5.61 F17 Lecture 24: Molecular Orbital Theory, Matrix Mechanics and Variational Principle

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 5.61 Physical Chemistry I Fall, 2016 Professors Robert W. Field Lecture 24: Molecular Orbital Theory, Matrix Mechanics and Variational Principle At this point, we have nearly completed our introduction to quantum mechanics and we’re finally ready to deal with the electronic structure of molecules. To begin with, we are going to treat what is absolutely the simplest molecule we can imagine: H+ 2 . This simple molecule will allow us to work out the basic ideas of what will become molecular orbital (MO) theory.  r We set up our coordinate system as shown at right, with the electron positioned at r, and the two nuclei positioned at points RA and RB , at a distance R from one another. The Hamiltonian is easy to express: RA e−  k Q
 Q Q
Q rA
Q rB Q
Q
Q QQ  
s - HB HA    H+ 2 R Coordinates RB 2 2 1 1 1 b = − 1 ∇2r − ∇A − ∇B − H − + b b b −R b B| 2M | 2{z } |2M {zA} | {zB} ||RA{z− r̂|} ||RB{z− r̂|} ||RA {z } Electron Kinetic Energy HA Kinetic Energy HB Kinetic Energy e− −HA Attraction e− −HB Attraction HA −HB Repulsion Now, just as was the case for atoms, we would like a picture where we can separate the electronic motion from the nuclear motion. For helium, we did this by noting that the nucleus was much heavier than the electrons and so we could approximate the center of mass coordinates of the system by placing the nucleus at the origin. For molecules, we will make a similar approximation, called the Born-Oppenheimer approximation. Here, we note again that the nuclei are much heavier than the electrons. As a result, they will move much more slowly than the light electrons. Thus, from the point of view of the electrons, 5.61 Lecture 24 Fall, 2017 Page 2 the nuclei are almost sitting still and so the moving electrons see a static field that arises from fixed nuclei. A useful analogy here is that of gnats flying around on the back of an elephant. The elephant may be moving, but from the gnats point of view, the elephant is always more or less sitting still. The electrons are like the gnats and the nucleus is like the elephant. The result is that, if we are interested in the electrons, we can to a good approximation fix the nuclear positions, RA and RB , and just look at the motion of the electrons in a molecule. This is the Born-Oppenheimer approximation, which is sometimes also called the clamped-nucleus approximation, for obvious reasons. Once the nuclei are clamped, we can make two simplifications of our Hamiltonian. First, we can neglect the kinetic energies of the nuclei because they are not moving. Second, because the nuclei are b A and R b B with the numbers RA and RB . Thus, our fixed, we can replace the operators R Hamiltonian reduces to 2 1 1 1 b el (RA , RB ) = − ∇r − H − + 2 |RA − r̂| |RB − r̂| |RA − RB | where the last term is now just a number — the electrostatic repulsion between two protons at a fixed separation. The second and third terms depend only on the position of the electron, r, and not its momentum, so we immediately identify those as a potential and write: 2 1 b el (RA , RB ) = − ∇r + V RA ,RB (r̂) + . H eff 2 |RA − RB | This Hamiltonian now only contains operators for the electron (hence the subscript “el”), and we write the Schrödinger equation for the electron as: b el (R)ψel (r) = Eel ψel (r) H where ψel is the wave function for the single electron in H2+ . [Note: here we use the shorthand R to denote both RA and RB .] However, this Schrödinger equation does not tell the whole story. Because the Hamiltonian depends on the nuclear positions, the electronic wavefunction will also depend on nuclear locations. For example, the figure below shows the difference between the effective potentials that the electron “feels” when the nuclei are close together versus far apart: R Small  ........................ ........ ....... ...... .... .... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. R Large ...................... .......... ....... ...... .... . . ... ... ... .... . .. .. .. .. ... .. .. .. ..  ........................ ........ - .... ... .... ... ... ... .... ... .... .. .. .. .. .. .. .. ... .. .. ... .. ... .. .. .. ?? ?? Veff (r) ....... ...... .... .... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. .............. ....... .......... .... .... .... ... .. ... . ... ... ... .. . ... .. . ... .. . ... .. . ... .. ... . ... .... ... ... . ... . ... ... ? ...................... .......... ....... ...... .... . . ... ... ... .... . .. .. .. .. ... .. .. .. .. N ? - 5.61 Lecture 24 Fall, 2017 Page 3 Because the electron feels a different potential at each bond distance R, the wavefunction will also depend on R. In the same limits as above, we will have: ... ... .... .... .. ... .. .. .. ... ... ... .. .. .... .... ... .... . . .. ... .. .... .. ... .. ... .. .. .. ... ... .. .. . .. ... . . ... .. .. ... ... .. .. .... . . . . . ... .... .. . . . . . . . .... ... .. . . . . .... . . ... . .. . ...... . . . . . . . ........... ...... .... . . . . ....... . .... . ........ . . . . . ......... ..... . . . . . . . ........... . ....... . . . .................. . . . . . . . . . . . ....................... ............................. ψel (r) .... ..... ...... ...... ...... .. . ... .... .. .. .... ... ... .... .. .. .. ... .. .. ... ... .. .. ... .... .. .... .. ... ... . .... ... ... .. ... ... ... .. ... .. ... ... .. . . . ... ... . . .. . . ... ... . .. . . .... . . ... . . . ..... . . . . . . .... ...... . .... . . . . . . ........ . . . .... .. ..... . . . ......... . . . . . . . . ......... ........ ............. ...... . . . . . . . . . . . . . . . . . . . ............................... . . ............................. R Small R Large Finally, because the electron eigenfunction, ψel , depends on R, then the eigen-energy of the electron, Eel (R), will also depend on the bond length. Thus a more precise form of the electronic Schrödinger equation would be: b el (R)ψel (r; R) = Eel (R)ψel (r; R) H where the additional dependence of everything on the value of R is made explicit. Mechanically, then, what we have to do is solve for the electronic eigenstates, ψel , and their associated eigenvalues, Eel (R), at many different fixed values of R. The way that these eigenvalues change with R will tell us about how the energy of the molecule changes as we stretch or shrink the bond. This is the central idea of the Born-Oppenheimer approximation, and it is really very fundamental to how chemists think about molecules. We think about classical point-like nuclei clamped at various different positions, with the quantum mechanical electrons whizzing about and gluing the nuclei together. When the nuclei move, the energy of the system changes because the energies of the electronic orbitals change as well. There are certain situations where this approximation breaks down, but for the most part the Born-Oppenheimer picture provides an extremely useful and accurate way to think about chemistry. How are we going to solve for these eigenstates? It should be clear that looking for exact solutions is going to lead to problems in general. Even for H2+ the solutions are extremely complicated and for anything more complex than H2+ exact solutions are impossible. So we have to resort to approximations again. The first thing we note is that if we look closely at our intuitive picture of the H2+ eigenstates above, we recognize that these