5.7 Marcus Theory II

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@theyneedacrane Күн бұрын

I'm struggling with the how reorganization energy is defined. In the first video, it's being defined in the Libby case as the change in energy between the equilibrium position of D+A and the intersection of the potential energy surface of D+/A-. In this video, we're now seeing a case where the potential energy surface of D+/A- can intersect directly with the equilibrium position (barrierless ET). Based on the previous definition, I would define the reorganization energy as zero. But at 6:40 in this video, it seems that reorganization energy is being defined in a different way than before, such that it is now the same as deltaG. Can you clarify this please?

@ksmin0514 2 жыл бұрын

Thank you so much. My old confusion is gone :)

@mariem448 3 жыл бұрын

thank you for this video, it's very intructive

@mattkunq 2 жыл бұрын

Is there a more intuitive explanation of why the inverted happens. Like what is happening at the molecular level?

@Tomas.Malina 2 жыл бұрын

The quantum mechanical reason for the slowing-down of the transfer in the inverted region is due to overlap between the two states. The farther apart you go, even though the reaction releases more energy, it isn't likely to run because the wavefunctions of the two states don't overlap (or, the overlap between them is smaller). In more detail: the rate of the energy transfer is proportional to two terms, the strength of the interaction between D and A (which is exponentially decreasing with distance between D and A), and the Franck-Condon factor, basically the overlap between the initial and final state electronic wavefunctions. A graphical representation of this is shown for example in Robert E. Blankenship's book Molecular Mechanisms of Photosynthesis (2002) in Chapter 6 on page 105 (Figure 6.4), if you are able to get your hands on that book (there is a newer edition from 2021 which may have a slightly different page numbering).

@mattkunq 2 жыл бұрын

@@Tomas.Malina Thanks for that, but then why don't they overlap. Why do certain reactions have inverted regions and other don't?

@Tomas.Malina 2 жыл бұрын

@@mattkunq The overlap depends on the positions of the atoms in the molecules and their vibrational motion. In short, the electron has a higher chance to be found at the "edges" of the eletronic potential curve (for a certain energy above the minimum - which is where the two potential curves intersect - the electron can be anywhere inside the curve). The probability of finding the electron inside this area is not equal at all places though, it is highest at the edges (explanation below, it is to do with the vibratinal motion if the nucleus). This is where you'll likely find the electron and it is also likely that this is where the electron will "want to be transferred to" (again, this "want" is due to the overlap of the wavefnctions). In more detail: The potential surfaces in this picture are representing the electronic energy, however, every molecule (of 2 atoms and more) has also internal degrees of freedom for various vibrational motions. If you calculate the electron densities for a pure electronic state (without any vibrational energy of the nuclei), the highest probability of finding an electron will be at a location where the energy of the electronic surface is lowest - this is where the nucleus is, the electron is somewhere close. However, when the vibrational motion of the nuclei is excited, the nuclei start to move around (this is a simplification, they vibrate even when the molecule is in its ground vibrational state, but the probability of finding an electron somewhere along the vibration coordinate is highest at the equilibrium). This means the electrons will move with the nuclei (in a classical analogy, an electron orbits around the nucleus at ~1000x the speed of the vibrational motion of the nucleus, so it will tightly follow the nucleus - at least in the adiabatic approximation, which is considered for most quantum chemistry calculations). The first order approximation for the nuclear motion is a harmonic oscillator - a weight on a spring moving back and forth (if we consider the simplest stretching vibration). If you take a time average of all the positions of the oscillating weight (nucleus) during its periodic motion, it will actually spend the least time at the lowest potential energy point (where the spring is not stretched/compressed, the weight moves at the highest velocity, therefore if you look at the oscillating weight at a random point in time, you will most likely see it near one of the turning points where the spring is most stretched/compressed, where the weight moves the slowest, therefore stays the longest). If the vibrational motion is excited to a higher vibrational state (higher than the first excited state), there will be some local maxima along the motion, depending on the energy of the vibrational state - this is unlike the analogy of a harmonic oscillator, which can work only for the first excited state. Nevertheless, even at higher vibrational excited states, there is a high chance that the electron will be found near the turning points. You can imagine that the turning points are at the location of the electronic energy surface - if you draw a horizontal line across it, the electron can move (as if on a spring, because the nucleus the electron belongs to moves like an oscillator) inside the potential curve - and due to the harmonic motion, it will have the highest probability to be found at the edges. If you google an image for "vibrational wavefunction" or "Franck-Condon principle", you'll see exactly what I mean. This probability of finding the electron at the sides is relevant also for the acceptor's state ("the initial and final state have to overlap") - the electron has the highest chance to move into a location at the edge of the potential curve. If the final state of the acceptor is the lowest state (usually it is, at room temperature, excited vibrational motions of nuclei are rare and up till now we have been talking about vibrational motion of the donor, not the acceptor - the acceptor is most likely "at rest"), it will have to want to move to the side of the potential curve as well. However, the position of this high probability for the acceptor state is not vertically accessible from the donor, since the electron is at the donor's edge of potential surface (and an excited vibrational state, so even farther from the acceptor). Now, the reason why we consider the donor is in a vibrationally excited state while the acceptor isn't is simply because if the donor is not vibrationally excited, it cannot even think about transferring the electron, because it cannot supply the energy to overcome the barrier (to move the electron at the position where the potential surfaces intersect). The (few) donors that do have the energy to do so will most likely encounter an acceptor that isn't excited (since most molecules aren't at 300 K). You may ask: why should the electron need to access the acceptor's state "vertically", why won't it just slide along the potential curve into the minimum? That is because of the speed of the electron. As I mentioned, the movement of the nucleus is about 1000x slower than the movement of the electron, therefore should the electron change states (from donor to acceptor), the nuclei of the donor and of the acceptor haven't moved at all during the time it took to make this change of state. The running along x axis in this graph (the horiontal line you imagined two paragraphs above) represents the movement of the nuclei (which oscillate at approximately 10^12 Hz), not electrons (which move around the nucleus at more like 10^15 Hz and closely follow it wherever the nucleus moves). Therefore, the electron needs to look for places to jump along the vertical axis - this is called the Franck-Condon principle. Edit: to address your second question: I don't know. If there indeed are reactions that the Marcus theory doesn't apply to, I'd guess that it has to do with the breaking of the adiabatic approximation - the electrons are unable to follow the nucleus closely because the speed of the nucleus is comparable with the speed of the electrons (in some cases this indeed happens), or the potential cannot be approximated by a parabolla (and ergo the vibration can't be approximated by a weight on a spring, this happens as well). But that is just a guess. Most of the things that "don't work" in quantum mechanics are due to some approximations you make being invalid :) ..Or, come to think of it, it could do with symmetries of the molecule. If the molecule is somehow symmetrical, that means that it has degenerate energy states (multiple configurations have the same energy). In such a case, you could actually reach another state without any transition (the electron doesn't need to "hop" into another state, but slowly "evolve" into it). If that was the case, the shape of both the potential and the electronic wavefunction would change along the x axis (the curves in the video would have a different shape for different position of the nucleus), therefore it would be possible for the electron to reach the acceptor not by jumping to it, but by a continuous change in superposition of the two states - at one moment, it would 100 % belong to D, at a later moment, it would 95 % belong to D and 5 % to A, etc. all the way until it reached a pure state of 100 % belonging to A. Then the speed won't depend on the overlap between the two states, because they are in superposition - they are a part of the same wavefunction.