefunctions The coefficients, c A an c B can also be obtaine. To o so, we woul substitute energy ε or ε from equation 7 into equations 6a an 6b an solve for the coefficients. For ε, we get c A = c B ; for ε, we get c A = -c B. Two wavefunctions are prouce: Ψ = c A φ A + φ B Ψ = c B φ A φ B The wavefunctions nee to be normalize. Ψ Ψ τ = First, multiply the functions Ψ Ψ = c A φ A + φ B c A φ A + φ B expans to Ψ Ψ = c A φa + c A φ A φ B + c A φb Put this expression into the normalization integral c A φa + c A φ A φ B + c A φb τ = c A + c A SAB = c A + S AB Create: November H+calck.mc page 7 of5

8 The integral equals, so: = c A + S AB Solve for c A has solution(s) ( + S AB ) ( + S AB ) Take the positive root. The normalize function is: ( φ A + φ B ) Ψ = + S AB Equation 8 Exercise: epeat the steps in this calculation to normalize Ψ. You shoul get: ( φ A φ B ) Ψ = S AB Equation 9 Energy of the Trial Function The energies ε an ε epen on the integrals H AA, H AB, an S AB. The expressions for these integrals can be foun in many quantum chemistry texts. They are: S AB ( ) := e + + Note that the integrals epen on the internuclear istance, AB for which we use the notation H AA ( ) := e ( + ) H AB ( ) := S AB ( ) e ( + ) Create: November H+calck.mc page 8 of5

9 Graphs of these integrals as a function of internuclear istance are shown here. SAB( ).5 HAB( ) HAA( ).5 5 Exercise: When woul the overlap integral, S AB equal? Create: November H+calck.mc page 9 of5

10 The energies, incluing the nuclear repulsion, /, are graphe as a function of ε ( ) ( H AA ( ) + H AB ( ) ) := + ε ( ) := + S AB ( ) ( H AA ( ) H AB ( ) ) ( S AB ( ) ) +.5 ε( ) ε( ) Exercise: Use the zoom an then the trace functions to fin the minimum in the re curve. What is your calculate issociation energy for H +? How o the energy levels of H + change with istance? In the following graph, you can vary an observe these changes. emember that is in bohrs, so 5 pm = bohr. := x :=.. Energy Levels of H+ e is E+ Energy/hartree Create: November H+calck.mc page of5

11 Exercise: What happens to the energy levels at very large istances ( try = 5, 7, bohr)? What is the energy at bohr? This energy correspons to the energy of what chemical species? Exercise: What happens to the energy levels at very small istances (try =.5,,.5,. bohrs)? What causes the change in energy that you observe? Graphs of the Molecular Orbitals To graph the molecular orbitals, set Z =, an = (this is the bon lenght in atomic units). We nee a range variable, i, that extens over the entire molecule ( - ). eefine the molecular orbital, Ψ in terms of these variables. The 6 in the function sets nucleus A at a istance of 4 bohr when i reaches. This is just a convenient point to place the atom. := i :=,... Z := Ψ () i := + S AB ( ) i 6 + i Ψ() i. Ψ() i. 5 i 5 i Note that in Ψ there is a builup of probability between the nuclei an hence Ψ is a boning molecular orbital. Create: November H+calck.mc page of5

12 Ψ () i := S AB ( ) i 6 i 6+ Here are the graphs for Ψ. In Ψ there is a noe between the nuclei an thus Ψ is an antiboning molecular orbital..5.5 Ψ() i.5 Ψ() i 5 i 5 i Create: November H+calck.mc page of5

13 Three imensional plots of the Wavefunctions In aition to the two-imensional graphs shown above, we can observe the three-imensional behavior of the wavefunctions. The three-imensional graphs show a more realistic spatial epiction of the orbitals To prouce a three-imensional plot in Mathca, we nee to assemble an array of the functions' values as we change the x an y coorinates. The graph will then epict the value of the function at any x,y point. These proceures require a calculation of the value of the function at each x, y point. In this case, the value of x an y is labele by the range variables i an j. The offset values change the locations of the function on the gri. We compute 4 x points an 4 y points for a 4x4 gri. More points result in a smoother graph but also take more calculation time. i :=.. j :=.. := i offset := j offset := sc :=.6 ( j j offset ) r Ai := sc i i offset + r Bi := sc i i offset +, j, j + j ( j offset + ) Ψ i :=, j + S AB ( ) ra i, j + rb i, j Ψ i :=, j S AB ( ) ra i, j rb i, j Boning MO Antiboning MO Ψ Ψ Create: November H+calck.mc page of5

14 Exercise: Convert the surface graphs to contour plots by ouble-clicking a graph an selecting contour. What changes o you notice? Can you relate the contour values to what you see on the surface plot? Exercise: Change the value of an observe the behavior of the graphs. What happens to the electron ensity in the boning MO when = 4? (You may have to change the y-axis values to see this.) Create: November H+calck.mc page 4 of5

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