molecular eigenstates look very much like the sum of the 1s atomic orbitals for the two hydrogen atoms. That is, we note that, to a good approximation, we should be able to write: ψel (r) ≈ c1 1sA (r) + c2 1sB (r) where c1 and c2 are constants. In the common jargon, the function on the left is called a 5.61 Lecture 24 Fall, 2017 Page 4 molecular orbital (MO), whereas the functions on the right are called atomic orbitals (AOs). If we write our MOs as sums of AOs, we are using what is called the Linear Combination of Atomic Orbitals (LCAO) approach. The challenge, in general, is to determine the “best” choices for c1 and c2 . For H+ 2 , it seems reasonable that the best choice for the ground state will be c1 = c2 . But how can we be sure this is really the best we can do? And what about the possibility that we want something other than the ground state? Or if we want to describe a more complicated molecule like HeH+2 ? The Variational Principle In order to make further progress, we will use the Variational Principle to predict a better estimate of the ground state energy. This method is very general and its use in physical chemistry is widespread. Assume you have a Hamiltonian (such as for H+ 2 ) but you do not know the ground state energy, E0 , and/or ground state eigenfunction, φ0 . Z D E Z ∗ b 0 = E0 φ0 =⇒ H b = φ Hφ b 0 dτ = φ∗ E0 φ0 dτ = E0 . Hφ 0 0 Now, let’s make a guess, ψ, at the ground state wavefunction, which we will call the trial wavefunction. Compute the average value of the energy for the trial wavefunction: R ∗ Z b ψ Hψdτ b R Eavg = = ψ ∗ Hψdτ (if ψ is normalized). ∗ ψ ψdτ The Variational Theorem tells us that Eavg ≥ E0 for any choice of the trial function ψ! This makes physical sense, because the ground state energy is, by definition, the lowest possible energy, so it would be nonsense for the average energy to be lower than the energy of the trial wavefunction. SIDEBAR: PROOF OF VARIATIONAL THEOREM Expand ψ (assumed normalized) as a linear combination of the unknown eigenstates, φn , of the Hamiltonian: ψ= X n a n φn . 5.61 Lecture 24 Fall, 2017 Page 5 Note that in practice you will not know these eigenstates. The important point is that no matter what function you choose you can always expand it in terms of the infinite set of b orthonormal eigenstates of H. Z Z X X X ∗ ∗ ψ ψdτ = an am φ∗n φm dτ = a∗n am δmn = |an |2 = 1 n,m n,m Z X b = ψ ∗ Hψdτ X = En |an |2 Eavg = Eavg n a∗n am En δnm n E0 = “1” · E0 = X |an |2 E0 using the definition of “1” above n ==⇒ Eavg − E0 = X n |an |2 En − X n |an |2 E0 = X |an |2 (En − E0 ) ≥ 0 since En > E0 n Notice that in the last line we have shown that Eavg ≥ E0 . It is important to note that the equal sign is only obtained if an = 0 for all eigenstates φn that have En > E0 . In this situation, ψ actually is the ground state of the system (or at least one component of the ground state, if the ground state is degenerate). The variational method does two things for us. First, it gives us a way to compare two different wavefunctions and to show which one is closer to the wavefunction of the ground state: the state that has a lower average energy is the better approximation. Second, if we define our trial wavefunction in terms of parameters, then variation of those parameters gives us a way to optimize the parameters. Assume that ψ depends on a set of parameters c — such as is the case for our LCAO wavefunction above. We’ll put the parameters in brackets −ψ[c] in order to distinguish them from things like positions that are inside of parentheses −ψ(r). Then the average energy will depend on these parameters: R b ψ[c]∗ Hψ[c]dτ Eavg (c) = R . ψ[c]∗ ψ[c]dτ Thus, we can solve for the optimal set of parameters without knowing anything about the exact