5.6 Physial Chemistry Leture #24-25 MOLECULAR ORBITAL THEORY- PART I At this point, we have nearly ompleted our rash-ourse introdution to quantum mehanis and we re finally ready to deal with moleules. Hooray! To begin with, we are going to treat what is absolutely the simplest moleule we an imagine: H +. This simple moleule will allow us to work 2 out the basi ideas of what will beome moleular orbital (MO) theory. We set up our oordinate system as shown at right, with the eletron positioned at r, and the two nulei positioned at points R A and R B, at a distane R from one another. The Hamiltonian is easy to write down: R A H A Ĥ = 2 r 2 A 2 2M A B 2 2M B ˆR A ˆr r A e- r ˆR B ˆr r B R H 2 + Coordinates H B + ˆR A ˆR B R B Eletron Kineti Energy H A Kineti Energy H B Kineti Energy e - -H A Attration e - -H B Attration H A -H B Repulsion Now, just as was the ase for atoms, we would like a piture where we an separate the eletroni motion from the nulear motion. For helium, we did this by noting that the nuleus was muh heavier than the eletrons and so we ould approximate the enter of mass oordinates of the system by plaing the nuleus at the origin. For moleules, we will make a similar approximation, alled the Born-Oppenheimer approximation. Here, we note again that the nulei are muh heavier than the eletrons. As a result, they will move muh more slowly than the light eletrons. Thus, from the point of view of the eletrons, the nulei are almost sitting still and so the moving eletrons see a stati field that arises from fixed nulei. A useful analogy here is that of gnats flying around on the bak 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 eletrons are like the gnats and the nuleus is like the elephant.
5.6 Physial Chemistry Leture #24-25 2 The result is that, if we are interested in the eletrons, we an to a good approximation fix the nulear positions R A and R B and just look at the motion of the eletrons in a moleule. This is the Born-Oppenheimer approximation, whih is sometimes also alled the lamped-nuleus approximation, for obvious reasons. One the nulei are lamped, we an make two simplifiations of our Hamiltonian. First, we an neglet the kineti energies of the nulei beause they are not moving. Seond, beause the nulei are fixed, we an replae the operators Rˆ A and Rˆ B with the numbers R A and R B. Thus, our Hamiltonian redues to ˆ 2 H R, R = r + el A B 2 R rˆ R rˆ R R A B A B where the last term is now just a number the eletrostati repulsion between two protons at a fixed separation. The seond and third terms depend only on the position of the eletron, r, and not its momentum, so we immediately identify those as a potential and write: Hˆ 2,, = r RA R R R + V B rˆ + el A B 2 eff R R A B This Hamiltonian now only ontains operators for the eletron (hene the subsript el ), and we write the Shrodinger equation for the eletron as: r Ĥ el ( R) ψ el ( r ) = E el ψ el where ψ el is the wave funtion for the single eletron in H 2 +. [Note: here we use the shorthand R to denote both R A and R B.] However, this Shrodinger equation does not tell the whole story. Beause the Hamiltonian depends on the nulear positions, the eletroni wavefuntion will also depend on where the nulei are. For example, the figure below shows the differene between the effetive potentials the eletron feels when the nulei are lose together versus far apart: R Small R Large V eff (r)
5.6 Physial Chemistry Leture #24-25 3 Beause the eletron feels a different potential at eah bond distane R, the wavefuntion will also depend on R. In the same limits as above, we will have: ψ el (r) R Small R Large Finally, beause the eletron eigenfuntion, ψ el, depends on R then the eigenenergy of the eletron, E el (R), will also depend on the bond length. Thus a more preise form of the eletroni Shrodinger equation would be: r;r = E r,r Ĥ el ( R) ψ el el ( R) ψ el where the additional dependene of everything on the value of R is made expliit. Mehanially, then, what we have to do is solve for the eletroni eigenstates, ψ el, and their assoiated eigenvalues, E el (R), at many different fixed values of R. The way that these eigenvalues hange with R will tell us about how the energy of the moleule hanges as we streth or shrink the bond. This is the entral idea of the Born-Oppenheimer approximation, and it is really very fundamental to how hemists think about moleules. We think about lassial point-like nulei lamped at various different positions, with the quantum mehanial eletrons whizzing about and gluing the nulei