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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Merp on August 09, 2013, 08:23:30 PM
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I'm slightly confused as to one of the justifications for Zaitsev's rule. I understand that a bond angle increase leads to more stability in bulky groups, but I don't understand how substituents donate electron density to the pi bond, or why this causes more stability. The book states "Alkyl groups are electron-donating, and they contribute electron density to the pi bond."
First of all, how do electrons move towards the pi bond? It seems rather counter-intuitive, since there's already a high electron density at the pi bond, one would expect the opposite. Secondly, how does the further concentration of electrons at the pi bond INCREASE stability? I thought one characteristic of stability was the dispersal of charge throughout the entire molecule. Before reading the exact explanation in the book, I inferred that this stabilizing effect was due to some kind of conjugation between the bonding nonhybridized P orbitals and the substituent hybridized orbitals (much like in carbocations), but the book makes no mention of this. If anyone could provide some further insight into this, that'd be great.
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Try reading these and see if that gets you started:
http://en.wikipedia.org/wiki/Zaitsev's_rule#Thermodynamic_considerations
http://en.wikipedia.org/wiki/Hyperconjugation
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Ok, this was informative and it also mentions hyperconjugation as a reason for the Zaitsev rule (which my book failed to mention), but it still separately mentions this: " Alkyl groups are electron donating, and increase the electron density on the pi bond of the alkene" and doesn't go into further detail on this point. If I understand correctly, any substituent will always be electron donating because of the net dipole moment (in a methyl group, for example, the carbon has a slight "advantage" and creates a slight dipole moment for each of the C-H bonds). Still, how does the creation of a partial negative charge at the pi bond and a partial positive charge at the substituents INCREASE stability? I was always taught that the dispersal of charge throughout an entire molecule makes it (generally) more stable. The only time I've seen any mention of partial charges increasing stability is in the presence of a solvent, but alkenes don't even dissolve in polar solvents for the most part.
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Sp2 hydrogens are electronegative because they have greater s character. It follows from sp2 hybridization (compared to the more common sp3 of a saturated molecule ) that sp2 has more s character, which allows for electrons to approach a nucleus more closely (a definition of electronegativity). Based on this, we see that a C-C double bond has greater electronegatovity than normal. This makes any neighboring sp3 bond electron donating.
The E2 reaction as dictated by Zaitsevs rule proceeds through a carbocation intermediate. This cation is stabilized by better electron donation, because the negative charge of electrons helps reduce the localized positive charge. It follows that a hydrogen atom, for example, has fewer electrons available to donate in this regard than, say, a carbon atom with a neutral formal charge (4 valence electrons surrounding it, plus bonds, as opposed to the one valence electron and bond of a hydrogen atom). Therefore, Zaitsebs rule prefers the pathway where the cafbocation intermediate will be most stabilized by electron donation.
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Hmm. Ok. Thanks for the response. I never thought of the whole s character thing, that's interesting. I don't mean to be excessively picky here, but I'd just like to get this 100% straight. About your second paragraph, how does an E2 mechanism have a carbocation intermediate if it's a concerted reaction (maybe you mean E1)? And even if this is the case... there is a fundamental energy difference between alkenes, not just a difference in the rate at which they're formed. This is proved through analysis of hydrogenation enthalpy. Therefore, one alkene must definitively be more stable than another, not just the intermediates or transition states. In an alkene, the SP2 carbon doesn't have a positive charge to begin with. I now understand how there's a movement of negative charge towards the double bond but I don't why this INCREASES stability.
OR are you saying that this effect increases the stability of the transition state (or carbocation in the E1 mechanism) and therefore decreases the activation energy and increases the rate of reaction?
Again, sorry for being picky.
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Hi Merp. Sorry for my unclear post (yes, I meant E1). I typed it on my cell phone and was a little rushed; I actually shouldn't have wasted either of our times putting together a half-assed post like that. But now that I'm here, I'll do my best to answer your question.
