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I have come across a picture online which talks about 1s,2s,2p orbitals of a sodium atom. enter image description here

Observation:In the picture we see some volume of space in common to two orbitals.

For example:The circled(red) region is common to both 1s and 2py orbitals

enter image description here

My opinion about the observation: The common volume to two orbitals makes sense even when we consider the 3-Dimensional structure of an atom.

How it makes sense?

The S orbital is spherical in shape and P orbital is in dumb-bell shape. So,some region of P orbital will lie in s orbital also(as both orbitals are present around the nucleus).

According to the knowledge I have in Quantum mechanics: An electron can occupy only one orbital at any given instant i.e for example,it can either be in 1s or 2p but definitely not in both the orbitals.This is because the two orbitals have quite different energies.

Question:

Please,explain the anomaly.

My level of knowledge: I can understand high school level Quantum mechanics and a little beyond.I can understand wave particle duality of matter and the concepts of probability densities.

Reference for the picture:

https://en.wikipedia.org/wiki/Atomic_orbital#/media/File:Schrodinger_model_of_the_atom.svg

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    $\begingroup$ Particles in quantum mechanics can exist in a superposition of states of different energies. Even if this weren't true, the three 2p orbitals all have the same energies in the absence of external electromagnetic fields (ignoring small corrections from a possible nuclear magnetic moment), so a particle with definite energy could still exist in a superposition of those anyway. $\endgroup$ – probably_someone Oct 19 '18 at 9:57
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    $\begingroup$ Possible duplicate of What does an orbital mean in atoms with multiple electrons? What do the orbitals of Helium look like? $\endgroup$ – StephenG Oct 19 '18 at 10:20
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    $\begingroup$ I suggested a duplicate and I'd particularly point you at Emilio Pisanty's answer where he says "orbitals are completely fictional concepts. That is, they are unphysical and they are completely inaccessible to any possible measurement.". That is, the whole state of all the electrons together has meaning, the state of individual electrons is meaningless. $\endgroup$ – StephenG Oct 19 '18 at 10:23
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    $\begingroup$ I disagree wholly with the statement that orbitals are fictional. You might as well say that atoms are fictional. And the slippery slope does not end there. $\endgroup$ – my2cts Oct 19 '18 at 11:07
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    $\begingroup$ The pictures you suggested are actually not the orbital itself. The orbitals are spread all over the spaces $\endgroup$ – ChoMedit Oct 19 '18 at 12:54
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There is a good answer here already, but there still seems to be some confusion, so I will give my input as well.

The regions in your diagrams correspond to the "most likely place to find the electron" if if was just in that state. Furthermore, the regions are defined by some threshold probability of finding the electron at that location. So technically, all orbitals "overlap", because the probability amplitude is defined for all of space. We just make these diagrams by only showing the most probably regions.

The issue you seem to be having is that the "location" of the electron determines which orbital it is in. The opposite is actually the case. The orbital that the electron is in determines it probability amplitude in the position basis.

Therefore, there are multiple reasons (probably more than I have just said) to be fine with "overlapping" orbitals

  1. The regions in the diagram correspond to single orbitals if the electron was only in that orbital (no superpositions)
  2. The regions in the diagram just show part of the probability density. Technically the probability density is defined in all space, so all orbitals "overlap" at all points in space
  3. The "location" of the electron does not determine which orbital it is in. If the electron is in some given orbital, we can then use the diagram to determine the most likely region the electron could be observed to be in.
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  • $\begingroup$ You say that all orbitals overlap at all points in space. Then, what does it, exactly, mean when one says that an electron is in 1s or any other orbital? $\endgroup$ – ayc Oct 19 '18 at 15:57
  • $\begingroup$ @ayc The number represents a definite energy. The letter represents a definite magnitude of angular momentum. $\endgroup$ – Aaron Stevens Oct 19 '18 at 17:09
  • $\begingroup$ In your third point you say that if the electron is in some given orbital,we can .... to in.This is what I think based on that point:when you want to know the position of an electron you take schrodinger equation and solve it to get some wave function and you then find the quantum numbers(n,l,m) and then you use these quantum numbers to find where the electron is around the nucleus.For ex:if you got quantum numbers as (2,1,0)it means that electron probability density around the nucleus is in the shape of the 2p orbital. $\endgroup$ – ayc Oct 23 '18 at 5:39
  • $\begingroup$ @ayc Yep you got it. $\endgroup$ – Aaron Stevens Oct 23 '18 at 6:56
  • $\begingroup$ So,if we are looking at an atom and lets say it has electrons in both 1S and 2S orbitals.Technically entire 1s probability distribution,or perhaps most, overlaps with that of a part of 2s's.Does this mean that I'm more likely to find an electron in that overlapping region i.e 2s ? $\endgroup$ – ayc Oct 24 '18 at 18:48
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An electron can occupy only one orbital at any given instant i.e for example,it can either be in 1s or 2p but definitely not in both the orbitals.This is because the two orbitals have quite different energies.