together. When the nulei move, the energy of the system hanges beause the energies of the eletroni orbitals hange as well. There are ertain situations where this approximation breaks down, but for the most part the Born-Oppenheimer piture provides an extremely useful and aurate way to think about hemistry. How are we going to solve for these eigenstates? It should be lear that looking for exat solutions is going to lead to problems in general. Even for H 2 + the solutions are extremely ompliated and for anything more omplex than H 2 + exat solutions are impossible. So we have to resort to approximations again. The first thing we note is that if we look losely at our intuitive piture of the H + 2 eigenstates above, we reognize that these moleular eigenstates look very muh like the sum of the s atomi orbitals for the two hydrogen atoms. That is, we note that to a good approximation we should be able to write: ψ el ( r) + s A ( r) s B ( r ) 2
5.6 Physial Chemistry Leture #24-25 4 where and 2 are onstants. In the ommon jargon, the funtion on the left is alled a moleular orbital (MO), whereas the funtions on the right are alled atomi orbitals (AOs). If we write our MOs as sums of AOs, we are using what is alled the linear ombination of atomi orbitals (LCAO) approah. The hallenge, in general, is to determine the best hoies for and 2 for H 2 + it looks like the best hoie for the ground state will be = 2. But how an we be sure this is really the best we an do? And what about if we want something other than the ground state? Or if we want to desribe a more ompliated moleule like HeH +2? THE VARIATIONAL PRINCIPLE In order to make further progress, we will use the Variational Priniple to predit a better estimate of the ground state energy. This method is very general and its use in physial hemistry is widespread. Assume you have a Hamiltonian (suh as H 2 + ) but you don t know the ground state energy E 0 and or ground state eigenfuntion φ 0. Hˆφ = E φ H ˆ = ˆ 0 0 0 φ 0 Hφ 0 dτ = φ 0 E 0 φ 0 dτ =E 0 Now, assume we have a guess, ψ, at the ground state wavefuntion, whih we will all the trial wavefuntion. Compute the average value of the energy for the trial wavefuntion: Hˆ dτ E = ψ ψ = ψ Hˆ avg ψdτ (if ψ normalized) ψ ψ dτ the Variational Theorem tells us that E avg E 0 for any hoie of the trial funtion ψ! This makes physial sense, beause the ground state energy is, by definition, the lowest possible energy, so it would be nonsense for the average energy to be lower. SIDEBAR: PROOF OF VARIATIONAL THEOREM Expand ψ (assumed normalized) as linear ombination of the unknown eigenstates, φ n, of the Hamiltonian: ψ = a n φ n n Note that in pratie you will not know these eigenstates. The important point is that no matter what funtion you hoose you an always expand it in terms of the infinite set of orthonormal eigenstates of Ĥ.
5.6 Physial Chemistry Leture #24-25 5 ψ ψ dτ = an am φ n φ m dτ = a n a m δ mn = a n = 2 nm, nm, g ψ Hˆψdτ a n a m n ˆ E av = = φ Hφ m dτ = an a m φ n E φ m m d τ nm, nm, = a a E δ = a 2 n m m mn n E n nm, n Now, subtrating the ground state energy from the average E E = E = a 2 0 0 n n E 0 2 2 avg E 0 an En an E 0 = n n = a 2 n E n E 0 n n 0 sine E n E Where, in the last line we have noted that the sum of terms that are nonnegative must also be non-negative. It is important to note that the equals sign is only obtained if a n =0 for all states that have E n >E 0. In this situation, ψ atually is the ground state of the system (or at least one ground state, if the ground state is degenerate). 0 The variational method does two things for us. First, it gives us a means of omparing two different wavefuntions and telling whih one is loser to the ground state the one that has a lower average energy is the better approximation. Seond, if we inlude parameters in our wavefuntion variation gives us a means of optimizing the parameters in the following way. Assume that ψ depends on some parameters suh as is the ase for our LCAO wavefuntion above. We ll put the parameters in brakets -ψ[] in order to differentiate them from things like positions that are inside of parenthesis -ψ(r).then the average energy will depend on these parameters: ψ [ ] Hˆ ψ [ ] dτ Eavg ( ) = ψ [ ] ψ [ ]dτ Note that, using the variational priniple, the best parameters will be the ones that minimize the energy. Thus, the best parameters will satisfy ψ [ ] E Hˆ ψ ave [ ] dτ = = 0 i i ψ [ ] ψ [ ]dτ Thus, we an solve for the optimal parameters without knowing anything about the exat eigenstates!