So, the more I think about the question, the more I suspect I might be in a little over my head here. Your question fundamentally, as I understand it, is: Why is a double bond stabilized via non-Hydrogen substituents?
I'll give my understanding of why this is, but again I will preface it by saying that I am not the best authority on this. If anyone else who knows better than I reads this, please feel free to correct me. Rather than try to sound smart, I'll just provide a lot of links in my explanation for further reading; hopefully, this will compensate for any vagueness in my explanation.
In my previous post, I attempted to explain why hybridizations with more s character (s > sp > sp2 > sp3) are more electronegative. The concept of s character comes up in a number of topics, but basically a higher s character means that electrons are able to come closer to the nucleus than a p orbital. This all translates to a higher electronegativity. Here (http://faculty.uml.edu/ndeluca/84.334/topics/fig0112c.gif) is a graph comparing the 2s and 2p orbitals for the electron probability vs. radial distance. Here (http://books.google.de/books?id=gY-Sxijk_tMC&pg=PA17&lpg=PA17&dq=why+is+sp2+more+electronegative+than+sp3&source=bl&ots=etyJd8GnX9&sig=-CUJwUQkmkA1IpmgerzCnedSvso&hl=en&sa=X&ei=WSsHUsOVCsWQtAbqmYDwCg&redir_esc=y#v=onepage&q=why%20is%20sp2%20more%20electronegative%20than%20sp3&f=false) is an excerpt from Modern Physical Organic Chemistry (2006) by Eric Anslyn (see pages 17-20) which talks a little bit about this phenomenon.
Alkyl substituents are considered electron donors because they have relatively low electronegativity. In general, any atom in the α position--even electronegative ones--with lone electron pairs are also considered donors. While alkyl substituents possibly alleviate carbocation intermediates in an E1 reaction, I think that Zaitsev's rule is explained a lot more fundamentally by hyperconjugation effects. The π* antibonding orbital in a double bond is empty, and this allows sigma bonds from substituents to "occupy," or delocalize, into these empty orbitals, resulting in a more stable system. There are some good pictures on this from these lecture notes here (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&ved=0CE8QFjAF&url=http%3A%2F%2Fwww.princeton.edu%2Fchemistry%2Fmacmillan%2Fgroup-meetings%2Fhyperconjugation.pdf&ei=NCQHUraUG4Wr4AT414BY&usg=AFQjCNGA-ECGRPw412PI9pf5J-7AKazf-g&sig2=C8GqavJ8KpCiYqSSSz8Ukg&bvm=bv.50500085,d.bGE). For additional reading on hyperconjugation, there's a recent (2011) paper here (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=10&ved=0CG8QFjAJ&url=http%3A%2F%2Fwww.researchgate.net%2Fpublication%2F229584193_Hyperconjugation%2Ffile%2F79e415023e1da52766.pdf&ei=NCQHUraUG4Wr4AT414BY&usg=AFQjCNHgPjbLjqaFEuVl3zw4spazqkUltg&sig2=zhK-W1zEMudd8kPUtIv9jA&bvm=bv.50500085,d.bGE). The second paper in particular has some interesting things to say, for example on page 127 it talks about hyper conjugation actually being a stronger force than "normal" π-conjugation from resonance in butadiyne.
A couple more sources (1 (http://onlinelibrary.wiley.com/doi/10.1002/1521-3773(20021004)41:19%3C3579::AID-ANIE3579%3E3.0.CO;2-S/abstract), 2 (http://onlinelibrary.wiley.com/doi/10.1002/anie.200901923/abstract)) go on to say that the dominant stabilizing interaction (even more than sterics) is hyperconjugation.
