I know that electrons can move from say 2s orbital to an unoccupied 2p orbital, as in Carbon atom which can form 4 bonds this way. But I want to know is it possible for an electron say in orbital 2p to switch its place with an electron in 2s? I know only 2 electrons are allowed in an orbital, but is it possible for one electron to oust another and take its place so the ousted electron would have to take the other one's place? Is this theoretically possible?
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2$\begingroup$ Your question digs very deep into the quantum mechanics, namely exchange interaction ( en.wikipedia.org/wiki/Exchange_interaction) of fermions, which are indistinguishable particles ( en.wikipedia.org/wiki/Identical_particles ) $\endgroup$– ssavecAug 24, 2014 at 19:23
3 Answers
The short answer to this is: according to our current understanding of quantum mechanics, we can't tell.
And now for the long answer:
First off, if you haven't already seen Richard Feynman's 60-second primer to the scientific method, you should definitely watch it now. It's short. I promise I won't go anywhere while you're watching it.
...
Alright. So the way science works is to test ideas by experiment. This means that in order to distinguish between which of two ideas is right, we'll need to come up with some difference in their behavior. This can sometimes be problematic in science because the difference in behavior between two models can be incredibly small. To take just one example, it wasn't until the 20th century that an apparent error in the orbit of Mercury was confirmed, which was eventually explained by relativity. Before that, if you had given someone the theory of relativity and the Newtonian theory of gravity and asked them to tell you which one was right, they would have shrugged their shoulders. There simply wasn't enough high-quality evidence to decide between one or the other.
On the other hand, what if there's a nightmare situation in which we simply can't tell the difference no matter how good our instruments are? Imagine that I give you a small cube which lights up when you shake it, but I've used magic so that no matter what experiments you try to use to figure out what's inside of the cube, you can't get any information. You can try to use x-rays to image the inside, or weigh it, or spin it around to see if the insides are unevenly distributed, or look at it with an IR camera...but every single time, the magic renders your results useless.
Now suppose I asked you how the box worked. Without being able to get any info, you're stuck! No matter how clever you are in making your guess or how many guesses you come up with, you can't really say anything about any of them, because you're being magically prevented from getting information about what's in the box.
This is a very important idea, so I want to make sure you really understand it. Without some sort of actual behavioral difference that can be looked at, science has no way of distinguishing between alternate ideas.1
Now, quantum mechanics is sometimes a lot like our magic box (to pretty much anyone who tries to study it, it is a magic box). Our current understanding of QM says that we have this wavefunction that tells us what the system is doing. We can't actually look at the wavefunction. The best we can do is to use some operators on it, which will give us the values of certain properties (don't worry if this all sounds very mysterious, it takes years of study to understand some of this stuff).
So how can we see if the wavefunction matches what we think it is? Well, we can look at what we should get from the operators, and what we actually do get from measuring it. For instance, by applying something called the Hamiltonian operator to the wavefunction, we get the total energy of the system. We can then measure the total energy of the system and see if theory and experiment agree. We can do that a whole bunch of times, measuring the momentum, and kinetic energy, and the average position of the system. If we do it enough times, we can say "eh, it's probably correct" and then move on to the next problem.
Now, to get back to your original question: can electrons ever switch places? Well, our current understanding of quantum mechanics is that, when electrons switch places, the wavefunction changes sign.2 "Aha!" I hear you say. "So we can watch the wavefunction and see if it changes sign, and if we ever see it flip, we can say that the electrons switched places!"
Well...not really. Because we can't look at the wavefunction directly, we can only measure things through the operators. And here's where the nasty trick comes into play: for all operators that we can actually observe, there won't be any change if the wavefunction flips its sign.
We're stuck! Just like our magic box, we can't get any meaningful information about what's going on inside, and so we can't really tell if the electrons switched places. Even if they did, it would look exactly the same to us (no matter how good our instruments are, we cannot see a difference, because there literally isn't one). It's an undecidable problem!
Now, if you're anything like me, this is a really crummy and disappointing answer. It took me a long time to accept the fact that there are some things in science that we just can't decide. So let's get a little creative: what would have to change in science for us to be able to tell?
First, and most directly, we would need an alternate theory to QM that allows us to distinguish between the two cases. Seeing how incredibly complex current theories are, I doubt that will happen anytime soon (though I could be pleasantly surprised). The second is that we would need some way to measure the wavefunction directly, instead of looking at it through operators. I've heard whispers of groups making progress on that front, though I honestly don't know if there's any truth to that or not. One way or the other, I don't expect we'll be able to answer this question tomorrow, or anytime relatively soon.
Some time back, I heard about a very clever paper that had to deal with this problem. There was an effect that could either depend on the temperature or density of water. Of course, you can't control the temperature and density of water independently (the stuff expands when you heat it up), and so everyone had just assumed that temperature was responsible. This group realized that, if you supercool water, it takes on densities similar to those found above 0C. By carefully supercooling the water and measuring this effect, they found that at similar densities of water, the effect was the exact same, even when the temperature differed by over 50C. It was density that was responsible all along!
This is one of the most mysterious things about all of physics. Even after two years of studying quantum mechanics, I have yet to find anyone who can offer a satisfying explanation for why fermions (half-integer spin particles) should flip their wavefunction upon exchange of (ostensibly identical) particles. If you know of a good read on the subject, please put it in the comments and I'll add it here!
