Spin of electrons within (x,y,z) 2P orbitals

[Salman2]

Question. Is there a quantum reason why a lone electron in any of the three available (x,y,z) 2P orbital energy states must always be spin-up or spin-down ?

Every figure I see always shows a single electron as spin-up for the three 2P energy states, but is there a quantum theory constraint that would not allow an electron in 2Px to be spin-up, while simultaneously, another in 2Py would be spin-down ?

 

[mfb]

An electron has spin 1/2. If it has spin up (or down) in one direction, its spin is aligned with that direction. It cannot be aligned with two different directions at the same time.

while simultaneously, another in 2Py would be spin-down ?

A second electron? I would expect that their spin coupling leads to trouble there, but I don't know.

 

[gadong]

Question. Is there a quantum reason why a lone electron in any of the three available (x,y,z) 2P orbital energy states must always be spin-up or spin-down ?

Every figure I see always shows a single electron as spin-up for the three 2P energy states, but is there a quantum theory constraint that would not allow an electron in 2Px to be spin-up, while simultaneously, another in 2Py would be spin-down ?

Taking carbon as an example (1s²2s²2p²), the electronic state with the 2p spins parallel is a distinct state from the one with the 2p spins opposed. The two states are not energetically equivalent. It turns out that the first state lies at the lowest energy, while the second state corresponds to an excited state of the carbon atom.

 

[DrDu]

The electrons don't have to be in an eigenstate of s_z. The pictures with the spin symbolized by an arrow pointing up or down is an (over-)simplified picture. However if you have two electrons it makes a difference whether the spins are paired into a triplet (parallel arrows) or singlet (anti-parallel arrows). A singlet/triplett will always be a singlet/ triplet independent of the choice of the quantization axis.

 

[Salman2]

Taking carbon as an example (1s²2s²2p²), the electronic state with the 2p spins parallel is a distinct state from the one with the 2p spins opposed. The two states are not energetically equivalent. It turns out that the first state lies at the lowest energy, while the second state corresponds to an excited state of the carbon atom.

Thanks. I have a question. Do we know the first spin parallel state is at lower energy using Hund's Rule, or experiment, and if the latter, do you have a reference ?

 

[Salman2]

The electrons don't have to be in an eigenstate of s_z. The pictures with the spin symbolized by an arrow pointing up or down is an (over-)simplified picture. However if you have two electrons it makes a difference whether the spins are paired into a triplet (parallel arrows) or singlet (anti-parallel arrows). A singlet/triplett will always be a singlet/ triplet independent of the choice of the quantization axis.

Thank you. I have a question. Can we really say the single electrons found in 2px and 2py are 'paired' since they occupy completely different orbitals along the x,y,z coordinates ? Seems to me that the singlet \uparrow\downarrowand triplet states \uparrow\uparrow or \downarrow\downarrowwould only apply when two spins are 'paired' with close packing within any single 2px or 2py or 2pz orbital within the second energy level beyond the core 1s energy level.

Consider the possibility that a single \uparrow e- is located in one lobe of the 2px orbit, and a second \downarrow e- is located in one lobe of the 2pz orbit at angle > 90 degrees opposite. There is no quantum reason such electrons must be in orbits next to each other, such as x-y or y-z, that is, x-z equally possible. So, how do we conclude that these two electrons are 'paired' and can form a singlet state when they are spatially greatly separated, as compared to two electrons within any specific 2p orbit, which clearly are 'paired' due to close packing within a given orbit ? Clearly I am confused.

 

[DrDu]

The situation for two electrons in different orbitals is more involved. If spin orbit coupling can be neglected, singlet and triplet states are different eigenstates of the hamiltonian, i.e. they have different energies. Thus any superposition of singlet and triplet is possible but will change in the course of time.
As p orbitals carry angular momentum, when spin orbit coupling is taken into account, the eigenstates of the hamiltonian will no longer be pure singlets and triplets

 

[Einstein Mcfly]

Thank you. I have a question. Can we really say the single electrons found in 2px and 2py are 'paired' since they occupy completely different orbitals along the x,y,z coordinates ? Seems to me that the singlet \uparrow\downarrowand triplet states \uparrow\uparrow or \downarrow\downarrowwould only apply when two spins are 'paired' with close packing within any single 2px or 2py or 2pz orbital within the second energy level beyond the core 1s energy level.

Consider the possibility that a single \uparrow e- is located in one lobe of the 2px orbit, and a second \downarrow e- is located in one lobe of the 2pz orbit at angle > 90 degrees opposite. There is no quantum reason such electrons must be in orbits next to each other, such as x-y or y-z, that is, x-z equally possible. So, how do we conclude that these two electrons are 'paired' and can form a singlet state when they are spatially greatly separated, as compared to two electrons within any specific 2p orbit, which clearly are 'paired' due to close packing within a given orbit ? Clearly I am confused.

