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Topic: Zaitsev's rule  (Read 10216 times)

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Offline Corribus

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Re: Zaitsev's rule
« Reply #15 on: September 05, 2013, 09:19:10 AM »
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I would have preferred a simpler argument such as the graph was simply wrong. I'm not sure that is what is being said. It’s the graph that I have trouble with. It is the graph that blaisem used to argue s-orbitals are closer to the nucleus.
The graph is correct. Blaisem's argument is wrong. An s-orbital electron is not closer, over all, to the nucleus than a p-orbital electron, as my table shows. Both the average nucleus-electron distance (<r>) and the most probable value of the nucleus-electron distance in the 2s orbital is larger than those of an electron in the 2p orbital. The 2s orbital does have a small 'bubble' of electron density close to the nucelus, but this is more than compensated by the main part of the orbital, which is farther out than the 2p orbital.

Keep in mind, though, that these mathematical treatments are for orbitals (and states) of a 1-electron atom. When you start putting in more electrons, the penetration effects lower the energy of the 2s orbital with respect to the 2p, because of shielding effects. The orbitals plotted above are no longer relevant for anything other than a hydrogen atom (or other one electron atoms, scaled by a Z-factor).

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I tried to argue in a way that everyone would easily understand. In my manuscript, I suggested a campfire, but the argument is the same. As the coals become cooler, the emissions shift to longer and longer wavelengths. For example, hydrogen has an emission at 7400 nm or 1351 cm-1 (Pfund series). This corresponds with the transition form n = 6 to 5. Now, wouldn't you think if you were seated by a campfire in which electrons are now in the n = 6 level, they wouldn't also fall to n = 1? When the fire (hydrogen) is hot, you see these emissions. When it cools, they no longer occur. Since emission and absorption are related, the only way one might think hydrogen could absorb light at 7400 nm is if electrons were already present at n =5 (25 x Bohr radius). Presumably, if hydrogen atoms were placed in a cold environment, they would continue to emit light at increasing wavelength (lower energy). These will correspond with higher and higher quantum numbers. (This is why I always doubted the Bohr model.)
Excited hydrogen atoms are different than a campfire. The origins of emitted photons are different.  One needs only plot out the intensity as a function of wavelength to see that hydrogen atom emission comes in discrete lines. Light from a campfire is white - it's a continuum of frequencies originating from a source other than well-resolved atomic electronic energy levels.  The impact of temperature on these two processes will be quite different.  Moreover, in simple spectroscopic systems, the primary role of temperature is to determine the statistical distribution of excited atomic/molecular states, not the allowedness or likelihood of photophysical processes themselves.  (This isn't the case in more complex molecular systems, where temperature can impact electronic state coupling and so-forth.)

As I mentioned in my last post, temperature in the case of a hydrogen emission spectrum would affect the average Boltzmann-weighted distribution of electronic states.  A higher temperature would tend to put more hydrogens in higher-n states, which would tend to increase intensity of lower-energy emission lines (or increase intensity of absorption).  This of course ignores any additional chemistry that happens as temperature increases. In short, a dilute sample of hydrogen gas is not a blackbody radiator.

More to the point, I'm not sure what any of this really has to do with the Bohr model.  The lines for hydrogen are correctly predicted by the Bohr model, but the lines of other elements are not.

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As a chemist, I see no reason to relate all emissions to changes in electron orbits.
Well, they're not. But any emission arising from discrete atomic or molecular electronic states certainly is.

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If Bohr's interpretation of the atomic emissions is incorrect, could it result in the confusing graph noted by blaisem? (The mathematics are correct, but the emissions don't correlate with the radius.)

There's nothing confusing about blaisem's graph. It is predicted exactly by quantum mechanics and is consistent with pretty much everything we know about atomic structure.  Bohr's model was an ad hoc modification of classical theory to account for experimental data.  Because it was still based essentially on classical physics, however, it still has a number of critical failings.  A true quantum mechanical picture based on the wave-nature of electrons, however, does not succumb to these failings.  Blaisem's graph has nothing really to do with the Bohr model.

I'm not sure what manuscript you are referring to at the end of your post, but I will excerpt a quote from your quote:

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We need not accept the Bohr atom as the true reason for the energy levels one finds in the emission spectra. On the other hand, there is no reason to reject all connections between the energy levels and atomic structure. Different spectra for each element show an atomic connection exists. The challenge is to connect the emissions to atomic events.

