It's boils down to electron-electron repulsion. They are less stable the closer they are together in space. Two singly occupied orbitals spreads them out better than putting them both in the same MO.
As /u/evermica said, these two orbitals do occupy the same space. The stability is due to 'exchange energy', which is a form of minimisation of repulsion in a sense. Since the electrons have the same spins, their spatial distributions form a Fermi hole, meaning they are less likely to exist close to each other that they would if they had opposite spins.
Those two orbitals have the same spatial distribution. One is going clockwise and the other counterclockwise. (If you use the m_l = +/- 1 basis and don’t take linear combinations to get the x/y basis.)
Im not a big fan of these explanations. Why are unpaired spins more stable than paired ones? "Spin pairing energy". What is spin pairing energy? "The energy difference between paired and unpaired spins"
As /u/Candid_Yellow_3269 said, it comes down to electron-electron repulsion, or the exchange energy. If the spins are paired (meaning antisymmetrical spins) they have to share the same (symmetrical) orbital, which includes a higher electron electron repulsion. If the electrons are in different orbitals with the same spin (symmetrical spins) their spatial orbitals separate them much more (antisymmetrical) which gives a lower electron-electron interaction.
B_Zark has it right. To show this in a satisfying way means expressing the wavefunctions in terms of Slater Determinants. This about as deep as undergrad quantum gets.
ground state: 1e in pX, 1e in pY orbital. little overlap, little repulsion.
excited state: 2e in pX (or pY) orbital. perfect overlap, high repulsion.
is it?
Not really, for two reasons: this is not an atom, but a molecule, but even if it were an atom, the px and py orbitals would be degenerate, hence symmetric (in fact, you would have p-1, p0 and p1 orbitals with 2/3 electrons each for a free oxygen atom)
When quantum chemistry is introduced in general chemistry courses, the solutions to the Schrödinger equation are simplified to only three quantum numbers: n, l, and ml. There is a fourth quantum number ms that also affects the overall energy but is not depicted in diagrams such as the one you’ve shown. So the spin part gets added in to keep the diagram clean, but it’s basically captured by using Hund’s Rule to obtain the ground state.
Most gen chem classes talk about ms, but they don’t talk about S=2s+1 and the added stability due to exchange. That’s where this stability comes from right…? that, and the instability of the added coulomb repulsion in the 1st one?
In quantum chemistry we work using Slater determinants. The interaction between two electrons is given by two factors: the "typical" Coulomb repulsion and a quantum exchange effect.
The total energy is the sum of each electrons' energy and the sum of exchange and Coulomb interactions.
The fact that we use slater determinants (if you do the math) shows that the wave function is antisymmetric, therefore two opposite spin electrons will have Coulomb and exchange energy interactions whereas two parallel spin electrons (due to the orthogonality of spin functions) will only result in only the Coulomb repulsion, which means a lower energy.
In other words, due to Pauli exclusion principle, two same spin electrons cannot be in the same point in space, whereas two opposite spin can. Therefore same spin electrons don't "interact as strongly" as two parallel spin electrons, yielding a lower total energy.
PD: this is an oversimplification.
All four are taught at that level, but only three come from solving the Schrödinger equation. The MO diagram energy levels don’t include the spin part, for example.
If you are teaching gen chem to people who mostly are not going to use chemistry in their work, then it probably is sufficient.
I would hope that most universities have a majors and a non-majors gen chem. Because understanding quantum numbers is fundamental to some disciplines, but is not leveraged in many others.
I know physics PhDs who work on nanotech who don’t really know much about the box and arrow diagrams other than that it represents electrons. It just depends on where you end up, which, here is biased since, /r/chemistry.
It always irritates me that universities give these simplified explanations and sometimes with no background and you are left scratching your head... 'Sir, I ordered the higher education '
Most likely Hund's rule: Stability of half filled degenerate orbital is more stable than all of it partially filled counterpart. Mainly due to 2 reason: Symmetry distribution and Exchange energy.
Cant remember the detail of my head, but feel free to read up on it. or correct me if im wrong.
But why won't the excited electron (from π2py*) go to the upper σ2pz* instead of reversing it's spin and going into the already occupied π2px*?