eigenstates! Let us apply this in the particular case of our LCAO-MO treatment of H2+ . Our trial wavefunction is: ψel [c] = c1 1sA + c2 1sB 5.61 Lecture 24 Fall, 2017 Page 6 where c = (c1 , c2 ). We want to determine the best values of c1 and c2 and the variational theorem tells us we need to minimize the average energy to find the best values. First, we compute the average energy. The numerator gives: Z Z ∗ b b 1 1sA + c2 1sB )dτ ψel Hel ψel dτ = (c1 1sA + c2 1sB )∗ H(c Z Z Z ∗ ∗ ∗ b b b el 1sA dτ = c1 c1 1sA Hel 1sA dτ +c1 c2 1sA Hel 1sB dτ +c2 c1 1sB H | | | {z } {z } {z } ≡H11 ≡H12 ≡H21 Z ∗ b el 1sB dτ + c2 c2 1sB H | {z } ≡ c∗1 H11 c1 + ≡H22 ∗ c1 H12 c2 + c∗2 H21 c1 + c∗2 H22 c2 . Note that H11 and H22 are not the orbital energies of 1s on an H atom. H11 and H22 depend on R because the H–atom 1s orbital is in the potential defined by both H atoms. The normalization integral gives (Sij are “overlap integrals”) Z Z ∗ ψel ψel dτ = (c1 1sA + c2 1sB )∗ (c1 1sA + c2 1sB )dτ Z Z Z ∗ ∗ ∗ = c1 c1 1sA 1sA dτ +c1 c2 1sA 1sB dτ +c2 c1 1sB 1sA dτ | {z } {z } {z } | | ≡S11 ≡S12 ≡S21 Z ∗ + c2 c2 1sB 1sB dτ {z } | ≡ c∗1 S11 c1 + ≡S22 ∗ c1 S12 c2 + c∗2 S21 c1 + c∗2 S22 c2 . So that the average energy takes the form: Eavg (c) = c∗1 H11 c1 + c∗1 H12 c2 + c∗2 H21 c1 + c∗2 H22 c2 . c∗1 S11 c1 + c∗1 S12 c2 + c∗2 S21 c1 + c∗2 S22 c2 We note that there are some simplifications we could have made to this formula: for example, since our 1s functions are normalized S11 = S22 = 1. However, by not making these simplifications, our final expressions will be a little more general and that will help us use them in more situations. Now, we want to minimize this average energy with respect to c1 and c2 . Taking the derivative with respect to c1 and setting it equal to zero [Note: when dealing with complex numbers and taking derivatives one must treat variables and their complex conjugates as 5.61 Lecture 24 Fall, 2017 Page 7 independent variables. Thus d/dc1 has no effect on c∗1 ]: ∂Eavg c∗1 H11 + c∗2 H21 =0= ∗ ∂c1 c1 S11 c1 + c∗1 S12 c2 + c∗2 S21 c1 + c∗2 S22 c2 c∗ H11 c1 + c∗1 H12 c2 + c∗2 H21 c1 + c∗2 H22 c2 ∗ ∗ − 1∗ 2 (c1 S11 + c2 S21 ) ∗ ∗ ∗ (c1 S11 c1 + c1 S12 c2 + c2 S21 c1 + c2 S22 c2 ) c∗ H11 c1 + c∗1 H12 c2 + c∗2 H21 c1 + c∗2 H22 c2 ∗ =⇒ 0 = (c∗1 H11 + c∗2 H21 ) − 1 ∗ (c1 S11 + c∗2 S21 ) ∗ ∗ ∗ c1 S11 c1 + c1 S12 c2 + c2 S21 c1 + c2 S22 c2 ∗ ∗ =⇒ 0 = (c1 H11 + c2 H21 ) − Eavg (c∗1 S11 + c∗2 S21 ) . Applying the same procedure to c2 : ∂Eavg c∗1 H12 + c∗2 H22 =0= ∗ ∂c2 c1 S11 c1 + c∗1 S12 c2 + c∗2 S21 c1 + c∗2 S22 c2 c∗ H11 c1 + c∗1 H12 c2 + c∗2 H21 c1 + c∗2 H22 c2 ∗ ∗ − 1∗ 2 (c1 S12 + c2 S22 ) ∗ ∗ ∗ (c1 S11 c1 + c1 S12 c2 + c2 S21 c1 + c2 S22 c2 ) c∗1 H11 c1 + c∗1 H12 c2 + c∗2 H21 c1 + c∗2 H22 c2 ∗ ∗ ∗ (c1 S12 + c∗2 S22 ) =⇒ 0 = (c1 H12 + c2 H22 ) − ∗ ∗ ∗ ∗ c1 S11 c1 + c1 S12 c2 + c2 S21 c1 + c2 S22 c2 ∗ ∗ =⇒ 0 = (c1 H12 + c2 H22 ) − Eavg (c∗1 S12 + c∗2 S22 ) . We notice that the expressions above look strikingly like matrix-vector operations. We can make this explicit if we define the Hamiltonian matrix: ! R R   b el 1sA dτ b el 1sB dτ H11 H12 1sA H 1sA H R H≡ ≡ R b el 1sA dτ b el 1sB dτ H21 H22 1sB H 1sB H and the Overlap matrix: R    R S11 S12 1s 1s dτ 1s 1s dτ A A A B R . S≡ ≡ R 1sB 1sA dτ 1sB 1sB dτ S21 S22 Then the best values of c1 and c2 satisfy the matrix eigenvalue equation:    
H11 H12
