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Unformatted text preview: Chem 120A 04/27/07 Spring 2007
READING: Review BornOpp., molecular bonding, H2 + , H2 Lecture 38 Ratner and Schatz: Ch. 10 (Ch. 11) Atkins and Friedman: Chapter 8 Review of the BornOppenheimer (BO) approximation In applying the BO approximation, we divided the molecular Schr¨ dinger equation into two equations o based on the adiabatic approximation. First solving for the electronic eigenstates and eigenenergies allows us to set up an effective Schr¨ dinger for the nuclear motion, o ˆ (TN + Eq (R))ψq (R) = E ψq (R), (1) ˆ ˆ ˆ ˆ where R is the nuclear coordinate, Eq (R) = Φq (R) Te + Vee + VeN (R) + VNN (R) Φq (R) is the electronic energy, and E is the total energy of the molecular state ψq (R). The function Eq (R) is known as the BornOppenheimer for the nuclei when the electrons are in state Φq . The success of the BO approximation is that we can approximately separate the nuclear and electronic degrees of freedom in order to approximately solve the manybody molecular Schr¨ dinger equation. The wavefunctions of the molecule are then a product o of the electronic and nuclear wavefunctions: Ψ({r }, R) ≈ ψ (R)Φ({r }, R). (2) The simplest molecule: H+ 2 Qualitatively, Eq (R) → ∞ as R → 0 due to the Coulomb repulsion of the nuclei. Also, as R → ∞, the energy is just a sum of the individual atom energies, Eq (R) → ∑ Eatoms since the cross terms vanish. There are two types of BO potential energies, bonding and antibonding (also called binding and antibinding.) The bound state exists because the total energy is lower than the sum of the energies of the individual atoms at some distances. The bonding state is associated with the molecular orbital Φ+ , with E+ ≡ V+ , and the antibonding state is associated with Φ− with E− ≡ V− . These states are also called the bonding and antibonding orbitals. The hydrogen molecular ion is the simplest case of a molecule. It consists of two hydrogen nuclei and only a single electron. The coordinate system is shown in Fig. 4. The Schr¨ dinger equation for H+ can be solved o 2 exactly in elliptical coordinates (see McQuarrie and Simon 9.39.6). Here, we will approximately solve the problem using the BO approximation, then ﬁnding a qualitative solution using LCAOMO theory. The ﬁrst step in solving the Schr¨ dinger equation for this molecule is to ﬁnd the energies Eq (RAB ), where RAB is the o distance between the two nuclei. The electronnuclear potential for H+ is shown in Fig. 5. Notice that this potential looks similar to a double2 well potential, for which we have previously found the eigenfunctions, 1 φ− = √ (φL − φR ) 2 and 1 φ+ = √ (φL + φR ), 2 (3) where φL and φR are the harmonic oscillator wavefunctions for each side of the double well. Given this, we
Chem 120A, Spring 2007, Lecture 38 1 antibonding bonding R eq Figure 1: The bonding and antibonding BO potentials of H+ . 2 AB Figure 2: Coordinate system for H+ . 2 can guess that the wavefunctions for the H+ potential are 2 1 ψ+ = √ (1sA + 1sB ) 2 and 1 ψ− = √ (1sA − 1sB ), 2 (4) where 1sA = φ1s (r − RA ) and 1sB = φ1s (r − RB ). A more general solution is to take a linear combination of 1sA and 1sB with variational coefﬁcients, ψ = c1 1sA + c2 1sB . (5) Due to the symmetry of the potential and our analogy to the doublewell potential, we expect that the ground state will have c1 = c2 and the ﬁrst excited state will have c1 = −c2 . In general, if we had more nuclei or we wanted even higher excited states we could take an even more general linear combination, ψ = ∑i,α ci,α φi (r − Rα ), where now there could be more orbitals included for each nucleus. This method is very powerful because it allows us to expand the eigenfunctions of a manybody Hamiltonian that we cannot solve exactly as a linear combination of eigenfunctions for the hydrogen atom. The coefﬁcients {ci } are variational linear parameters that are chosen so that the ground state energy is minimized. For H+ 2 there are only two variational parameters. Because the superposition is linear in the variational parameters,
Chem 120A, Spring 2007, Lecture 38 2 H2+ Electronnucleus interaction potential A