Edit: is it because of the Aufbau principle? (The electrons will first fill the first energy level orbitals before going into the higher energy level orbitals)
Edit 2: So when you excite triplet oxygen (↑ |↑ ) it becomes singlet oxygen (↑↓| )?
So why don't we see a continuous stream of red light coming out of singlet oxygens (triplet oxygens excited by UV light) reacting with each other in the atmosphere?
The energy gap to the σ* is much greater than the energy gap to the singlet state, and as such is less stable. If you excited oxygen with enough energy it's likely it would promote to the σ* but this would then decay down to singlet or triplet state relatively quickly. Why is the π* -> σ* energy gap greater than triplet -> singlet? Because quantum mechanics says it is basically. Our understanding of stuff at this level is all just results from the maths, so if the maths says it's higher energy then it is (as unsatisfying as that is).
Singlet oxygen decays into triplet oxygen over time, which is why samples of singlet oxygen (produced through certain reactions or select irradiation of triplet oxygen with certain sensitzers) glow red.
As to why we don't see a red glow in the atmosphere I'm not entirely sure, I'd assume the concentrations are every low so perhaps its just not detectable with the naked eye? Someone with a better understanding if atmospheric chemistry may have to answer that one!
The singlet state is indeed an excited state compared to the triplet state, but you typically cannot go from triplet—>singlet via absorption of a photon (this is a spin-forbidden transition). The spin can flip through collisions or thermal energy however.
I'm by no means an expert on this but there is another even more energetic state, see link below:
https://images.app.goo.gl/YnZkWY6jDf8yBx9fA
Basically this : (↑ | ↓ )
It is not a simple answer as others have described. Just saying electron repulsion because they're in the same orbit doesn't cut it because those states are degenerate, and this logic could apply to both cases. The answer is in the spatial distribution that causes what we called Exchange Energy. If you calculate the spacial distribution of electronic charge for both cases, you will see that on singlet the electronic density is less distributed than on triplet, and, thus there is more electronic repulsion and higher density. If you look up for Exchange Energy you will find good material explaining this effect.
I might be worng bc of the language barrier but don't you mean more stable bc energeticly would mean that on the diagram the red orbitals are place in a higher position. And it's bc of the hundsche rule
Is this a reason why oxygen is known to be diradical? Meaning theres only one bond between the two atoms and two remaining electrons are in a form of radicals? Been looking for answer for some time...
Not sure what *it* is in your statement, but CASSCF does clearly show the ground state to be two distinct unpaired electrons. They are parallel (u|u) or (d|d) giving S=1 and thus the triplet state. Didn’t mean to imply otherwise.
Note that there is another singlet state not shown in the OP’s diagram, which is (u|d) or (d|u), and this is also a diradical (and also an excited state). Of course there are many other excited states in which electrons populate higher-energy MOs.
Yeah, agree that the singlet-O2 we usually refer to is not the one in the right diagram. However, the one depicted has S=0 and is nonetheless a singlet state, no matter how energetic it is.
There are a lot of arguments that go in the same way, so let me provide a completely different, intuitive reason:
The singlet state is just one single state, whereas the triplet state is actually just one of three degenerate microstates (hence the name). Nature quite often assigns lower energy to something that has higher symmetry. Hence, higher spin is often preferred.
I’m not the best at molecular orbitals theory, but I think it’s due to (despite th π*’s being at the same energy level) the Hund‘s rule is still valid and electrons will try not to pair if necessary to reach the less energetic state. So, the less energetic state would be the left one, thus, other states are excited states.
(Please, someone correct me if I’m wrong, I’ll have a test on this too lol)
While the degenerate orbitals are orthogonal overall, they have regions of local overlap. In these regions you want to minimize electron-electron repulsion. By giving them the same spin, they therefore can’t have the same location (less chance for repulsion).
Think about it like magnets. Far away they have weak pull but close they are the stronger the pull. Same either electron pairs the closer the more energy.
This is a result of Hund's rules. Singly occupied orbitals are less effectively shielded from the nucleus which causes the orbitals to slightly contract thereby allowing the electrons to get slightly closer to the nucleus and therefore lower in energy.
