Why these four 2p orbital electrons have higher energy state configuration?

In summary, the picture shown is from XPS and AES and depicts four electrons located at the 2p orbital with energy state L3 having higher energy configurations than those below at the 2p orbital. The higher energy configuration may be attributed to the orientation of the three p orbitals, with two adopting a side configuration and one adopting an outward configuration. Additionally, the total angular momentum quantum number j for each level must be considered, with the j=3/2 state having a degeneracy of 4 and the j=1/2 state having a degeneracy of 2. It is possible that the magnetic field plays a role in the energy configuration, but further research is needed.
  • #1
boladore
10
0
quantum.PNG

This picture is from XPS and AES.

Hi. sorry for my poor English first.

As you can see from above picture, four electrons located at 2p orbital with energy state L3 have higher energy configuration than those of below 2p orbital.

I'm wondering why these 4 electrons have higher energy. as you know, ordinary 2p orbital electrons all have same energy.

regards.
 
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  • #2
boladore said:
This picture is from XPS and AES.

Hi. sorry for my poor English first.

As you can see from above picture, four electrons located at 2p orbital with energy state L3 have higher energy configuration than those of below 2p orbital.

I'm wondering why these 4 electrons have higher energy. as you know, ordinary 2p orbital electrons all have same energy.

regards.

Well, I am not sure how it works for a solid state sample, but in a molecule, those would be the pi-orbitals created by parallel side-by-side overlap of two pairs of p-orbitals on adjacent atoms (the px and py orbitals by convention). The lower state would by the p-p sigma orbital, created from end-to-end overlap of the remaining two p-orbitals (typically the pz orbitals).

I am guessing that there is an analogous situation in the solid state sample that causes similar energy ordering of the orbitals, but that is only a guess.
 
  • #3
I would also assume that it has to do with the orientation of the three p orbitals.

Two of the orbitals adopt a side configuration (left-right and up-down) and one adopts a outward configuration (front-back). Perhaps the front-back orbital is lower energy for some reason.

Hmm... Are three dimensions the restriction for electrons? Or can they go into higher... Maybe 7 or some weird number...
 
  • #4
SpectraCat said:
Well, I am not sure how it works for a solid state sample, but in a molecule, those would be the pi-orbitals created by parallel side-by-side overlap of two pairs of p-orbitals on adjacent atoms (the px and py orbitals by convention). The lower state would by the p-p sigma orbital, created from end-to-end overlap of the remaining two p-orbitals (typically the pz orbitals).

I am guessing that there is an analogous situation in the solid state sample that causes similar energy ordering of the orbitals, but that is only a guess.


One of my professor said it is possible that if certain molecule is located in magnetic field, their 2p orbital electrons can have lower energy configuration.

Thanks for your opinion.
 
  • #5
Char. Limit said:
I would also assume that it has to do with the orientation of the three p orbitals.

Two of the orbitals adopt a side configuration (left-right and up-down) and one adopts a outward configuration (front-back). Perhaps the front-back orbital is lower energy for some reason.

Hmm... Are three dimensions the restriction for electrons? Or can they go into higher... Maybe 7 or some weird number...

This picture is about XPS(x-ray photoelectron spectroscopy).
and its measure range is 1~10nm deep from the top of the sample measured.
Those electrons are located in atoms at the surface of sample.

Thanks for your reply.
 
  • #6
boladore said:
One of my professor said it is possible that if certain molecule is located in magnetic field, their 2p orbital electrons can have lower energy configuration.

Thanks for your opinion.

Actually, I think my initial reply is probably incorrect (sorry, it was late at might here). It might sort of make sense for a covalent network solid, but I don't think it makes any sense for a crystalline metal. The magnetic field explanation also seems incorrect, since that should split the energy levels of all 3 p-orbitals, according to the Zeeman effect.

What I failed to notice last night is the labels on the orbitals in the other diagrams. Those indicate the total angular momentum quantum number j for the different levels, once coupling of the orbital and spin angular momenta have been considered for each individual electron. This is called the j-j coupling picture for angular momenta. For an electron in a 2p orbital, l=1, and s=1/2. These can couple to give a total angular momentum of j=(1+1/2)=3/2, or j=(1-1/2)=1/2. The degeneracy of one of these states, that is, the number of equivalent electrons having the same j-quantum number, is just given by 2j+1. Thus, the degeneracy of the j=3/2 state is 4, and the degeneracy of the j=1/2 state is 2, which matches up with what is shown in the first picture for XPS.

Anyway, I think that is a more complete/correct explanation of what is going on in the picture from your book. Sorry for any confusion .. hope this helps.
 
  • #7
SpectraCat said:
Actually, I think my initial reply is probably incorrect (sorry, it was late at might here). It might sort of make sense for a covalent network solid, but I don't think it makes any sense for a crystalline metal. The magnetic field explanation also seems incorrect, since that should split the energy levels of all 3 p-orbitals, according to the Zeeman effect.

What I failed to notice last night is the labels on the orbitals in the other diagrams. Those indicate the total angular momentum quantum number j for the different levels, once coupling of the orbital and spin angular momenta have been considered for each individual electron. This is called the j-j coupling picture for angular momenta. For an electron in a 2p orbital, l=1, and s=1/2. These can couple to give a total angular momentum of j=(1+1/2)=3/2, or j=(1-1/2)=1/2. The degeneracy of one of these states, that is, the number of equivalent electrons having the same j-quantum number, is just given by 2j+1. Thus, the degeneracy of the j=3/2 state is 4, and the degeneracy of the j=1/2 state is 2, which matches up with what is shown in the first picture for XPS.

Anyway, I think that is a more complete/correct explanation of what is going on in the picture from your book. Sorry for any confusion .. hope this helps.

Thanks. I was thinking about that too. actually, I didn't show my picture to my professor. What I asked was just "Why these~~~" with just electron configuration picture on the blackboard. I think my professor was confused that time.
 

1. Why do these four 2p orbital electrons have higher energy state configuration?

The four 2p orbital electrons have higher energy state configuration because they occupy the 2p subshell, which is one of the higher energy subshells in an atom. Electrons in higher energy subshells have more energy and are further away from the nucleus, making them more unstable and easier to excite.

2. How does the number of electrons in a subshell affect its energy state?

The number of electrons in a subshell directly affects its energy state. As the number of electrons in a subshell increases, the repulsion between them also increases, making the subshell more unstable and higher in energy.

3. Can the energy state of an electron change?

Yes, the energy state of an electron can change. This can happen through excitation, where the electron absorbs energy and jumps to a higher energy level, or through relaxation, where the electron releases energy and falls to a lower energy level.

4. What factors influence the energy state of an electron?

The energy state of an electron is influenced by several factors, including the number of electrons in the subshell, the distance from the nucleus, and the shielding effect of other electrons. Other factors such as the presence of external fields or the type of atom can also affect an electron's energy state.

5. How does the energy state of an electron impact chemical bonding?

The energy state of an electron plays a crucial role in chemical bonding. Atoms with similar energy levels can easily bond as their electrons can be shared or transferred to achieve a more stable configuration. On the other hand, electrons with significantly different energy levels may not form stable bonds, leading to less reactive or inert elements.

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