6 4 2 0 B
2 4 6 V(r) 100 r D oublewell v(x), phi(x) potential phi+ phi 6 4 2 0 x 2 4 6 Figure 3: The H+ electronnuclear potential and the doublewell potential with the ﬁrst two eigenfunctions 2 shown. minimizing the ground state energy to ﬁnd the coefﬁcients is equivalent to setting up and solving the secular ˆ determinant to ﬁnd the energies and {ci }. The Schr¨ dinger equation, H ψ = E ψ , is used to set up the secular o ˆ determinant. When ψ is an eigenfunction of the Hamiltonian, we know that ψ H ψ d τ = E ψ 2 d τ , where d τ is for all of the electron coordinates. Since we have a linear combination of functions which are not eigenfunctions of the Hamiltonian, we have instead E = ˆ ψ Hψ dτ ψ 2 d τ (6) ˆ f1 H f2 d τ + c2 c1 ˆ f2 H f1 d τ + c2 2 ˆ f2 H f2 d τ (7) ˆ fi H f j d τ . ˆ ψ H ψ d τ = c1 2 ˆ f1 H f2 d τ + c1 c2 ≡ c1 2 H11 + c1 c2 H12 + c2 c1 H21 + c2 2 H22 . where f1 and f2 are atomic orbitals, we have assumed c1 and c2 are real for simplicity, and Hi j ≡ Similarly, the denominator in Eq. 6 is ψ 2 d τ = c1 2 S11 + c1 c2 S12 + c2 c1 S21 + c2 2 S22 , where Si j ≡ overlap integral ≡ fi f j d τ . (8) (9) The overlap integral goes like Si j ∼ e−RAB , which is exponentially small and quickly goes to zero for large RAB . This helps us to regain the atomic limit as the nuclei become far apart. The task is to minimize the numerator of Eq. (6) subject to the constraint that the denominator is equal to one. A rigorous derivation of this by means of the Lagrange method of undetermined coefﬁcients is given in Szabo and Ostlund, in ”Modern Quantum Chemistry”: the relevant pages are included in the supplementary material for this lecture. This minimization provides a variational determination of the coefﬁcients c1 and c2 and results in a set of linear equations that must be solved for the eigenvalues (energies) and coefﬁcients. This set of linear equaChem 120A, Spring 2007, Lecture 38 3 tions is known as the secular equation (also known as solving the secular determinant) A simpler derivation of this is outlined in Ratner and Schatz, Ch. 10, p. 148149. The secular equation is H ˜ ˜ c1 c2 = ES ˜ ˜ c1 c2 , (10) where H and S are 2 × 2 matrices for the H+ system. Rearranging Eq. 10, we have 2 ˜ ˜ ˜ ˜ c1 [H − ES] = 0, c2 ˜ ˜ ˜ ˜ (11) which has nontrivial solutions when the determinant of the matrix H − ES is equal to zero: ˜ ˜ ˜ ˜ H − ES = 0. (12) ˜ ˜ ˜ ˜ ˆ To ﬁnd the eigenvalues and eigenfunctions of H , we ﬁnd the values of E that make the determinant in Eq. 12 equal to zero, then use these energies in Eq. 10 to solve for c1 and c2 subject to the normalization constraint, c1 2 + c2 2 = 1. Note that the two solutions are not linearly independent. This is due to the fact that the solutions must be orthogonal. Our variational wavefunctions are linear combinations of atomic orbitals (LCAO) which form molecular orbitals (MO). This type of approximate solution to a molecular Schr¨ dinger equation is called LCAOMO. o Recall that the H+ Hamiltonian under the BornOppenheimer approximation is 2 1 1 1 ∇2 ˆ , H= − − − + 2 rA rB RAB (13) where the coordinate system is shown in Fig. 1. The variational wavefunctions are of the form ψ = c1 1sA + AB Figure 4: Coordinates for the H+ system. 2 c2 1sB . Because there are two different atomic wavefunctions in our linear combination (1sA and 1sB ), the secular determinant is 2 × 2. Expanding out the determinant, H11 − ε S11 H12 − ε S12 H21 − ε S21 H22 − ε S22
Chem 120A, Spring 2007, Lecture 38 = 0. (14)
4 Atomic orbitals psi(r) 1sA(r) 1sB(r) A r B Figure 5: The two atomic orbitals for nuclei A and B. We must evaluate each of the relevant Hamiltonian and overlap matrix elements in order to solve for ε . For the overlap matrix, we have S11 = S22 = S12 = S21 = 1sA 1sA d τ = 1sA 1sB d τ . 1sB 1sB d τ = 1, (15) Note that for the offdiagonal elements these are twocenter integrals and d τ signiﬁes the two volume elements: one which is centered at nucleus A and a second that is centered at nucleus B. Since S12 is a measure of the overlap of the two atomic orbitals, we expect that S12 ∝ e−RAB (see, for example Fig. 2). Solving for S12 we ﬁnd S12 ≡ S = e−RAB (1 + RAB + R3 AB ). 