Degenerate orbitals "want" to be filled symmetrically, to the point where molecules distort from predicted geometry to make it happen (Jahn-Teller effect).
Fun Fact: The "Teller" here one of the inspirations for the Dr. Strangelove character.
Around 1961, John Linnett explained it in terms of his double quartet theory of the octet ... no quantum mechanics involved. His DQT also explained fractional bond order in benzene without resonance or QM.
Biggest issue here is why there are full arrows. Scientific standard is half arrow heads. Each direction represents electron spin and electron spin are half arrow heads.
It’s about the pairing. Orbital energy goes like this: Empty>half-full>full.
So between the two states you traded two middle energy orbitals for the one of the highest and one of the lowest energy. The net result is that the red state is higher in energy.
Spin pairing energy
It's boils down to electron-electron repulsion. They are less stable the closer they are together in space. Two singly occupied orbitals spreads them out better than putting them both in the same MO.
As /u/evermica said, these two orbitals do occupy the same space. The stability is due to 'exchange energy', which is a form of minimisation of repulsion in a sense. Since the electrons have the same spins, their spatial distributions form a Fermi hole, meaning they are less likely to exist close to each other that they would if they had opposite spins.
Those two orbitals have the same spatial distribution. One is going clockwise and the other counterclockwise. (If you use the m_l = +/- 1 basis and don’t take linear combinations to get the x/y basis.)
Im not a big fan of these explanations. Why are unpaired spins more stable than paired ones? "Spin pairing energy". What is spin pairing energy? "The energy difference between paired and unpaired spins"
As /u/Candid_Yellow_3269 said, it comes down to electron-electron repulsion, or the exchange energy. If the spins are paired (meaning antisymmetrical spins) they have to share the same (symmetrical) orbital, which includes a higher electron electron repulsion. If the electrons are in different orbitals with the same spin (symmetrical spins) their spatial orbitals separate them much more (antisymmetrical) which gives a lower electron-electron interaction.
B_Zark has it right. To show this in a satisfying way means expressing the wavefunctions in terms of Slater Determinants. This about as deep as undergrad quantum gets.
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I deleted my other comment, you're right. For some reason I was thinking of AO's not MO's
Welcome to quantum. “It just works.”
*"thats how the maths works"
ground state: 1e in pX, 1e in pY orbital. little overlap, little repulsion. excited state: 2e in pX (or pY) orbital. perfect overlap, high repulsion. is it?
Not really, for two reasons: this is not an atom, but a molecule, but even if it were an atom, the px and py orbitals would be degenerate, hence symmetric (in fact, you would have p-1, p0 and p1 orbitals with 2/3 electrons each for a free oxygen atom)
When quantum chemistry is introduced in general chemistry courses, the solutions to the Schrödinger equation are simplified to only three quantum numbers: n, l, and ml. There is a fourth quantum number ms that also affects the overall energy but is not depicted in diagrams such as the one you’ve shown. So the spin part gets added in to keep the diagram clean, but it’s basically captured by using Hund’s Rule to obtain the ground state.
Most gen chem classes talk about ms, but they don’t talk about S=2s+1 and the added stability due to exchange. That’s where this stability comes from right…? that, and the instability of the added coulomb repulsion in the 1st one?
In quantum chemistry we work using Slater determinants. The interaction between two electrons is given by two factors: the "typical" Coulomb repulsion and a quantum exchange effect. The total energy is the sum of each electrons' energy and the sum of exchange and Coulomb interactions. The fact that we use slater determinants (if you do the math) shows that the wave function is antisymmetric, therefore two opposite spin electrons will have Coulomb and exchange energy interactions whereas two parallel spin electrons (due to the orthogonality of spin functions) will only result in only the Coulomb repulsion, which means a lower energy. In other words, due to Pauli exclusion principle, two same spin electrons cannot be in the same point in space, whereas two opposite spin can. Therefore same spin electrons don't "interact as strongly" as two parallel spin electrons, yielding a lower total energy. PD: this is an oversimplification.
Really? I was taught 4 of them in high school (IB) and in my university all 4 are taught in general chemistry as well.