5.6 Physial Chemistry Leture #24-25 6 Let us apply this in the partiular ase of our LCAO-MO treatment of H 2 +. Our trial wavefuntion is: ψ el [ ] = sa + 2s B where =( 2 ). We want to determine the best values of and 2 and the variational theorem tells us we need to minimize the average energy to find the right values. First, we ompute the average energy. The numerator gives: ψ Hˆ ψ d ˆ el el el τ = ( sa + 2 s B ) H ( s A + 2 s B ) dτ = s AĤe ˆ l s A dτ + 2 s A H el s B dτ + 2 s Ĥ B τ + ˆ el s A d 2 2 s B H el s B d τ H H 2 H 2 H 22 + H H + 2 2 2 H + H 2 2 22 2 while the normalization integral gives: ψ ψ τ = s + el el d 2 ( A s B ) ( s 2 ) A + s B dτ = s A s A dτ + 2 s s d A τ + B 2 s B s d A τ + 2 2 s s dτ B B S S 2 S 2 S 22 S + S + S + S 2 2 2 2 2 22 2 So that the average energy takes the form: H + H + E 2 2 2 H 2 + 2 = H 22 2 avg S + + S 2 2 S + S 2 2 2 22 2 We note that there are some simplifiations we ould have made to this formula: for example, sine our s funtions are normalized S =S 22 =. By not making these simplifiations, 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 respet to and 2. Taking the derivative with respet to and setting it equal to zero [Note: when dealing with omplex numbers and taking derivatives one must treat variables and their omplex onjugates as independent variables. Thus d/d has no effet on ]:
5.6 Physial Chemistry Leture #24-25 7 E avg = 0 = H + H 2 2 S + S + S + S 2 2 2 2 2 22 2 H + H + 2 2 H + H 2 2 2 22 2 ( 2 ( S + ) S 2 2 S + S + 2 2 S + 2 2 S 2 22 2 ) H + H + H + H 0 = H + 2 2 2 2 2 22 2 H S + S 2 2 S + S + S + 2 2 2 2 S 2 2 2 22 2 0 = ( ) H + H E 2 2 avg ( ) S + S 2 2 Applying the same proedure to 2 : E avg H 2 = = + H 2 22 0 S + 2 S + S + S 2 2 2 2 2 22 2 H H + 2 2 + H + H 2 2 2 22 2 2 ( S 2 + 2 ( S 22 ) S + + S 2 2 2 S 2 + S 2 22 2 ) H 0 = H + H + + H 2 2 2 H 2 + H 2 22 2 2 2 22 S + S ( S + + S 2 2 22 ) 2 2 2 S2 + 2 S 22 2 0 = ( H + 2 2 H 22 ) E avg ( S 2 + 2 S 22 ) We notie that the expressions above look strikingly like matrix-vetor operations. We an make this expliit if we define the Hamiltonian matrix: H s ˆ H dτ s A Ĥ el s A H els A B dτ 2 H H 2 H 22 s ˆ B H els A dτ B els B dτ s Ĥ and the Overlap matrix: sasa s dτ τ S S A s B d 2 S S2 S22 s s B A dτ s B s B dτ Then the best values of and 2 satisfy the matrix eigenvalue equation: Whih means that: ( ) S 2 S 22 H H2 S E S 2 2 = avg 2 H2 H22 E avg = 0 ih = Eavg is Eq. 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 respet to
5.6 Physial Chemistry Leture #24-25 8 and 2 instead of and 2, we would have gotten a slightly different equation: H H2 S S 2 = Eavg H2 H22 2 S 2 S 22 2 or E avg = 0 Hi = EavgSi Eq. 2 Taking the derivatives with respet to and 2 is mathematially equivalent to taking the derivatives with respet to and 2 [beause we an t hange without hanging its omplex onjugate, and vie versa]. Thus, the two matrix equations (Eqs. &2) above are preisely equivalent, and the seond version is a little more familiar. We an make it even more familiar if we think about the limit where s A and s B are orthogonal (e.g. when the atoms are very, very far apart). Then we would have for the Overlap matrix: sasa dτ s A s B dτ 0 S = = s s τ τ 0 B A d s s d B B Thus, in an orthonormal basis the overlap matrix is just the identity matrix and we an write Eq. 2 as: E avg = 0 Hi = E avg Now this equation is in a form where we ertainly reognize it: this is an eigenvalue equation. Beause of its lose relationship with the standard eigenvalue equation, Eq. 2 is usually alled a Generalized Eigenvalue Equation. In any ase, 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 and 2 is equivalent to finding the lowest eigenvetor of H. Matrix Mehanis The variational priniple illustrates what is atually a very general and powerful way of thinking and omputing known as matrix mehanis (MM).