At any rate, I am not entirely sure why sp2C-H, which can also delocalize into the antibonding π-orbital, is less stable than sp2C-CH3. Possibly this might have to do with the hydrogen having greater s character (as a pure s-orbital), which could hinder delocalization into the π* antibonding orbital of the double bond. For non-alkyl substituents, I also suspect substituents with lone pairs like halogens would be stabilized by lone pair hyperconjugation with the pi antibonding orbital, and potentially a properly-oriented empty orbital from a geminal substituent.
Well, I hope that helped at least somewhat!
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Ok, wow thank you. I really appreciate the thoroughness, and the stuff you linked was quite interesting (but perhaps a lot of topics are beyond me at the moment). Maybe I'm the one who's in over his/her head, since I haven't really done any quantum/physical classes. You've answer my question and then some, so thank you. After reading the stuff you linked, I agree that hyperconjugation seems to be at the heart of this effect as I suspected. I suppose it's reasonable that my book didn't mention it (it's just a basic organic textbook, so obviously the physical analysis of organic reactions is downplayed).
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Sorry for bumping this up and hijacking the topic, especially as I found the thread awkward to begin with in that I thought zaitsev's rule might have been about the propensity of E2 eliminations to give the more substituted alkene, http://en.wikipedia.org/wiki/Zaitsev's_rule.
None the less, in the ensuing discussion, the amount of s-character was raised. This included the link noted in red. In my manuscript, I have been rather critical of the Bohr atom, so that linked graph was interesting.
In my previous post, I attempted to explain why hybridizations with more s character (s > sp > sp2 > sp3) are more electronegative. The concept of s character comes up in a number of topics, but basically a higher s character means that electrons are able to come closer to the nucleus than a p orbital. This all translates to a higher electronegativity. Here (http://faculty.uml.edu/ndeluca/84.334/topics/fig0112c.gif) is a graph comparing the 2s and 2p orbitals for the electron probability vs. radial distance. Here (http://books.google.de/books?id=gY-Sxijk_tMC&pg=PA17&lpg=PA17&dq=why+is+sp2+more+electronegative+than+sp3&source=bl&ots=etyJd8GnX9&sig=-CUJwUQkmkA1IpmgerzCnedSvso&hl=en&sa=X&ei=WSsHUsOVCsWQtAbqmYDwCg&redir_esc=y#v=onepage&q=why%20is%20sp2%20more%20electronegative%20than%20sp3&f=false) is an excerpt from Modern Physical Organic Chemistry (2006) by Eric Anslyn (see pages 17-20) which talks a little bit about this phenomenon.
Since I am old and not that bright to begin with, let me see if I am understanding this graph correctly. The graph shows the greatest electron density for a 2p orbital is at ~4x Bohr radius and a 2s orbital is at ~5x Bohr radius.
I have thought about this for some time, but I must confess that I am just not smart enough to interpret this correctly. Furthermore, when I look at the graph, I am inclined to interpret the graph to show 2p orbitals are closer to the nucleus. Okay, please educate me.
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Hyperconjugation is caused when p-orbitals from carbon atoms single bonded to each other overlap to increase the stability of the bond. That is the reason why tertiary carbocations are more stable than secondary, primary or methyl ones.
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Hyperconjugation is caused when p-orbitals from carbon atoms single bonded to each other overlap to increase the stability of the bond. That is the reason why tertiary carbocations are more stable than secondary, primary or methyl ones.
Isn't the hyperconjugation you are speaking of σ(C-C) --> P(carbocation) type interaction? porbitals overlapping is a π bond....
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Sorry for bumping this up and hijacking the topic, especially as I found the thread awkward to begin with in that I thought zaitsev's rule might have been about the propensity of E2 eliminations to give the more substituted alkene, http://en.wikipedia.org/wiki/Zaitsev's_rule.
None the less, in the ensuing discussion, the amount of s-character was raised. This included the link noted in red. In my manuscript, I have been rather critical of the Bohr atom, so that linked graph was interesting.