Are you perhaps confusing the way that electrons are CALCULATED in open-shell singlet calculations (where up and down spins are allowed to "be" in different spatial orbitals even though their are the same number of up and down spins) with the fact that they are paired or not paired? The first is a model of nature meant to give more accurate total energy differences, the second is an experimentally testable fact (does the system have an overall magnetic moment or not).

 

[Salman2]

Are you perhaps confusing the way that electrons are CALCULATED in open-shell singlet calculations (where up and down spins are allowed to "be" in different spatial orbitals even though their are the same number of up and down spins) with the fact that they are paired or not paired? The first is a model of nature meant to give more accurate total energy differences, the second is an experimentally testable fact (does the system have an overall magnetic moment or not).

Thanks, very useful comment to include magnetic moment to help determine if two unpaired electrons in 2p are paired or not. So, take C-12 isotope for example. It has two single e- within two separate 2p orbitals, 2px \uparrowand 2py\uparrow. It is known experimentally that C-12 has 0.0 magnetic moment. Therefore, we know via experiment (not Hund's rule) that the two e- in 2p energy level ARE NOT PAIRED, since, if they were paired, C-12 would have a magnetic moment > 0.0. Experimental data on magnetic moments takes priority over Hund's Rule to help understand pairing or not of the 2p electrons in C-12 isotope.

Thus, concerning my OP question, given the experiment data on C-12 that it has magnetic moment = 0.0, this means the two single electrons in the 2px and 2py orbits of C-12 could be either \uparrow\uparrow OR \downarrow\downarrow OR \uparrow\downarrow, that is, BECAUSE THEY ARE NOT PAIRED, all quantum spin states are equally possible for each individual 2p orbital.

Do I have this all correct ?

==

I think the above is incorrect.

The 2p electrons in C-12 must be PAIRED to have a magnetic moment = 0.0. See this link for C-12 magnetic moment: http://www.webelements.com/carbon/isotopes.html. Elements with unpaired e- are called paramagnetic, but clearly C-12 is diamagnetic since magnetic moment = 0.0.

Of interest is that two e- of C-12 are not paired within the same x,y,z orbit of 2p, but they are paired in two different 2p orbits. From comments below by Gadong, using NIST table reference, C-12 has the two e- in 2p in triplet state \uparrow\downarrow in ground state, and this triplet state, from PAIRED e- found in two different 2p orbits, results in C-12 having a 0.0 magnetic moment.

What is very confusing is that I have read that CARBON as an element is paramagnetic (thus magnetic moment > 0.0) because it has two UNPAIRED electrons in 2p...but this is factually incorrect for the C-12 isotope given that experimental data shows magnetic moment is 0.0.

 

[gadong]

Thanks. I have a question. Do we know the first spin parallel state is at lower energy using Hund's Rule, or experiment, and if the latter, do you have a reference ?

A Google search for {singlet carbon atom} provided the following reference:
http://www.tcm.phy.cam.ac.uk/~mdt26/downloads/hongo_hund.pdf

Normally I would check the ground states of atoms using the NIST periodic table at http://www.nist.gov/pml/data/periodic.cfm which is a respectable source, and can be used as a reference. The carbon ground state is a triplet. By implication, if a state is not the ground state, it is an excited state.

 

[Salman2]

A Google search for {singlet carbon atom} provided the following reference:
http://www.tcm.phy.cam.ac.uk/~mdt26/downloads/hongo_hund.pdf

Normally I would check the ground states of atoms using the NIST periodic table at http://www.nist.gov/pml/data/periodic.cfm which is a respectable source, and can be used as a reference. The carbon ground state is a triplet. By implication, if a state is not the ground state, it is an excited state.

Thanks, but I am confused. The first reference indicates the ground state for carbon is singlet, but the NIST table has it being triplet...which is correct ?

Also, how does any of this help explain the fact that C-12 isotope has a magnetic moment = 0.0 while C-13 isotope has magnetic moment = 0.7024 ? This would mean C-13 isotope is paramagnetic and thus has the e- in 2p being UNPAIRED, while C-12 isotope is diamagnetic with the two e- in 2p being PAIRED...how the heck does Hund's Rule explain the different experimental magnetic moments for C-12 and C-13 isotopes ?

See this link for C-12 and C-13 magnetic moment: http://www.webelements.com/carbon/isotopes.html

 

[DrDu]

A Google search for {singlet carbon atom} provided the following reference:
http://www.tcm.phy.cam.ac.uk/~mdt26/downloads/hongo_hund.pdf

Also this article finds the triplet state as the ground state and the singlet as an excited state.