Nobody accepts the Bohr atom as the true reasons for energy levels in emission spectra (in general). Even in the hydrogen atom the Bohr model fails to account for many of the intricacies of high resolution spectra (spin-orbit coupling, quantum electrodynamic effects, etc.). However there really is no challenge to connect emissions to atomic events. Quantum mechanics predicts the emissions of most atoms very well, even if exact solutions are impossible.  This has been verified through nearly a century of experimentation.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline orgopete

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Re: Zaitsev's rule
« Reply #16 on: September 05, 2013, 05:45:27 PM »
The graph is correct. Blaisem's argument is wrong. An s-orbital electron is not closer, over all, to the nucleus than a p-orbital electron, as my table shows.
In that case, I concede.

Quote from: PW
I tried to argue in a way that everyone would easily understand. In my manuscript, I suggested a campfire, but the argument is the same. As the coals become cooler, the emissions shift to longer and longer wavelengths. For example, hydrogen has an emission at 7400 nm or 1351 cm-1 (Pfund series). This corresponds with the transition form n = 6 to 5. Now, wouldn't you think if you were seated by a campfire (oops, I forgot to repeat coals) in which electrons are now in the n = 6 level, they wouldn't also fall to n = 1? When the fire (hydrogen) is hot, you see these emissions. When it cools, they no longer occur. Since emission and absorption are related, the only way one might think hydrogen could absorb light at 7400 nm is if electrons were already present at n =5 (25 x Bohr radius). Presumably, if hydrogen atoms were placed in a cold environment, they would continue to emit light at increasing wavelength (lower energy). These will correspond with higher and higher quantum numbers. (This is why I always doubted the Bohr model.)
Quote from: Corribus
Excited hydrogen atoms are different than a campfire...
It appears I am not explaining this well. I have assumed emissions and absorptions occur at the same wavelengths. Hence if H2 has an emission at 7400 nm, it should also absorb at 7400 nm. This is a pretty low energy level, it is in the infrared region. This is an n=5 to 6 transition. I didn't calculate where the lower energy emissions are predicted from the higher energy transitions, but they should become increasingly smaller in energy. The Balmer emission at 383 nm is for n = 9 to 2. Hence a 9 to 8 transition should also exist, but as you may note, it will be a very low energy transition. I had hoped everyone might note a paradox here. The highest energy transitions occur between the highest and lowest quantum numbers and the lowest occur between the highest quantum numbers. These are very low in energy. Consequently, if we apply the energy is n2 x Bohr radius, we get large radii for low energy transitions.

I had assumed a Boltzmann distribution of emissions, but I did not include a graph. I only have intensity data for the Balmer series of lines and there it was reported the red lines (lowest energy, n=3->2) were the most intense. However, I would think this would be temperature dependent. A higher temperature would shift the intensity of the lines toward the shorter wavelength. Therefore, as hydrogen, or any atom, becomes cooler, the emissions shift toward the longer wavelengths. I presume the Balmer series was discovered first because they are in the visible region. The Lyman, Paschen, Brackett, and Pfund series are emissions are higher or lower in energy and not in the visible region. It appears the Balmer emissions look very much like a black body emission, except it is lines and not continuous. Surely hydrogen can absorb radiation at low energy without first absorbing UV light to promote electrons to high energy levels.

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There's nothing confusing about blaisem's graph... Bohr's model was an ad hoc modification of classical theory ... however, it still has a number of critical failings.  A true quantum mechanical picture based on the wave-nature of electrons, however, does not succumb to these failings. 

Nobody accepts the Bohr atom as the true reasons for energy levels in emission spectra (in general). Even in the hydrogen atom the Bohr model fails to account for many of the intricacies of high resolution spectra (spin-orbit coupling, quantum electrodynamic effects, etc.). However there really is no challenge to connect emissions to atomic events. Quantum mechanics predicts the emissions of most atoms very well, even if exact solutions are impossible.  This has been verified through nearly a century of experimentation.
The point I was trying to make is that any idea that is sort of okay, may not be. The larger the number of rationalizations, the more we should question it. However, I am not  going to be able to convince everyone or anyone that the transition of 7400 nm in the H2 emission does not correspond with electrons shifting from 36 to 25 x the Bohr radius. The 7400 nm is data. It is correct. The mathematical correlation works with an n = 6->5 transition. The Bohr model gives the radius. The Bohr model gives the paradoxical emission order, highest energy with lowest quantum numbers and lowest energy with highest quantum transitions. Could the Bohr model be wrong? (I'm out.)
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