In my previous post, I attempted to explain why hybridizations with more s character (s > sp > sp2 > sp3) are more electronegative. The concept of s character comes up in a number of topics, but basically a higher s character means that electrons are able to come closer to the nucleus than a p orbital. This all translates to a higher electronegativity.
Here is a graph comparing the 2s and 2p orbitals for the electron probability vs. radial distance.
By the responses, it is clear that I have high jacked this topic. Since there was no response, I asked someone familiar with this graph.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Ffaculty.uml.edu%2Fndeluca%2F84.334%2Ftopics%2Ffig0112c.gif&hash=28026f764b44cd534e7a57cb260fb678667e142b)
Thanks for contacting me. Your confusion is understandable, but there is no error in the graph.First of all, the Bohr approach to atomic theory was thrown out nearly 100 years ago. Keep in mind that it only worked for 1 electron atoms or ions. The Bohr radius, which is for the hydrogen atom, has been confirmed (or supported) by using quantum mechanics and graphs of radial probability functions for the 1s orbital.
Now for your question. The key to the answer is in the profile of the 2s orbital versus the 2p orbital. Yes, on average, the electrons in the 2p orbital are closer to the nucleus than those in the 2 orbital. The important feature is the "blip" of electron density near the nucleus for the 2s (blue) curve. As the electron density for the p orbital approaches zero, there is a significant probability of finding an electron in the 2s orbital near the nucleus. This is known as "penetrating ability." Since the 2s orbital shows greater density near the nucleus, the energy of the orbital is lower than that of a 2p.
Lastly, don't confuse most probable nuclear distance of an electron in an orbital with bond length. Bonds involve two nuclei essentially competing for electrons to help shield or minimize nuclear-nuclear repulsion while optimizing electron-nuclear attractions, and minimizing electron-electron repulsions. Very different than a wave function for an atomic orbital and a single nucleus.
I would have missed this had Blaisem not linked it (and I actually read it and clicked the link, doesn't always happen). I find this reply hypocritical as it asks both to ignore the data, but still presents it. If it were to be ignored, don't rationalize it. For example, we are told that despite what might look like a probability of electron distribution at a much greater distance for a 2s electron, it really is at the much shorter distance due to "penetrating ability".
I believe this is similar to fluoride having the "greatest ionic content" yet HF is still a weaker acid than HBr. You can find many rationalizations because HBr is the stronger acid. The ionization argument is interesting as I would consider acidity as perhaps the best studied and best understood of any measure of ionization. In the case of fluoride, this all goes back to Pauling and electronegativity. Electronegativity is based upon the premise that bond energy is the sum of ionic and covalent content. It is this premise that is in error and leads to many ensuing paradoxes, including many using hyperconjugation to explain why carbon is an electron donor. (Hyperconjugation certainly takes place and is a consequence of carbon holding its electrons relatively weakly, just as N, O, & F hold them stronger, and BH4(-) holds them weaker still.) I find dividing bond character paradoxical. F2 (18p,18e) is covalent and NaF (20p,20e) is ionic. I find it difficult to imagine CH3F (18p,18e) as containing partial ionic and covalent character. I believe I can explain the data Pauling was attempting to explain without resorting to his bond energy premise, but I cannot do so here.
This leaves us with Bohr. Should it be thrown as as quoted above? Where does that leave us? Does the confusion reveal a weak theory? As an organic chemist, I had always been confused by the Bohr energy levels. I am familiar with high energy breaking bonds, infrared stretching bonds, and microwaves causing molecular motion. It escapes me these low energy transitions are related to high energy levels (principle quantum numbers) and large electron radii. Just sayin'…
I don't know if I'm going to pick up several more flags for my skepticism, but I must thank blaisem for bringing this to my attention.
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@orgopete
I haven't followed the whole thread, but your most recent post drew my attention. Can you explain more directly what your point of confusion is with respect to the Bohr model. Maybe I can help.
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@orgopete
I haven't followed the whole thread, but your most recent post drew my attention. Can you explain more directly what your point of confusion is with respect to the Bohr model. Maybe I can help.