That is quite correct for orbitals defined by the principle quantum number n, identified with the energy eigenvalue in the simple quantum mechanical solutions for the hydrogen atom, and mathematically extrapolated to electrons in atoms and molecules.

Note that $n$ is connected with a wavefunction, it is the complex conjugate square of the wavefunction that defines the orbitals, i.e. orbitals are a probability distribution, giving a value on how probable it is to find the electron at a particular (r,theta,phi).

hydrogen orbitals

Note that all s states, i.e zero angular momentum, have a probability of existing around r=0. This does not mean that they have the same energy, because the quantum mechanical solution for a potential just give probabilities.

In the classical planetary model, the energy of an orbit depends on the distance from the center of force/potential. In a quantum mechanical model, the energy is an eigenvalue of the solution of the quantum mechanical equation with the potential, but the solutions are not connected to space coordinates, except through probabilities.

In proton rich nuclei, electron capture can happen, because in such nuclei there is a large enough probability for s state electrons to overlap with the probability distribution of a proton, and the proton changes into a neutron and a different nucleus . It is a complicated process which exists through the probabilistic nature of quantum mechanics.

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  • $\begingroup$ All S states have the probability of existing around r=o,I agree.And this doesn't mean that they have same energy,I agree.Could you elaborate the statement:" This does not mean that they have the same energy, because the quantum mechanical solution for a potential just give probabilities, the energy is an eigenvalue of the solution, not connected to space coordinates, except through probabilities". $\endgroup$ – ayc Oct 19 '18 at 11:24
  • $\begingroup$ Is this understanding of mine correct:when you want to know the position of an electron you take schrodinger equation and solve it to get some wave function and you then find the quantum numbers(n,l,m) and then you use these quantum numbers to find where the electron is around the nucleus.For ex:if you got quantum numbers as (2,1,0)it means that electron probability density around the nucleus is in the shape of the 2p orbital. $\endgroup$ – ayc Oct 23 '18 at 6:40
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    $\begingroup$ yes, but you cannot know at what (x,y,z) the electron is , only the probability distribution of finding it there $\endgroup$ – anna v Oct 23 '18 at 9:16
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Okay, now I get you. I'll try to explain your question in the comment in simple terms.

If we agree to the fact that 1S and 2P orbitals have different energies then the volume of space common to those two orbitals cannot exist as they have quite different energies,but from the picture and 'my opinion about the observation' it seems like they do.That's the anomaly.I'm just asking how do you explain the probability of finding an electron in the volume of space common to two orbitals.

Note, however, that it is extremely difficult (and "dangerous") to try to epxlain it to someone who doesn't know much more than high school QM, but I'll try.

I'll go one by one.

If we agree to the fact that 1S and 2P orbitals have different energies

Yes. Well, we don't just "agree", that is just true, and it can be shown. That's tright.

then the volume of space common to those two orbitals cannot exist as they have quite different energies

What? How comes a volume cannot exist? The volume is a real 3D volume. It might be empty, but it exists.

What's more, there's nothing wrong with 2 electrons together, having different energies. Pauli's exclusion principle forbids that two electrons share all their quantum numbers, but if the electrons have different energies, then it's okay to be in teh same place.

I'm just asking how do you explain the probability of finding an electron in the volume of space common to two orbitals.

This is more complex, and it requires further knowledge of QM.

I'll try to make an intuitive analogy.

In QM, let's say that we just don't know things until we measure them. We don't know where an electron is until we measure its position. This is different than classic mechanics, where you can estimate positions just gazing roughly, without perturbating the system. Not here; in QM, you don't know anything until you measure it.

So, how do we measure things? Regardless of how scientist manage themselves to develope the appropiate instruments, measuring is "revealing" some unknown information. For example: where's the electron? Take a picture and find out. "Oh, it was here".