3 (16) For the Hamiltonian matrix, we have H11 = H22 = H12 = H21 = where J≡ and K≡ 1sA − S 1 1 + − e−RAB (1 + RAB ). 1sB d τ = rB RAB RAB (18) 1sA − 1 1 + 1sA d τ = e−2RAB (1 + RAB), rB RAB 1sA − 1sA ∇2 1 1 1 −−+ 1sA d τ ≡ E1s + J, 2 rA rB RAB 1 1 1 ∇2 1sB d τ ≡ E1s S + K , −−−+ 2 rA rB RAB (17) The integral J is called the Coulomb integral. It is like the Coulomb interaction between the electron which is localized on nucleus A and nucleus B. It should go to zero as RAB approaches inﬁnity, and we see that it does. The integral K is called the exchange integral. Deﬁning E0 = E1s , we can rewrite the secular determinant as E0 S + K − ε S E0 + J − ε E0 + J − ε E0 S + K − ε S = 0. (19) Chem 120A, Spring 2007, Lecture 38 5 antibonding bonding R eq Figure 6: Energies of the bonding and antibonding MOs of H+ . The equilibrium bond length (Req ) is about 2 2.5 a0 . The energy at this bond length is about 0.5 a.u less than the energy of the dissociated hydrogen atom plus proton.
Coulomb and Exchange Integrals E/a.u J/(1+S) K/(1+S) R_AB Figure 7: The Coulomb and exchange integrals. Note the Coulomb integral is always positive while the K exchange integral is negative at some distances. The minimum value of 1+S is about 0.05 a.u. The two solutions to this equation are ε+ = E 0 + J+K 1+S and ε− = E 0 + The energy ε+ is lower than the energy of a dissociated hydrogen atom plus a proton and corresponds to the bonding MO, while ε− is the energy of the antibonding MO. These two energies are shown as a function of RAB in Fig. 6. Bonding is achieved for R ∼ Req because the exchange integral is negative at these distances (Fig. 7), and the overall energy is lower than the energy of the dissociated system. This indicates that the exchange integral is responsible for the chemical bond. This simple approximation gives ε+ = E0 − 1.7 eV, compared to an experimental value of E0 − 2.6 eV. One could improve the estimate of the ground state energy by taking a more sophisticated variational wavefunction. For example, we could include more atomic orbitals in the linear combination, or make the charge of the nuclei a parameter. (If this is done, the effective charge will increase above one.) Alternatively, we could change the functional form of the variational wavefunction in order to reduce the energy. See Professor Harris’ lectures last week or Atkins/Friedman Ch. 8.2, p. 252 for a qualitative analysis of the factors determining the formation of the bond. Back to our current variational treatment, in order to ﬁnd the coefﬁcients c1 and c2 , we put the energies back into the individual secular equations and ﬁnd the coefﬁcients, subject to the normalization condition. Doing J−K . 1−S (20) Chem 120A, Spring 2007, Lecture 38 6 B onding MO Antibonding MO psi(r) psi(r) A B A r B r Figure 8: The bonding and antibonding molecular orbitals of H+ . The two nuclei are shown for reference. 2
+ + + + Figure 6: Combination of two 1s orbitals to form a σg bonding orbital. this we ﬁnd that for ε+ , c1 = c2 , whereas for ε− , c1 = −c2 . The eigenfunctions are therefore 1 ψ+ = √ (1sA + 1sB ) 2 and 1 ψ− = √ (1sA − 1sB ). 2 (21) These two wavefunctions correspond to the bonding and antibonding molecular orbitals for H+ . These MOs 2 are shown in Fig. 8. Notes: • Each MO is a linear combination of atomic orbitals (LCAO). The MOs reﬂect the symmetry of the molecule (i.e. they reﬂect the symmetry of the positions of the nuclei.) • Notation: We give MOs names to denote symmetry. We have seen that the lowest energy MO is a symmetric sum of two 1s orbitals. This situation is shown in Fig. . We label this type of orbital as σg . The σ label denotes the fact that there are no nodes in the MO along the bond axis and the label g denotes the fact that the orbital has inversion symmetry (i.e. if we swap all points such that x → −x, y → −y, and z → −z, the orbital still looks the same.) These labels can be determined for MOs that are LCAOs for any diatomic molecule. In order to determine what the labels are for a given set of MOs, we must ﬁrst consider what states the valence electrons of the atoms in the diatomic were in (i.e. s, p, d , etc.) Similar to other situations we have seen, having a higher number of nodes makes a given MO higher in energy, so we label according to the number of nodes. For example, if a MO has one node along the bond axis then this MO is a π orbital. We make these assignments because the l quantum numbers of the electrons in the atomic orbitals are not good quantum numbers for the MOs. However, because a diatomic molecule possesses symmetry with respect to rotation about the bond axis, the zprojection of the angular momentum should be conserved. In order to determine the label of a molecule, we simply sum the absolute values of the ml quantum numbers for the electrons in their atomic orbitals to ﬁnd the total ML for the MO. The total ML is equal to the number of nodes along the bond axis in the MO, and it is used directly to determine the label for the MO. For example, if we have two s
Chem 120A, Spring 2007, Lecture 38 7 orbitals combining to form and MO, then ML = 0, and this is a σ orbital. For two p orbitals, we can have ML = 0, 1, corresponding to one σ orbital and two π orbitals. For two d orbitals, we can have ML = 2 which corresponds to a δ orbital. The label g or u is added afterward to indicate whether or not the orbital has inversion symmetry. The simplest multielectron molecule: H2 LCAOMO for H2 Here we will follow a variational approach similar to that which we used for H+ in order to ﬁnd the MOs 2 and corresponding energies for the hydrogen molecule. Let’s try using the H+ MOs to describe H2 . The 2 lowest energy conﬁguration would be to place both of the electrons into the 1σg orbital. Due to the Pauli exclusion principle, the electrons must have opposite spin if they are both in the same orbital, so the resulting variational wavefunction is ψ = σg (1)σg (2)[α (1)β (2) − α (2)β (1)], (22) where α and β indicate the spin up or spin down states and the labels 1 and 2 label the electrons. Notice that this wavefunction is a singlet state, which means that the total spin angular momentum is zero, due to the antisymmetry of the spin part of the wavefunction. Expanding the spatial part of the wavefunction, σg (1)σg (2) =
= 1 [1sA (1) + 1sB (2)] [1sA (2) + 1sB (1)] 2 + 2S 1 [1sA (1)1sB (2) + 1sB (1)1sA (2) + 1sA (1)1sA (2) + 1sB (1)1sB (2)] , 2 + 2S (23) where S is the overlap integral between 1sA and 1sB . In the limit that the internuclear separation is inﬁnite, RAB → ∞, σg (1)σg (2) → 2φ (HA )φ (HB ) + φ (H− )φ (H+ ) + φ (H+ )φ (H− ), B B A A (24) where φ is the wavefunction of the species given in parenthesis. Notice here that the ﬁrst term corresponds to having two separated H atoms, while the second two terms correspond to ionic conﬁgurations where both electrons are associated with one nucleus, yielding the H− and H+ ions. This indicates that in the limit of large internuclear separation our variational wavefunction predicts that the ionic conﬁgurations are just as probable as having two neutral H atoms, so the total energy in this limit is E = 1/2[2EH + EH + + EH − ] > 2EH . Let’s now consider how we evaluate the energy from our variational wavefunction. The total molecular Hamiltonian is 1 1 e2 −1 ˆ ˆ ˆ − + , H = HA (1) + HB (2) + 4πε0 rA2 rB1 r12 (25) where the ﬁrst two terms are just the hydrogen atom Hamiltonians for each isolated atom and the terms in brackets are the attraction of electron 2 to nucleus A, the attraction of electron 1 to nucleus B, and the repulsive interaction between the two electrons, respectively. Given this Hamiltonian, we ﬁnd that the total energy of our variational wavefunction is E= 1 2E1s (1 + S2 ) + J + K , 1 + S2 (26) Chem 120A, Spring 2007, Lecture 38 8 Figure 9: Energies calculated using the LCAOMO for H2 . where J∼ ∼ and K∼ 1sA (1)1sB (2) −1 1 1 − + 1sA (2)1sB (1)d τ1 d τ2 . rA2 rB1 r12 (27) (1sA (1)1sB (2)) −1 1 1 − + (1sA (1)1sB (2))d τ1 d τ2 , rA2 rB1 r12 −1 1 1 1sA (1)2 1sB (2)2 − + d τ1 d τ2 rA2 rB1 r12 The integral J is called the Coulomb integral, while K is called the exchange (or resonance) integral. Notice that since J depends on the the square of the amplitudes of the wavefunctions for both electrons integrated over all space it is essentially