Well, perhaps it's philosophic. I looked at the graph and thought about the argument that s orbitals are closer to the nucleus. The graph looked to me as though the p orbital were closer, or at least the electrons were closer.
Read the comment I received, the Bohr atom was thrown out 100 years ago, but let me explain…
I had been trying to place the Bohr atom in context. If you compare atomic emissions, longer wavelength emissions correspond with larger quantum number transitions. I understand that to indicate that as an element cools, it cools to higher and higher quantum numbers. Seems wrong to me.
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Hi Pete,
I apologize for the delay in replying to your post. I have a couple of thoughts for you.
First, I take issue with a number of points in the email/message you quoted, in which it was stated that Bohr's model was "thrown out 100 years ago". I wouldn't say Bohr's model was thrown out. Such a statement shows a disregard for the historical context of Bohr's work in this area. Bohr was a smart guy and clearly understood that his model didn't solve every problem with the classical approach to atomic structure. What Bohr was after was the solution of one particular problem: atomic spectral lines. Rydberg was the first to identify a mathematical pattern to the various sets of spectral lines (Balmer, Paschen, etc.) that had been known for some time, but although Rydberg found this pattern, nobody could explain its origin. Certainly not with a classical model of the hydrogen atom. Bohr found a "solution" by restricting electrons to certain orbital distances from the nucleus. Using classical Coulombic energy differences and rigid circular orbits, he was able to show that the resulting energy differences matched the Rydberg formula. This aspect of quantization of electron position was an important step toward full-blown quantum mechanics. But Bohr's model was still fundamentally based on classical physics - that electrons behave like waves was not incorporated into the model at all - and thus it violates many of the principles of QM as we know them today. As such it fails pretty dramatically to give a true picture of atomic structure and - more importantly - doesn't even do a good job of predicting spectral lines in atoms other than hydrogen.
The email/message you quoted also appears to be incorrect in a few places - although maybe I'm missing some of the context. For example: "Since the 2s orbital shows greater density near the nucleus, the energy of the orbital is lower than that of a 2p." This is not true. The energy of hydrogen atomic orbitals depend only on n, the principle quantum number. The 2S and 2P orbitals have exactly the same energy in the hydrogen atom. It is only in a multielectron system where the 2S orbtial becomes lower in energy than the 2P orbital, the so-called shielding effect. (If you want to be really correct, in hydrogen the 2P is actually slightly lower in energy than the 2S due to the Lamb effect, but this has nothing to do with penetration and the effect is quite small - <5 micro-eV - so for all purposes the hydrogen 2S and 2P have exactly the same energy, and this is what is predicted by most elementary QM models.)
Second, the Bohr model correctly predicts the spectral lines of hydrogen, and also correctly predicts the most probable distance of the electron from the nucleus... but, for transitions originating from the lowest energy states, anyway, only in orbitals involved in the allowed spectral transitions of the atom. This latter point is a bit historically fortuitous, admittedly, but it is important to point this out to fully understand what's going on. Consider: the Bohr model predicts electron-nuclear distances in hydrogen for the n = 1, 2, and 3 states to be 1, 4 and 9 in units of the Bohr radius. This n^2 dependence is mirrored in the energies of these states as well. However, as you've pointed out with your plot above, an electron in the 2S differs in its most probable distance (obtained by finding the roots of the derivative of the radial probability distribution) from an electron in the 2P. In fact, if we calculate these, we get the following for the maxima of the n = 1, 2 and 3 levels/sublevels*:
| n | Orbital | Rmax | | | <r> |
| 1 | S | 1.0 | | | 1.5 |
| 2 | S | 5.24 | 0.76 | | 6.0 |
| | P | 4.0 | | | 5.0 |
| 3 | S | 13.07 | 4.19 | 0.74 | 13.5 |
| | P | 12.0 | 3.0 | | 12.5 |
| | D | 9.0 | | | 10.5 |
For completeness I also calculated the mean distance or expectation value, <r>. All distances are in units of the Bohr radius.