5.6 Physial Chemistry Leture #24-25 9 Matrix mehans is atually ompletely equivalent to the wave mehanis we ve been disussing so far, but it has two major benefits. First, it plaes emphasis on the global struture of the problem, allowing us to make preditions about the eigenstates of abstrat systems with a modest amount of effort and often without doing a single integral(!). Seond, beause MM involves linear algebra rather than symboli alulus, it is muh easier to program a omputer to solve problems in MM than in wave mehanis. For the purposes of this lass, we an think of MM as a different language for quantum mehanis. We know how to write down many expressions in wave mehanis, and it is just a question of translating those expressions into MM before we do omputations. Toward that end, we summarize a lexion for MM expressions in terms of their wave ounterparts below. All of these equivalenes an be proven mathematially, but we will rarely have ause to worry about those details. Wave Mehanis Matrix Mehanis ψ= onjugate φ + 2 φ 2 + 3 φ 3 + ψ = 2 (φ transpose i are basis funtions) 3 ψ= φ + 2 φ 2 + 3 φ 3 + ψ = 2 3 ˆ O (operator) O (matrix); φ ˆ O ij = i O φ j dτ ˆ O is Hermitian O = O ψ O ˆ χdτ ψ O χ? S (overlap); S ij = φ i φ j dτ ψ χdτ ψ S χ O ˆ ψ = εψ O ψ = ε S ψ We ve already enountered some of these rules above in dealing with the variational priniple; the Hamiltonian beame a matrix, the overlap matrix
5.6 Physial Chemistry Leture #24-25 0 ropped up, wavefuntions beame row and olumn vetors, the eigenvalues of the operator were represented by the eigenvalues of the matrix. Mathematially, the overlap matrix is an annoyane; it omes from the fat that we haven t hosen our basis funtions so that they are orthonormal. If they were orthonormal, the overlap matrix would beome the identity matrix (unity along the diagonal, zeroes off) and we ould just ignore it. We illustrated this above for the ase of H + 2. If the basis funtions are orthonormal, we also have a very nie interpretation for the oeffiients, i. Speifially, i 2 is the probability of finding the system in state i. This last point is very useful in interpreting MM and will ome up frequently in the PSets. Let s go ahead and apply what we ve learned to obtain the MO oeffiients and 2 for H 2 +. At this point we make use of several simplifiations. The off-diagonal matrix elements of H are idential beause the Hamiltonian is Hermitian and the orbitals are real: s ˆ A H s ˆ ˆ el B dτ = s B H el s A dτ = s B H el s A dτ V 2 Meanwhile, the diagonal elements are also equal, but for a different reason. The diagonal elements are the average energies of the states s A and s B. If these energies were different, it would imply that having the eletron on one side of H 2 + was more energetially favorable than having it on the other, whih would be very puzzling. So we onlude s ˆ A H el s A dτ = s BĤel s B dτ ε Finally, we remember that s A and s B are normalized, so that sasa dτ = s B s B dτ = and beause the s orbitals are real, the off-diagonal elements of S are also the same: S2 = s = = A s B dτ s B s A dτ S 2. Inorporating all these simplifiations gives us the generalized eigenvalue equation: ε V2 S 2 = E V ε avg 2 2 S 2 2 All your favorite mathematial programs (Matlab, Mathematia, Maple, MathCad ) are apable of solving for the generalized eigenvalues and eigenvetors, and for more ompliated ases we suggest you use them.
