Chem120a : Exam 3 (Chem Bio) Solutions

Transcription

1 Chem10a : Exam 3 (Chem Bio) Solutions November 7, 006 Problem 1 This problem will basically involve us doing two Hückel calculations: one for the linear geometry, and one for the triangular geometry. We ll start with the linear geometry. In this case, atom 1 is next to atom, atom is next to atoms 1 and 3, and atom 3 is next to atom. That gives us the following Hückel secular determinant: α E = α E 0 α E = Now we make the standard definition x, and expand the matrix. x x x = ( 3 x 3 x ) = 3 ( x ) (x) Since this whole thing has to be equal to 0, we know that either x = 0 or x = 0. Respectively, these cases give us the energies x = α E = 0 = E = α x = α E = ± = E = α ± Remembering that < 0, this gives us the energy ordering α + < α < α Now let s treat the triangular state. In this state, each hydrogen atom is next to the other two hydrogens, so the secular determinant becomes α E = α E α E =

2 Again, we make the variable transformation x and expand our determinant: x x x = ( 3 x 3 + 3x ) = 3 (x + ) ( x x + 1 ) = 3 (x + ) (x 1) Hopefully, you remember how to factor polynomials. If not, those of you with a graphing calculator should be able to quickly see that you have a single root at x = and a double root at x = 1. If you don t have a graphing calculator, hopefully you remember polynomial division, and once you identify one of the roots (by inspection) you can divide out that root and factor the quadratic that s left for you. Even if you forget the fundamental theorem of algebra, and only find roots (say, by using a solver on your calculator), you ll get the right energies (but you ll miss the multiplicity of the higher energy). Anyway, however you manage to get that solution, you then find the following energies: x = α E = 1 = E = α (with twofold degeneracy) x = α E = = E = α + Again, we can put these into the order α + < α = α So now we compare the energy of these two states. Since we only have two electrons in H + 3, we only need to compare the lowest energy level in each conformation (since it will be filled by the two electrons, and all the other levels will be empty). Remembering that is negative, we see that α + < α +, which means that the triangular state is more stable. Finally, we sketch the wavefunctions. To judge the relative stability, we only need to look at the ground states, although the excited states are also sketched for linear conformation is figure 1 and for the triangular conformation in figure. Looking at the ground state wavefunctions, we can see that there is more overlap in the triangular state than in the linear state, which would lead us to intuitively expect that its energy should be lower (and that the triangular state should be more stable).

3 Figure 1: Qualitative molecular wavefunctions for the linear case, increasing energy from left to right. The overlaid curves show the particle-in-a-box states associated with each level. Figure : Qualitative wavefunction for the triangular case. The state with one node has a twofold degeneracy (the node could equivalently go through one of the other atomic orbitals). Problem We all know that intuitively that He is not a very stable molecule. Why? well from a LCAO-MO picture we would think of the 4 electrons filling molecular orbitals that looked something like: 1σ u 1σ g 3

4 In this picture there is no energetic advantage for the electrons in He atoms to populate molecular orbitals. In the bond order picture we would calculate a bond order of: B.O. = 1 ( ) = 0 So He doesn t form a very good bond. If the LCAO picture is extended to include linear combinations of the s orbitals of He, we would get a He ground state looking like: σ u s s σ g 1σ u 1σ g Now, by using a dipole allowed transition to promote one of the 1σ u electrons, we could end up with two different possibilities: σ u σ u s s s s σ g σ g 1σ u 1σ u 1σ g 1σ g 4

5 Where an electron was promoted from the ground state to an excited state observing the selection rule S = 0. Either way we can write the electronic configuration of these states as (1σ g ) (1σ u) 1 (σ g ) 1 With a term symbol for the total electronic state of 1 Σ u. Because the selection rule S had to be observed in making this state from the ground state, the total spin of the molecule is S = 0 with the excited electron being S = ± 1. Now, because we consider the σ g MO to be a bonding orbital, we calculate the Bond Order of this state as: B.O. = 1 (3 1) = 1 and the symmetry of the electronic state is given by: Problem 3 g g u g = u (a) Whenever we talk about spectroscopy, we re looking at a change in energy between initial and final states (that change in energy corresponds to the energy of the photon, which is what we actually measure). In this case, we re looking at absorption, so the energy of the final state must be greater than that of the initial state. We can separate the energy into rotational and vibrational components: E = E f E i = (E f,rot + E f,vib ) (E i,rot + E i,vib ) = E vib + E rot Now we ll calculate the changes in each type of energy separately. We start with the vibrational energy shift, which is the same as it is for our normal selection rules. Since this is absorption spectroscopy, and the spacing between vibrational levels is much larger than the spacing for rotational levels, we can assume that only the v = +1 rule is being used in this process. ( E vib = ω v f + 1 ) ( ω v i + 1 ) (( = ω v ) ( v + 1 )) = ω For the rotational energies, we need to apply the new rotational selection rules: J = ±. E rot = BJ f (J f + 1) BJ i (J i + 1) = B ((J ± ) (J ± + 1) J (J + 1)) 5