What do we notice here? As already stated above, the energy is the same for the 2S and 2P levels, and for the 3S, 3P and 3D levels. The Bohr model correctly predicts these. The Bohr model also correct predicts the most probable distances for the 1S, 2P and 3D levels. But not the 2S, 3S and 3P levels. As you may have correctly guessed, what the 1S, 2P and 3D level probability distributions have in common is that they have no nodes - only a single maximum value where the Bohr model predicts it should be (1, 4, and 9 Bohr radii). Not coincidentally, these are also some of the primarily levels between which spectral transitions are allowed.
Here's what I mean. Take only the first two levels for simplicity. 1S, 2S and 2P. 1S and 2P have electron-nuclear most probable distances predicted perfectly by the Bohr model and 2S does not. The Bohr model was formulated primarily to match hydrogen spectral transitions. The 1S --> 2P transition is allowed. The 1S --> 2S transition is not allowed. Granted, the latter would have the same energy as the former, so it's seemingly a trivial point, but because the Bohr model doesn't incorporate the wave properties of electrons at all, it should be no surprising that the model's predictions of distance only match those orbitals which have no nodes. Same can be said for the n = 2 to n = 3 levels. The 2P --> 3D transition is allowed and both orbitals have no nodes, so this transition is correctly predicted in both energy and most probable electron distance by the Bohr model. 2P --> 3P transition is not allowed. The 2P --> 3S transition is allowed, but its energy is the same as the 2P --> 3D transition, so Bohr's correct prediction of this energy gap is coincidental, and it does not correctly predict the most probable distance(s) of electrons in the 3S orbital. For each primary shell (n value), there is one orbital that has no nodes, and so there is always an orbital in each shell that conforms to both the energy and most probable distance predicted by the Bohr model.
I hope that maybe clarifies some of the strengths and weaknesses of the Bohr model. It certainly isn't right, but it does offer a correct "scaffold" for the energy level structure of the hydrogen atom, it is useful for students as an introductory model and it also has profound historical significance. Therefore to say it has been "thrown out" seems like something of an unfair statement to my mind.
Third, you do seem to have some area of confusion about atomic and molecular structure. Hybridization is a bonding concept related to molecular orbital theory and as such it is no replacement for atomic structure models, even an obsolete one like the Bohr model. Hybridization as a molecular model has itself come into question in many circumstances and atomic orbitals certainly don't hybridize without some energetic payoff in the form of a molecular bond. So it's not really relevant here at all. Atomic spectral lines are completely and quantitatively explained - including fine and hyperfine structure - by a full quantum mechanical treatment (in the case of hydrogen, without the need of approximations). At this point improvements will likely only come in the form of better approximations, but in the case of atomic spectra the models are so good that instrumental measuring limitations are probably a more likely source of line position ambiguity (I could be wrong here). Certainly models are good enough to be able to specify the composition of stars light years away, so it's hard to see what the need is of further improvement.
Finally,
I had been trying to place the Bohr atom in context. If you compare atomic emissions, longer wavelength emissions correspond with larger quantum number transitions. I understand that to indicate that as an element cools, it cools to higher and higher quantum numbers. Seems wrong to me.
Here you seem to be confusing black body radiation with an emission spectrum of a dilute, pure gas. A container of hydrogen gas would obey statistical mechanics: the strength of an emission line would be proportional to the fraction of hydrogen in the origin state at the time of the experiment, which would be dependent on temperature. At higher temperatures, higher energy states tend to be populated, so emission will occur from these states to lower energy states, and these transitions, as you've noted, would tend to have longer wavelength emission. As the temperature decreases, only lower lying states would be populated, and emission from these states would have shorter wavelengths. Therefore as the temperature cools, you would tend to see higher frequency (bluer) emission from the gas. This is different from a black body radiator, where there is a continuum of allowed states. When you cool a black body, the mean radiation frequency red-shifts to lower frequencies.
