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

[boladore]

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.

 

[SpectraCat]

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 p<sub>x 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.</sub>

 

[Char. Limit]

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...

 

[boladore]

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 p<sub>x 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.</sub>

<sub>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.</sub>

 

[boladore]

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.

 

[SpectraCat]

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.

 

[boladore]

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.