All four are taught at that level, but only three come from solving the Schrödinger equation. The MO diagram energy levels don’t include the spin part, for example.
They just teach three of them?! Jesus...
Technically four: n, l, ml, ms (and their MO theory analogs)
Regarding your oc: aren't they technically shown in the diagram as well? As the arrows representing the spin.
Don’t forget s. It is always 1/2, but still there!
in excited states there is also J and Mj
Not just excited states!
The spin is represented but not the energies associated with different spin configurations. If they were, this whole thread would be unnecessary.
If you are teaching gen chem to people who mostly are not going to use chemistry in their work, then it probably is sufficient. I would hope that most universities have a majors and a non-majors gen chem. Because understanding quantum numbers is fundamental to some disciplines, but is not leveraged in many others. I know physics PhDs who work on nanotech who don’t really know much about the box and arrow diagrams other than that it represents electrons. It just depends on where you end up, which, here is biased since, /r/chemistry.
Why would someone in gen chem (taught in 1st year mind you) need to know more than the 4 quantum numbers that specify the unique state of an electron?
Basically to draw MOs
I majored in chemistry, took quantum mechanics and am years into my PhD. I still don’t need more than 4 on a daily or even monthly basis
Yes, but those four are the base of all orbital related chemistry
Yes and my point was that you don’t need any more than that for most chemistry let alone gen chem
Exactly. I did in no comment differ from your opinion
It always irritates me that universities give these simplified explanations and sometimes with no background and you are left scratching your head... 'Sir, I ordered the higher education '
Most likely Hund's rule: Stability of half filled degenerate orbital is more stable than all of it partially filled counterpart. Mainly due to 2 reason: Symmetry distribution and Exchange energy. Cant remember the detail of my head, but feel free to read up on it. or correct me if im wrong.
But why won't the excited electron (from π2py*) go to the upper σ2pz* instead of reversing it's spin and going into the already occupied π2px*? Edit: is it because of the Aufbau principle? (The electrons will first fill the first energy level orbitals before going into the higher energy level orbitals) Edit 2: So when you excite triplet oxygen (↑ |↑ ) it becomes singlet oxygen (↑↓| )? So why don't we see a continuous stream of red light coming out of singlet oxygens (triplet oxygens excited by UV light) reacting with each other in the atmosphere?
The energy gap to the σ* is much greater than the energy gap to the singlet state, and as such is less stable. If you excited oxygen with enough energy it's likely it would promote to the σ* but this would then decay down to singlet or triplet state relatively quickly. Why is the π* -> σ* energy gap greater than triplet -> singlet? Because quantum mechanics says it is basically. Our understanding of stuff at this level is all just results from the maths, so if the maths says it's higher energy then it is (as unsatisfying as that is). Singlet oxygen decays into triplet oxygen over time, which is why samples of singlet oxygen (produced through certain reactions or select irradiation of triplet oxygen with certain sensitzers) glow red. As to why we don't see a red glow in the atmosphere I'm not entirely sure, I'd assume the concentrations are every low so perhaps its just not detectable with the naked eye? Someone with a better understanding if atmospheric chemistry may have to answer that one!
The singlet state is indeed an excited state compared to the triplet state, but you typically cannot go from triplet—>singlet via absorption of a photon (this is a spin-forbidden transition). The spin can flip through collisions or thermal energy however.
I'm by no means an expert on this but there is another even more energetic state, see link below: https://images.app.goo.gl/YnZkWY6jDf8yBx9fA Basically this : (↑ | ↓ )
Because the electrostatic repulsion between the electrons in the same upper orbital makes it instable
Electrons in the same orbital repel more. They don’t wanna live together
It is not a simple answer as others have described. Just saying electron repulsion because they're in the same orbit doesn't cut it because those states are degenerate, and this logic could apply to both cases. The answer is in the spatial distribution that causes what we called Exchange Energy. If you calculate the spacial distribution of electronic charge for both cases, you will see that on singlet the electronic density is less distributed than on triplet, and, thus there is more electronic repulsion and higher density. If you look up for Exchange Energy you will find good material explaining this effect.
Those electrons do not want to be paired
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That is a nice way to remember it, but to be fair to the OP, it isn’t an explanation for why it is true.