S11 S12 ∗ ∗ ∗ ∗ c1 c2 = Eavg c1 c2 . H21 H22 S21 S22 Which means that: ∂Eavg = 0 ⇐⇒ c† · H = Eavg c† · S. ∂c (Eq. 1) This equation doesn’t look so familiar yet, so we need to massage it a bit. First, it turns out that if we had taken the derivatives with respect to c∗1 and c∗2 instead of c1 and c2 , we would have gotten a slightly different equation:       H11 H12 c1 S11 S12 c1 = Eavg H21 H22 c2 S21 S22 c2 5.61 Lecture 24 Fall, 2017 Page 8 or ∂Eavg = 0 ⇐⇒ H · c = Eavg S · c (Eq. 2) ∂c∗ Taking the derivatives with respect to c∗1 and c∗2 is mathematically equivalent to taking the derivatives with respect to c1 and c2 [because we can’t change c1 without changing its complex conjugate, and vice versa]. Thus, the two matrix equations (Eqs. 1 and 2) above are precisely equivalent, and the second version is a little more familiar. We can make it even more familiar if we think about the limit where 1sA and 1sB are orthogonal (e.g. when the atoms are very, very far apart). Then we would have for the Overlap matrix: R R    1sA 1sA dτ R 1sA 1sB dτ 1 0 R S≡ = = 1. 1sB 1sA dτ 1sB 1sB dτ 0 1 Thus, in an orthonormal basis the overlap matrix is just the identity matrix and we can write Eq. 2 as: ∂Eavg = 0 ⇐⇒ H · c = Eavg c. ∂c∗ Now this equation is in a form where we certainly recognize it: this is an eigenvalue equation. Because of its close relationship with the standard eigenvalue equation, Eq. 2 is usually called a Generalized Eigenvalue Equation. In any case, we see the quite powerful result that the Variational theorem allows us to turn operator algebra into matrix algebra. Looking for the lowest energy LCAO state is equivalent to looking for the lowest eigenvalue of the Hamiltonian matrix, H. Further, looking for the best c1 and c2 is equivalent to finding the eigenvector of H that belongs to the lowest energy eigenvalue of H. Matrix Mechanics The variational principle illustrates what is actually a very general and powerful way of thinking and computing known as matrix mechanics (MM). Matrix mechanics is actually completely equivalent to the wave mechanics we’ve been discussing so far, but it has two major benefits. First, it places emphasis on the global structure of the problem, allowing us to make predictions about the eigenstates of abstract systems with a modest amount of effort and often without evaluating a single integral(!). 5.61 Lecture 24 Fall, 2017 Page 9 Second, because MM involves linear algebra rather than symbolic calculus, it is much easier to program a computer to solve problems in MM than in wave mechanics. Wave Mechanics Matrix Mechanics  ψ = c 1 φ1 + ψ = c 2 φ2 + ψ = c 3 φ 3 + . . . (φi are basis functions)  c1   ~ =  c2  ψ  c3  ... ψ ∗ = c∗1 φ∗1 + c∗2 φ∗2 + c∗3 φ∗3 + . . . ~ † = c∗ c∗ c∗ . . . ψ 1 2 3 b (operator) O O (matrix);Oij = R b is Hermitian O O = O† b ψ ∗ Oχdτ R Sij = φ?i φj dτ ~† · O · χ ψ ~ R R ψ ∗ χdτ b = εψ Oψ conjugate transpose
b j dτ φ∗i Oφ S (overlap) ~† · S · χ ψ ~ ~ = εS · ψ ~ O·ψ We’ve already encountered some of these rules above in dealing with the variational principle; the Hamiltonian became a matrix, the overlap matrix cropped up, wavefunctions became row and column vectors, the eigenvalues of the operator were represented by the eigenvalues of the matrix. . . Mathematically, the overlap matrix is an annoyance; it comes from the fact that we haven’t chosen our basis functions so that they are orthonormal. If they were orthonormal, the overlap matrix would become the identity matrix (unity along the diagonal, zeroes off) and we could just ignore it. We illustrated this above for the case of H2+ . If the basis functions are orthonormal, we also have a very nice interpretation for the coefficients, ci . Specifically, |ci |2 is the probability of finding the system in state i. This last point is very useful in interpreting MM and will come up frequently in the Problem Sets. Let’s go ahead and apply what we’ve learned to obtain the MO coefficients c1 and c2 for H+ 2. At this point we make use of several simplifications. The pair of off-diagonal matrix elements between 1sA and 1sB of H are identical because the Hamiltonian is Hermitian and 5.61 Lecture 24 Fall, 2017 Page 10 the orbitals are real: Z ∗ Z Z ∗ b b b el 1sA dτ ≡ V12 . 