6 If we separate the plus and minus cases, we get the following two solutions: E + rot = B (4J + 6) E rot = B ( 4J + ) If we really want to combine these into one solution, we can do that by writing it as E rot = B (±4J + 4 ± ) We should note that when we take the minus case, our initial J can not be 0 or 1 (because we can t be going into a J = 1 state, since J is always a nonnegative integer). Putting the whole thing together for the rotational-vibration spectrum, we find that the transition energies are given by E + = ω + B (4J + 6) E = ω + B ( 4J + ) or, combined together: E = ω + B (±4J + 4 ± ) (b) In this part, we are to sketch the spectrum we would get out of this transition. If you directly plot the results of the above, that of course gives you the right result, as shown in figure 3. However, there are a few specific points we were expecting to see. First, the spacing between peaks has changed. When J = ±1, the spacing between peaks is (to the accuracy of the rigid-rotor harmonic oscillator approximation) B. Now the spacing is 4J, as we can easily show: E J+1,J = ω + B (±4J + 4 ± ) ( ω + B (±4(J + 1) + 4 ± )) = ±4B Also, we note that the two branches are not centered about ω, which is clear both from the equations and from figure 3. Problem 4 This problem aims to underline the relationship between the shape of the electronic potential curve for the nuclei and the corresponding vibrational energy levels by having you do a little bit of number crunching. The formula for the vibrational levels with a correction for anharmonicity included is ( E v = ω e v + 1 ) ( ω e x e v + 1 ) The first two parts (a and b) of the problem ask that we find the energies associated with the 0 1 and 1 vibrational transitions. These energies will be ( E v v = ω e (v v) ω e x e v + 1 ) ( + ω e x e v + 1 ) 6

7 Figure 3: Rotational-vibrational spectrum for J = ± where v is the final level and v is the initial level. We are told that ω e = 3000 cm 1 and ω e x e = 50 cm 1, so we may plug in those values and our desired quantum numbers to find E 0 1 = ω e 9 4 ω ex e ω ex e = ω e ω e x e = 900 cm 1 E 1 = ω e 5 4 ω ex e ω ex e = ω e 4 ω e x e = 800 cm 1 We are asked in part c to give a qualitative reason for why the transition energies are different. This essentially comes down to the fact that the electronic potential curve for a diatomic molecule is less like a harmonic oscillator than a Morse potential, as shown in Figure 4. As we discussed earlier in the course (see your notes for September 8th), the equal spacing of energy levels for the harmonic oscillator has to do with how much it widens as it grows. The particle in the box doesn t widen at all and grows infinitely quickly, leading to a set of energy levels whose spacings diverge. The Coulomb potential, on the other hand, levels off asymptotically such that it becomes infinitely wide, leading to a set of energy levels whose spacings vanish. The harmonic 7

8 Figure 4: Comparison between a harmonic oscillator and a Morse potential with the same equilibrium bond length and force constant. The first four energy levels are shown for each. oscillator, then, widens just enough as energy grows so that its energy levels will be spaced equally. The Morse potential is like the Coulomb potential, insofar as it widens out asymptotically, so we would expect its energy level spacings to decrease. Another way of understanding this decrease when anharmonicity is included is to recall that the harmonic oscillator / rigid rotor model is based on a Taylor expansion in the potential, V (R) V (R e ) + V (R e ) (R R e ) + V (R e ) (R R e ) which we choose to truncate to second order. This approximation suffers at larger values of R, as can be seen in the picture. This suggests that we treat the third-order term as a perturbation to our HORR vibrational energies, thereby introducing anharmonicity something that s not harmonic, but cubic. Our energy correction will be given by E (1) v = V (R e ) φ v (R R e ) 3 φ v 6 where φ v (R) is the vibrational wavefunction for the v th energy level. V (R e ) is always negative for a Morse potential, and (R R e ) 3 will only be negative for a very small portion of R s range of [0, ). We expect, then, that E v (1) will be negative as well. Furthermore, its magnitude should grow as the vibrational wavefunctions are nonzero over larger ranges in R. We therefore expect vibrational energy spacings to decrease. 8

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