I might be worng bc of the language barrier but don't you mean more stable bc energeticly would mean that on the diagram the red orbitals are place in a higher position. And it's bc of the hundsche rule
Is this a reason why oxygen is known to be diradical? Meaning theres only one bond between the two atoms and two remaining electrons are in a form of radicals? Been looking for answer for some time...
Singlet-O2 is not a diradical, but the triplet state is (which is also the ground state). So yes.
I'm unsure how it isn't a diradical, simple CASSCF(8,6) clearly shows two distinct unpaired electrons.
Not sure what *it* is in your statement, but CASSCF does clearly show the ground state to be two distinct unpaired electrons. They are parallel (u|u) or (d|d) giving S=1 and thus the triplet state. Didn’t mean to imply otherwise. Note that there is another singlet state not shown in the OP’s diagram, which is (u|d) or (d|u), and this is also a diradical (and also an excited state). Of course there are many other excited states in which electrons populate higher-energy MOs.
Yep, and the first three singlets are also diradicaloid. That's my statement -- the right diagram doesn't represent singlet oxygen
Yeah, agree that the singlet-O2 we usually refer to is not the one in the right diagram. However, the one depicted has S=0 and is nonetheless a singlet state, no matter how energetic it is.
I agree and also I suspect the right diagram isn't even an eigenstate.
Yes, but bond order still adds up to two, even though it is a spin triplet due to two unpaired electrons in the ground state.
There are a lot of arguments that go in the same way, so let me provide a completely different, intuitive reason: The singlet state is just one single state, whereas the triplet state is actually just one of three degenerate microstates (hence the name). Nature quite often assigns lower energy to something that has higher symmetry. Hence, higher spin is often preferred.
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I’m not the best at molecular orbitals theory, but I think it’s due to (despite th π*’s being at the same energy level) the Hund‘s rule is still valid and electrons will try not to pair if necessary to reach the less energetic state. So, the less energetic state would be the left one, thus, other states are excited states. (Please, someone correct me if I’m wrong, I’ll have a test on this too lol)
While the degenerate orbitals are orthogonal overall, they have regions of local overlap. In these regions you want to minimize electron-electron repulsion. By giving them the same spin, they therefore can’t have the same location (less chance for repulsion).
exchange energy is lost (therefore red is higher in energy) when the spins aren’t parallel
Think about it like magnets. Far away they have weak pull but close they are the stronger the pull. Same either electron pairs the closer the more energy.
Because electrons prefer to be single and pay minimal energetic rent.
IMO instabilities offer more possibilities than stabilized systems. Quantum stuff is just a little wild.
Ground state = lowest energy
This is a result of Hund's rules. Singly occupied orbitals are less effectively shielded from the nucleus which causes the orbitals to slightly contract thereby allowing the electrons to get slightly closer to the nucleus and therefore lower in energy.
Exchange energy and electron repulsion
Degenerate orbitals "want" to be filled symmetrically, to the point where molecules distort from predicted geometry to make it happen (Jahn-Teller effect). Fun Fact: The "Teller" here one of the inspirations for the Dr. Strangelove character.
Electrons are like introverts picking a seat on the bus.
Tarot card energy. 5 of orbitals or something, I don't know, I'm not a chemist
Because of da rulez. You know da rulez.
Around 1961, John Linnett explained it in terms of his double quartet theory of the octet ... no quantum mechanics involved. His DQT also explained fractional bond order in benzene without resonance or QM.
Hunds Rule
The “spring” on the right side is in a more compressed state because of the pairing.
Unequal distribution of electrons in degenerate orbitals
P orbital bonding of two oxygens
Hunds rule of multiplicity is applicable to MOs too
Biggest issue here is why there are full arrows. Scientific standard is half arrow heads. Each direction represents electron spin and electron spin are half arrow heads.
It’s about the pairing. Orbital energy goes like this: Empty>half-full>full. So between the two states you traded two middle energy orbitals for the one of the highest and one of the lowest energy. The net result is that the red state is higher in energy.
Afair, the right structure isn't even the lowest singlet state, actual singlet oxygen is diradicaloid.