5.6 Physial Chemistry Leture #24-25 However, this ase is simple enough that we an solve it by guess-and test. Based on our physial intuition above, we guess that the orret eigenvetor will have = 2. Plugging this in, we find: ε V 2 S 2 = Eavg V 2 ε S 2 ( ε + V 2 ) = E avg ( ε + V 2 ) ( + ) ε + V E avg = 2 E + S 2 σ + S 2 S 2 Thus, our guess is orret and one of the eigenvetors of this matrix has + = 2. This eigenvetor is the σ-bonding state of H 2, and we an write down the assoiated orbital as: ψ σ el = sa + 2s B = sa + s s +s B A B where in the last expression we have noted that is just a normalization onstant. In freshman hemistry, we taught you that the σ-bonding orbital existed, and this is where it omes from. We an also get the σ -antibonding orbital from the variational proedure. Sine the matrix is a 2x2 it has two unique eigenvalues: the lowest one (whih we just found above) is bonding and the other is antibonding. We an again guess the form of the antibonding eigenvetor, sine we know it has the harateristi shape +/-, so that we guess the solution is =- 2 : ψ σ (r) ψ σ (r)
5.6 Physial Chemistry Leture #24-25 2 ε V 2 S 2 = E avg V 2 ε S 2 ( ε V2 ) = E ( ε V 2 ) ( ) S 2 avg S 2 ε V2 Eavg = = Eσ S2 so, indeed the other eigenvetor has =- 2. The orresponding antibonding orbital is given by: ψ σ el = sa + 2s B = sa s s s B A B where we note again that is just a normalization onstant. Given these forms for the bonding and antibonding orbitals, we an draw a simple piture for the H 2 + MOs (see above). We an inorporate the energies obtained above into a simple MO diagram of H 2 + : ε V E 2 σ = S2 E sa =ε E sb =ε ε + V E σ = 2 + S2 On the left and right, we draw the energies of the atomi orbitals (sa and sb) that make up our moleular orbitals (σ and σ) in the enter. We note that when the atoms ome together the energy of the bonding and antibonding orbitals are shifted by different amounts: ε V2 ε V2 ε( S 2 ) εs2 E E = ε = = V 2 σ s S2 S2 S 2 S 2 ε + V2 ε( + S2 E E = ε = ) ε +V 2 εs 2 V 2 s σ = + S2 + S2 +S 2 +S 2
5.6 Physial Chemistry Leture #24-25 3 Now, S 2 is the overlap between two s orbitals. Sine these orbitals are never negative, S 2 must be a positive number. Thus, the first denominator is greater than the seond, from whih we onlude εs2 V2 εs2 V2 Eσ E s = > = E s S + S E σ 2 2 Thus, the antibonding orbital is destabilized more than the bonding orbital is stabilized. This onlusion need not hold for all diatomi moleules, but it is a good rule of thumb. This effet is alled overlap repulsion. Note that in the speial ase where the overlap between the orbitals is negligible, S 2 goes to zero and the two orbitals are shifted by equal amounts. However, when is S 2 nonzero there are two effets that shift the energies: the physial interation between the atoms and the fat that the sa and sb orbitals are not orthogonal. When we diagonalize H, we aount for both these effets, and the orthogonality onstraint pushes the orbitals upwards in energy. Now, the rather miraulous thing about this simple MO treatment of H 2 + is that it atually predits the formation of a hemial bond between two atoms! To see this, we remember that the energies of the σ and σ will depend on the distane between the nulei. This is a diret result of the Born-Oppenheimer approximation we made at the start of this setion. At some expense of time, the expliit forms of S 2 (R), ε(r) and V 2 (R) an be worked out using the expliit forms for the hydrogen atomi orbitals. For more about how these integrals are evaluated, you an look at MQ Problems 0.6-0.. The integrals themselves are not interesting, but the results are: S = e R ( + R + 3 R 2 ) H 2 = e R 7 R R 2 R 2 6 6 2R ε = 2 + e + R Given that all the onstituent integrals are R-dependent, it should be lear that the MO energies, E σ and E σ, will both depend on R. The resulting energies are plotted below. The lear onlusion is that, in the σ state, the energy is typially lower when the two atoms are lose together than when they are far apart the two atoms are bound to one another! Further, there is a harateristi distane that the atoms like to be relative to one another
5.6 Physial Chemistry Leture #24-25 4 the energy is lowest at about.3 Angstrom. At this point, the one-eletron bond is worth about.7 ev. Both of these numbers are quite lose to physial reality, where the true bond length is about.0 Angstrom and the binding energy is about 2.8 ev. Thus, from this model we would onlude that hemial bonds must exist, even if we didn t know from experiene that they do. Note that, as you might have guessed, the antibonding orbital is unbound at all separations, R. E σ -E s E σ -E s
MIT OpenCourseWare http://ow.mit.edu 5.6 Physial Chemistry Fall 203 For information about iting these materials or our Terms of Use, visit: http://ow.mit.edu/terms.