1sA Hel 1sB dτ = 1sB Hel 1sA dτ = 1sB H Note that V12 is positive at small R and goes to zero as R → ∞ (see page 13). Meanwhile, the diagonal elements are also equal, but for a different reason. The diagonal elements are the average energies of the states 1sA and 1sB . If these energies were different, it would imply that having the electron on one side of H2+ was more energetically favorable than having it on the other, which would be very puzzling. So we conclude Z Z b el 1sB dτ ≡ ε. b el 1sA dτ = 1sB H 1sA H Note that ε depends on R and, as R → ∞, ε → − 21 . ε depends on R because the electron in a 1s orbital centered on nucleus A is attracted by nucleus B at a distance R from nucleus A. Finally, we remember that 1sA and 1sB are normalized, so that Z Z 1sA 1sA dτ = 1sB 1sB dτ = 1 and because the 1s orbitals are real, the off-diagonal elements of S are also the same: Z Z S12 = 1sA 1sB dτ = 1sB 1sA dτ = S21 . Incorporating all of these simplifications gives us the generalized eigenvalue equation:       c1 c1 1 S12 ε V12 . = Eavg V12 ε c2 S12 1 c2 All of your favorite mathematical programs (Matlab, Mathematica, Maple, MathCad,. . . ) are capable of solving for the generalized eigenvalues and eigenvectors, and for more complicated cases we suggest you use them. However, this case is simple enough that we can solve it by guess–and–test. Based on our physical intuition above, we guess that the correct eigenvector will have c1 = c2 . Plugging this in, we find:       ε V12 c1 1 S12 c1 = Eavg V12 ε c2 S12 1 c2     (ε + V12 )c1 (1 + S12 )c1 =⇒ = Eavg (ε + V12 )c1 (1 + S12 )c1 ε + V12 =⇒= Eavg = ≡ Eσ . 1 + S12 5.61 Lecture 24 Fall, 2017 Page 11 Thus, our guess is correct and one of the eigenvectors of this matrix has c1 = c2 . This eigenvector is the σ–bonding state of H2+ , and we can express the associated orbital as: ψelσ = c1 1sA + c2 1sB = c1 1sA + c1 1sB ∝ 1sA + 1sB where in the last expression we have noted that c1 is just a normalization constant. In freshman chemistry, we taught you that the σ–bonding orbital existed, and this is where it . comes from. ..... . ∗ . . . We can also get the σ –antibonding orbital .......... ....... . . from the variational procedure. Since the . . .. ..... ....... ....... . . . .... ψσ (r) . . . ....... ... matrix is a 2×2 it has two unique eigenval...... ........... . . . . . . . . . . . . . . . . ........... . . . . . . . ..... ues: the lowest one (which we just found ......................................................... ........................... ..... .... ............. . . above) is bonding and the other is anti. . ... . . ... .... bonding. We can again guess the form of ... ...... ... .. the antibonding eigenvector, since we know . . ψσ (r) .... ... it has the characteristic shape +/−, so that .... ..... we guess the solution is c1 = −c2 :     c1 c1 1 S12 = Eavg −c1 −c1 S12 1     (ε − V12 )c1 (1 − S12 )c1 = Eavg ⇒ −(ε − V12 )c1 −(1 − S12 )c1  ε V12 V12 ε ⇒ Eavg =  ε − V12 = Eσ ∗ 1 − S12 so, indeed the other eigenvector has c1 = −c2 . The corresponding antibonding orbital is given by: ∗ ψelσ = c1 1sA + c2 1sB = c1 1sA − c1 1sB ∝ 1sA − 1sB where we note again that c1 is just a normalization constant. Given these forms for the bonding and antibonding orbitals, we can draw a simple picture for the H2+ MOs (see above). We can incorporate the energies obtained above into a simple MO diagram that looks like this: 5.61 Lecture 24 Fall, 2017 Eσ ∗ = Page 12 ε − V12 1 − S12 E1SA = ε E1SB = ε Eσ = ε + V12 1 + S12 On the left and right, we draw the energies of the atomic orbitals (1sA and 1sB ) that make up our molecular orbitals (σ and σ ∗ ) in the center. We note that when the atoms come together the energy of the bonding and antibonding orbitals are shifted by different amounts: ε − V12 −ε= 1 − S12 ε + V12 E1s − Eσ = ε − = 1 + S12 Eσ∗ − E1s = ε − V12 ε(1 − S12 ) εS12 − V12 − = 1 − S12 1 − S12 1 − S12 ε(1 + S12 ) ε + V12 εS12 − V12 − = 1 + S12 1 + S12 1 + S12 Now, S12 is the overlap between two 1s atomic orbitals. Since these atomic orbital wavefunctions are never negative, S12 must be a positive number. Thus, the first denominator is greater than the second, from which we conclude Eσ∗ − E1s = εS12 − V12 εS12 − V12 > = E1s − Eσ . 