an interaction between two charge distributions centered on nuclei A and B, plus the interactions of each of these distributions with the other nucleus. This is analogous to the situation of the H+ molecule where J contained the interaction terms for the single electron. In the integral K since, 2 notice that the labels A and B are switched on the orbitals on either side of the interaction terms. This is like the electrons have exchanged places. You may recognize the similarity of this term to the interference terms that we saw in the double slit experiments. There is no classical analog for this behavior since it comes from the addition of amplitudes; this is a purely quantum mechanical feature. The energies resulting from our choice of this LCAOMO variational wavefunction along with the exact energy of the H2 molecule are shown in Fig. 9. Note that at small distances the energies are close, but the LCAOMO model fails to accurately predict the energy for large internuclear separations. These energies are also shown for a larger range of internuclear separation in Fig. 12.11 of Engel. This failure to accurately predict the energies, however, is not a failure of the variational method, but only a failure of our choice of a variational wavefunction. In the next section we will consider how to improve our wavefunction. An improvement on LCAOMO: conﬁguration interaction for H2 We can improve our results by doing a true variational calculation where we actually have variable parameters in the wavefunction. In the basis of the H+ MOs, we can deﬁne a new variational wavefunction as a 2 ∗ linear combination of all possible conﬁgurations of two electrons in the σg and σu orbitals. The spatial part Chem 120A, Spring 2007, Lecture 38 9 Figure 10: Energies from a conﬁguration interaction treatment of H2 . of the wavefunction is then
∗ ∗ ∗ ∗ ψ = c1 σg (1)σg (2) + c2 σg (1)σu (2) + c3 σu (1)σg (2) + c4 σu (1)σu (2), (28) where c1 , c2 , c3 , and c4 are variational parameters to be determined by minimizing the ground state energy. Since the wavefunction has to satisfy the Pauli exclusion principle, each term in ψ must be multiplied by an appropriate spin wavefunction. When we add the spin wavefunctions, we get six terms with corresponding symmetry labels (term symbols): Σg : σg (1)σg (2)[α (1)β (2) − α (2)β (1)] Σu ∗ ∗ σu (1)σu (2)[α (1)β (2) − α (2)β (1)] ∗ ∗ : [σg (1)σu (2) + σu (1)σg (2)][α (1)β (2) − α (2)β (1)] ∗ ∗ [σg (1)σu (2) − σu (1)σg (2)]α (1)α (2) ∗ ∗ [σg (1)σu (2) − σu (1)σg (2)]β (1)β (2) ∗ ∗ [σg (1)σu (2) − σu (1)σg (2)][α (1)β (2) + α (2)β (1)]. We can use all of these terms to set up a secular determinant to solve for the eigenenergies in terms of the variational parameters. We would then minimize the energies in order to solve for the variational parameters in each eigenstate. We will not do this here, but the resultant energies are shown in Fig. 10. Note that the ground state is a 1 Σ+ state, which is a symmetric bonding state with the electrons paired in a single molecular g orbital. This is consistent with our original idea for the LCAOMO ground state variational wavefunction. The energy for the ground state calculated here is much closer to the true ground state energy because the contributions from the ionic conﬁgurations are 0 in the ground state. The two ionic states are 1 Σ+ and u 3 Σ+ . The state with both electrons in the antibonding orbital is the 1 Σ+ state, which has the highest energy, g u
Chem 120A, Spring 2007, Lecture 38 10 consistent with what we would expect. This treatment is called conﬁguration interaction because we have chosen a fully variational wavefunction that depends on all possible conﬁgurations of two electrons in our chosen basis. This is a powerful method that is widely applicable in quantum chemistry. This treatment is described in greater detail in Chapter 8.5 of Atkins, Molecular Quantum Mechanics, Third Edition. Chem 120A, Spring 2007, Lecture 38 11 ...
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 Spring '07
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 Physical chemistry, Mole, pH

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