 

[Einstein Mcfly]

Thanks, but I am confused. The first reference indicates the ground state for carbon is singlet, but the NIST table has it being triplet...which is correct ?

Also, how does any of this help explain the fact that C-12 isotope has a magnetic moment = 0.0 while C-13 isotope has magnetic moment = 0.7024 ? This would mean C-13 isotope is paramagnetic and thus has the e- in 2p being UNPAIRED, while C-12 isotope is diamagnetic with the two e- in 2p being PAIRED...how the heck does Hund's Rule explain the different experimental magnetic moments for C-12 and C-13 isotopes ?

See this link for C-12 and C-13 magnetic moment: http://www.webelements.com/carbon/isotopes.html

C-13 contains an extra neutron, which is a spin 1/2 particle.

 

[Salman2]

Thanks, but I am confused. The first reference indicates the ground state for carbon is singlet, but the NIST table has it being triplet...which is correct ?

Also, how does any of this help explain the fact that C-12 isotope has a magnetic moment = 0.0 while C-13 isotope has magnetic moment = 0.7024 ? This would mean C-13 isotope is paramagnetic and thus has the e- in 2p being UNPAIRED, while C-12 isotope is diamagnetic with the two e- in 2p being PAIRED...how the heck does Hund's Rule explain the different experimental magnetic moments for C-12 and C-13 isotopes ?

See this link for C-12 and C-13 magnetic moment: http://www.webelements.com/carbon/isotopes.html

==

Edit: OK, I see from comment below by DrDu that both references put triplet state \uparrow\uparrow as ground state for C-12 for 2p orbits. Also, I now see that differences in magnetic moment between C-12 and C-13 is due to addition of the neutron in C-13.

 

[Salman2]

Thanks, very useful comment to include magnetic moment to help determine if two unpaired electrons in 2p are paired or not. So, take C-12 isotope for example. It has two single e- within two separate 2p orbitals, 2px \uparrowand 2py\uparrow. It is known experimentally that C-12 has 0.0 magnetic moment. Therefore, we know via experiment (not Hund's rule) that the two e- in 2p energy level ARE NOT PAIRED, since, if they were paired, C-12 would have a magnetic moment > 0.0. Experimental data on magnetic moments takes priority over Hund's Rule to help understand pairing or not of the 2p electrons in C-12 isotope.

Edit: Removed confused comments

Do I have this all correct ?

==

I think the above is incorrect.

The 2p electrons in C-12 must be PAIRED to have a magnetic moment = 0.0. See this link for C-12 magnetic moment: http://www.webelements.com/carbon/isotopes.html. Elements with unpaired e- are called paramagnetic, but clearly C-12 is diamagnetic since magnetic moment = 0.0.

Edit: Removed confused comments

 

[Salman2]

Thanks, but I am confused. The first reference indicates the ground state for carbon is singlet, but the NIST table has it being triplet...which is correct ?

Also, how does any of this help explain the fact that C-12 isotope has a magnetic moment = 0.0 while C-13 isotope has magnetic moment = 0.7024 ? This would mean C-13 isotope is paramagnetic and thus has the e- in 2p being UNPAIRED, while C-12 isotope is diamagnetic with the two e- in 2p being PAIRED...how the heck does Hund's Rule explain the different experimental magnetic moments for C-12 and C-13 isotopes ?

See this link for C-12 and C-13 magnetic moment: http://www.webelements.com/carbon/isotopes.html

Edit: OK, from comment below the difference in magnetic moments between C-12 and C-13 is due to added neutron.

 

[Salman2]

Note to readers of thread. Getting back to my OP question, and all the replies, here are my conclusions, using isotope C-12 for example:

1. The two e- in 2p energy level are in triplet state, symbolically \uparrowin state 2px orbit and \uparrowin 2py

2. Because C-12 has 0.0 magnetic moment, the two \uparrow\uparrow are paired, but they are paired in two different 2p orbits, not within the same orbit.

If this is correct we can end the discussion.

 

[Einstein Mcfly]

Note to readers of thread. Getting back to my OP question, and all the replies, here are my conclusions, using isotope C-12 for example:

1. The two e- in 2p energy level are in triplet state, symbolically \uparrowin state 2px orbit and \uparrowin 2py

2. Because C-12 has 0.0 magnetic moment, the two \uparrow\uparrow are paired, but they are paired in two different 2p orbits, not within the same orbit.

If this is correct we can end the discussion.

1 is right and 2 is not. In 2 you're confusing the nuclear magnetic moment with the number of unpaired electrons. A free neutral carbon atom is a spin triplet and has two electrons of the same (unpaired) spin. That's what a triplet means. A neutral and free Carbon 13 atom is also a spin triplet for the exact same reason.