Unlike classical mechanics (CM), Quantum mechanics (QM) are random. That means position won't be always the same, even in the same system. There's essential randomness.


So let's get to the point. Imagine you want to measure the "state". That means measuring the quantum numbers of the electron, so that you can say "oh, this is a $1s$ orbital" (let's neglect the spin now).

So you have an atom, you use the whatever-meter and you find out that the electron was in $1s$.

You can measure another electron being in $2p$, with $m=0$, for example.

In QM, it is possible to have a "linear combination" of $1s$ and $2p$. In that case, you'll sometimes measure $1s$ and sometimes $2p$. It is random, but there is a well defined probability of measuring each one. The probability depends on the concrete entanglement you have.

So that's a more elaborate answer to the first thing: the same electron can be found in $1s$ or $2p$, depending on your luck. This is weird to newbies, but it is basic QM.

This is just to set up some ideas. However, this was not your question. If I understood well, your question was how can an electron be in a "shared volume", right?

Well, that's because the orbital is not exclusive of one level, as you can see in the picture.

The orbital you see there is "the volume in which it is likely to measure $2p$".

You must associate the quantum numbers to the electron, not to the volume. The volume is free land, and the state (Quantum numbers) belong to the electron. So an electron can move around a $1s$ orbital. That only means that "the electron is in a volume in which $1s$ electrons are likely to be there".

So it is like a tourist visiting another country, or a land shared by two countries. The thing is that, when we make a measurement, we're asking the electron for its passport, and then we see if it was a $1s$ or a $2p$.


To sum up:

  • The orbitals you're seeing there are not like Bohr levels. In Bohr levels, an electrno being in a level authomatically implied having the energy of that level.
  • Nevertheless, this is not a level representation. This is a picture of the "volumes that each nationality uses to visit".
  • Some sub-volumes can be visited by several "nationalities", but that doesn't mean that visitors change their nationality.
  • However, QM allows being "a combination of nationalities". This is, showing you different passports, randomly. This happens in entangled states.

(To be precise, after showing you one, it will remain being that one forever. After a measurement, the "system" picks only one and stays with it. The surprising thing is that identical systems, strictly identical, can show any of those passports, randomly.)

So this is all. Hope I helped.

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  • $\begingroup$ I have got some idea now,thank you.Could you explain what it,exactly,means when one says that an electron is in 1s or any other orbital? $\endgroup$ – ayc Oct 19 '18 at 15:58
  • $\begingroup$ It means that, when you measure it, you get those quantum numbers. That is to say that energy is n=1, angular momentum is l=0, m=0, and so on. $\endgroup$ – FGSUZ Oct 19 '18 at 16:08
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The figure that you give should not be taken too literally. It plots the orbitals as having a constant value in space. In reality there are positive and negative, in general complex valued regions. Two orbitals are different if they are orthogonal, that is, if the integral over all space of their product is zero. Since this one grand can have any sign, thus does not require that one orbital is zero wherever another is nonzero.

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  • $\begingroup$ I don't know if I can ask this,but I'm asking:could you explain your answer in simpler terms,if possible? $\endgroup$ – ayc Oct 19 '18 at 11:30
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    $\begingroup$ I am pretty sure these orbitals are showing the probability amplitude of finding the electron at that position. Technically this would be positive and real in all space, so you determine some threshold to determine the region where an electron is "most likely to be found". These are the regions shown in diagrams like these. $\endgroup$ – Aaron Stevens Oct 19 '18 at 11:40
  • $\begingroup$ @Aaron Stevens An orbital does not represent the probability amplitude. It is its absolute square that does this. $\endgroup$ – my2cts Oct 21 '18 at 10:03
  • $\begingroup$ @ayc Feel free to ask any question :-). Unfortunately, I cannot make this any simpler. Atomic orbitals are in general complex valued functions of space and time. They are basically waves. You cannot understand their orthogonality and hence the Pauli principle, atomic structure etc. on the basis of the corresponding probability distributions alone. $\endgroup$ – my2cts Oct 21 '18 at 10:06
  • $\begingroup$ I understand this. I'm saying the "orbitals" in the diagram shows a thresholded probability amplitude. I should have been more specific. $\endgroup$ – Aaron Stevens Oct 21 '18 at 12:10

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