1 − S12 1 + S12 Thus, the antibonding orbital is destabilized more than the bonding orbital is stabilized. This conclusion need not hold for all diatomic molecules, but it is a good rule of thumb. This effect is called “overlap repulsion”. Note that in the special case where the overlap between the orbitals is negligible, S12 goes to zero and the two orbitals are shifted by equal amounts. However, when is S12 nonzero there are two effects that shift the energies: the physical interaction between the atoms and the fact that the 1sA and 1sB orbitals are not orthogonal. When we diagonalize H, we account for both of these effects, and the orthogonality constraint pushes the anti-bonding orbital upward in energy. Now, the rather miraculous thing about this simple MO treatment of H+ 2 is that it actually predicts the formation of a chemical bond between two atoms! To see this, we remember 5.61 Lecture 24 Fall, 2017 Page 13 that the energies of the σ and σ ∗ orbitals will depend on the distance between the nuclei. This is a direct result of the Born-Oppenheimer approximation we made at the start of this section. At some expense of time, the explicit forms of S12 (R), ε(R) and V12 (R) can be worked out using the explicit forms for the hydrogen atomic orbitals. For more about how these integrals are evaluated, you can look at McQ Problems 10.6-10.11. The integrals themselves are not interesting, but the results are:   1 2 −R 1+R+ R S=e 3   1 1 7 1 2 −R H12 = e − − R− R R 2 6 6   1 1 ε = − + e−2R 1 + 2 R Given that all of the constituent integrals are R-dependent, it should be clear that the MO energies, Eσ and Eσ∗ , will both depend on R. The resulting energies are plotted below. The clear conclusion is that, in the σ state, the energy is typically lower when the two atoms are close together than when they are far apart — the two atoms are bound to one another! Further, there is a characteristic distance that the atoms like to be relative to one another — the energy is lowest at about R = 1.3 Ångstrom. At this point, the one-electron bond is worth about 1.7 eV. Both of these numbers are quite close to physical reality, where the true bond length is about 1.0 Ångstrom and the binding energy is about 2.8 eV. Thus, from this model we would conclude that chemical bonds must exist, even if we didn’t know from experience that they do. Note that, as you might have guessed, the antibonding orbital is unbound (repulsive) at all separations, R. 5.61 Lecture 24 10 8 6 E 4 2 0 Fall, 2017 Page 14 ... ... ... ... ... ... ... ... ... ... ... ...  ... Eσ∗ − E1s ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... .... ... .... ... .... ..... ... ..... ... ..... ... ..... ... ..... ... ..... ... ...... ...... ...  ....... ... E − E ........ 1s σ ... ........... ... ............. ... ............... ... .................... .... ................................ ... ..... ............................................................................. . . . . . . . . . . . . . . . . . . ..... . . . . . . . . . . . . . . . . . . . . ..... 1 ................. 2 4 5 ....... .................... 3 . . . . . . . . . . . . . . ............ . . . . . . . . . . . . . ................................................ R