*I couldn't find these anywhere so I had to calculate them myself. It's been a while since I did this. Errors may therefore be present.
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...First, I take issue with a number of points in the email/message you quoted, in which it was stated that Bohr's model was "thrown out 100 years ago". I wouldn't say Bohr's model was thrown out. ... But Bohr's model was still fundamentally based on classical physics - that electrons behave like waves was not incorporated into the model at all - and thus it violates many of the principles of QM as we know them today. As such it fails pretty dramatically to give a true picture of atomic structure and - more importantly - doesn't even do a good job of predicting spectral lines in atoms other than hydrogen.
This stands for itself.
Here's what I mean. Take only the first two levels for simplicity. 1S, 2S and 2P. 1S and 2P have electron-nuclear most probable distances predicted perfectly by the Bohr model and 2S does not. The Bohr model was formulated primarily to match hydrogen spectral transitions. The 1S --> 2P transition is allowed. The 1S --> 2S transition is not allowed. Granted, the latter would have the same energy as the former, so it's seemingly a trivial point, but because the Bohr model doesn't incorporate the wave properties of electrons at all, it should be no surprising that the model's predictions of distance only match those orbitals which have no nodes. ...
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.
Third, you do seem to have some area of confusion about atomic and molecular structure. Hybridization is a bonding concept related to molecular orbital theory and as such it is no replacement for atomic structure models, even an obsolete(really?) one like the Bohr model. Hybridization as a molecular model has itself come into question in many circumstances and atomic orbitals certainly don't hybridize without some energetic payoff in the form of a molecular bond. So it's not really relevant here at all. Atomic spectral lines are completely and quantitatively explained - including fine and hyperfine structure - by a full quantum mechanical treatment.
You are correct. I find it difficult to distinguish atomic spectra independently from molecular spectra.
I had been trying to place the Bohr atom in context. If you compare atomic emissions, longer wavelength emissions correspond with larger quantum number transitions. I understand that to indicate that as an element cools, it cools to higher and higher quantum numbers. Seems wrong to me.
Here you seem to be confusing black body radiation with an emission spectrum of a dilute, pure gas. A container of hydrogen gas would obey statistical mechanics: the strength of an emission line would be proportional to the fraction of hydrogen in the origin state at the time of the experiment, which would be dependent on temperature. At higher temperatures, higher energy states tend to be populated, so emission will occur from these states to lower energy states, and these transitions, as you've noted, would tend to have longer wavelength emission. As the temperature decreases, only lower lying states would be populated, and emission from these states would have shorter wavelengths. Therefore as the temperature cools, you would tend to see higher frequency (bluer) emission from the gas. This is different from a black body radiator, where there is a continuum of allowed states. When you cool a black body, the mean radiation frequency red-shifts to lower frequencies.
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.)
As a chemist, I see no reason to relate all emissions to changes in electron orbits. I am fine that UV effect electrons and bonding, IR, bond stretching and bending, and microwave, molecular motion. I understand that any explanation for the Balmer/Rydberg formulae is enticing. I don't know that I raise enticing to proof. I don't know what is correct, but relating the emission spectra to centripetal and Coulombic forces seems very misleading. 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.)
I appreciate everyone's efforts to help me from falling into a very deep hole.
Let me end with a quote form my manuscript.
If we reflect on the Bohr atom, we should also reflect on a philosophical note about it. The Bohr atom is not a single isolated idea about atomic structure. Even if we should find some concepts are not completely correct, we should be cautious of the impact on related ideas. 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. The calculations are correct. What the calculations mean in terms of a model for atomic structure may be different. For example, we could agree with Gilbert Lewis in thinking n is simply a quantum value.
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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).
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.
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.
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:
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.
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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.
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.)
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.
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.)