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1$\begingroup$ The short answer to this is: the question is meaningless. :D And you actually do not need the wave function formalism to explain why. $\endgroup$– WildcatSep 27, 2014 at 8:05
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$\begingroup$ I believe your answer was the one I was originally going for. This is why I shouldn't be allowed to write responses past midnight. $\endgroup$ Sep 27, 2014 at 15:55
Firstly, you seem to have a misconception about atomic orbitals and molecular orbitals. When carbon takes part in bonding within a molecule the electrons don't move from its 2s orbital into an unoccupied 2p orbital. I think what you are referring to is hybridization: You take the 2s and 2p orbitals and build linear combinations from them to get a set of e.g. four energetically equivalent $\ce{sp^3}$ orbitals and then you use these orbitals to form bonds between carbon and, say, the 1s orbitals of four hydrogen atoms thus getting methane $\ce{CH4}$. But the problem is that the carbon $\ce{sp^3}$ or hydrogen 1s orbitals are not actually present in a physical sense in the $\ce{CH4}$ molecule. Those orbitals are merely a mathematical tool, a basis, used to describe the actual molecular orbitals of the methane molecule. A more in-depth explanation of that can be found here.
Ok, so there is no moving electrons from one orbital to another when forming bonds. But what about your question concerning the switching of electrons? Well, electrons are indistinguishable particles. That means if you have two electrons, one on the left and one on the right, you could exchange them without ever noting a difference in the whole system: it will still have the same energy, same momentum, same wavefunction, etc. Let's take the example of lithium: lithium has two 1s electrons and one 2s electron, so three electrons in total. If you'd label those electrons with the names Peter, Paul, and Mary you couldn't say whether Peter and Paul are hanging around in the 1s orbital while Mary has a good time in the 2s orbital or whether Peter and Mary have fun in the 1s orbital while Paul sits lonely in the 2s orbital because electrons are indistinguishable. Both situations would look absolutely the same to you. So, even if two electrons would switch their orbital "positions", you really couldn't tell.
A point that might raise confusion is that there is still the Pauli exclusion principle, saying that two electrons having the same spin can't occupy the same orbital. But that is a completely different matter that has nothing to do with Peter, Paul and Mary - to stay in the picture. You can't say Peter and Paul are both spin-up and Mary is spin-down, so Peter and Paul can never be together in the 1s orbital and one of them is always paired up with Mary. The only thing you can say is that among Peter, Paul, and Mary you must have two with spin-up and one with spin-down (two spin-down and one spin-up would be equivalent). But they can't all be spin-up (at least not in the ground state of $\ce{Li}$) because then you would have two electrons with the same spin in the 1s orbital which is forbidden by the Pauli exclusion principle.
Is this theoretically possible?
Theoretically your question from the title is meaningless so science would not even answer it and that is why.
Electrons are indistinguishable, or identical, particles in a sense that they all have exactly the same physical properties so they cannot be distinguished from one another, even in principle. Thus, there is no observable difference between the two situations you are trying to distinguish:
- Electron-one is at $\ce{2s}$ orbital while electron-two is at $\ce{2p}$ orbital.
- Electron-two is at $\ce{2s}$ orbital while electron-one is at $\ce{2p}$ orbital.
You can not tell these two different configuration apart. Imagine that we play the following simple game: I show to you (how do I do so is irrelevant) an atom with one electron at $\ce{2s}$ orbital and another one at $\ce{2p}$ orbital. I then hide it from you and either do nothing, or exchange the electrons. Finally, I show you the atom once again and ask you, did I exchange electrons or not? Would you be able to answer that question? Surely, no! And this is essentially what makes your question from the title meaningless.
I know that electrons can move from say $\ce{2s}$ orbital to an unoccupied $\ce{2p}$ orbital, as in carbon atom which can form 4 bonds this way.
Of course, here you can clearly tell the difference between the following two situations
- Two electrons at $\ce{2s}$ orbital and two electrons at $\ce{2p}$ orbital.
- One electron at $\ce{2s}$ orbital while electron-one is at $\ce{2p}$ orbital.
by just counting electrons on diferent orbitals (again, how do you do so is irrelevant).
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$\begingroup$ I see your point, but there's a way to detect that switch if it happened. I think electrons had to jump an energy barrier to switch orbitals and probably change their spin. So if we could detect that energy change, then it would mean that it has happened. Like if two identical twins switch their places in a room, one may not be able to tell, but actually they have burned some energy while changing their places. Even if we assume they have exactly the same physical properties. We are not talking about the "states" here. The initial and final states may not be different at all. $\endgroup$– WiseOct 21, 2014 at 15:05
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$\begingroup$ So when I say is it theoretically possible, I mean is the barrier passable by electrons? Is there a state in an orbital no matter how much unstable and transient that 3 electrons exist in one orbital? $\endgroup$– WiseOct 21, 2014 at 20:32
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$\begingroup$ @Wise, of course, we are talking abous states here. And the thing you do not understand is that there is no such state as "e-1 on 2s orbital and e-2 on 2p orbital", as well there is no state "e-1 on 2p orbital and e-2 on 2s orbital". Electrons are indistinguishable, and consequently, there is only state "e-1 or e-2 on 2s orbital and e-2 or e-1 on 2p orbital". And this state is obviously the same as the state "e-2 or e-1 on 2s orbital and e-1 or e-2 on 2p orbital". $\endgroup$– WildcatOct 22, 2014 at 15:08
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$\begingroup$ @Wildcat You do not understand the crux of the question. Wise (and now, myself as well) wishes to know whether such interactions can happen. We do not wish to know whether we can distinguish the final state of the system and the initial state of the system; that is beside the point. $\endgroup$ Jun 16, 2017 at 3:00