Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: davidenarb on June 21, 2014, 06:09:45 PM
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Hello!
We know that a pi bond is formed via an overlap of two p orbitals. The most stable MO of the pi bond is when the two phases of the p bond overlap constructively ( positive lobe with a positive lobe, and negative lobe with the negative lobe)
My question is: Can a MO of a pi bond adopt a conformation where the positive lobe overlap with a negative lobe, and a negative lobe with the positive lobe) ?
I know that this MO is called the anti-bonding orbital, but I am not sure if I can consider this conformation as a "normal" pi bond as the term "antibonding" misleads my perception.
Thank you
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So, in other words, can these two p orbitals form a pi bond even though it is not the most stable state?
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This combination of p-orbitals would form an antibonding molecular orbital, which has a higher energy than an isolated atomic p-orbital. An electron placed in this molecular orbital would tend to push the two nuclei apart - that is, it would weaken the overall strength of the bond between the two nuclei (decrease the overall bond order). So no, strictly speaking this combination would not lead to formation of a pi-bond.
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This combination of p-orbitals would form an antibonding molecular orbital, which has a higher energy than an isolated atomic p-orbital. An electron placed in this molecular orbital would tend to push the two nuclei apart - that is, it would weaken the overall strength of the bond between the two nuclei (decrease the overall bond order). So no, strictly speaking this combination would not lead to formation of a pi-bond.
So why it is happening here
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Is it worth considering what would make an electron shell an entity of stability. That is the electron that takes you from 9 electrons to 10 electrons in a molecule could have been the one that took you from 8 to 9?
And the proton the same?
Why is it that electrons in outer shells feel the attraction of protons through the negative and thus repulsion of inner shells?
Is it because the electrons in inner shells while probably in a particular spot evens throughout the shell but instantaneously having particular location means electrons in outer shells can find the influence of a protons positive charge in the gaps like a back row spectator at the World Cup seeing the game by peering through the gaps left by the guys in front?
Could we therefore propose the electrons in outer shells will always have location driven by the momentary location of electrons in a inner shell?
Does that take us anywhere regards the current question?
Cheers
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If the antibonding pi-MO is the only interaction (whatever this should mean), the term "Bond" would simply be wrong because nothing would bond. (The energy would only get higher for smaller distances).
What you show in the coloured picture is the product from a DA-Reaction. The Interaction between the to p-orbitals is indeed destabilizing, however it's obviously not the only one. There are also two sigma-bonding interaction within this MO, leading to an overall bonding character.
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Is it because the electrons in inner shells while probably in a particular spot evens throughout the shell but instantaneously having particular location means electrons in outer shells can find the influence of a protons positive charge in the gaps like a back row spectator at the World Cup seeing the game by peering through the gaps left by the guys in front?
Could we therefore propose the electrons in outer shells will always have location driven by the momentary location of electrons in a inner shell?
I don't think we come very far when we only consider the classical picture. And then, an electron always "sees" the nucleus, this does not depend on the "inner electrons" (electrons are indistinguishable! ). This can be seen in the Hamiltonian. And also electrons do not have a "particular location" as in classical physiscs. There is, however, such a thing as instantaneous electron correlation, the holy grail of quantum chemistry. Also exchange interaction does not exist in the classical picture.
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If the antibonding pi-MO is the only interaction (whatever this should mean), the term "Bond" would simply be wrong because nothing would bond. (The energy would only get higher for smaller distances).
What you show in the coloured picture is the product from a DA-Reaction. The Interaction between the to p-orbitals is indeed destabilizing, however it's obviously not the only one. There are also two sigma-bonding interaction within this MO, leading to an overall bonding character.
Do you mean that, in some cases, pi bonds can be formed through the overlap of 2 p orbitals with opposite signs?
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What if the two atoms have more bonds than the pi? One strong bond could keep the atoms together despite the pi is antibonding, perhaps momentarily.
To synthesize cyclobutane, one path is to excite one ethylene molecule by UV light so it reacts with the other. Isn't that exactly the case of an antibonding pi where an other bond keeps both carbons together?
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Why is it that electrons in outer shells feel the attraction of protons through the negative and thus repulsion of inner shells?
Is it because the electrons in inner shells while probably in a particular spot evens throughout the shell but instantaneously having particular location means electrons in outer shells can find the influence of a protons positive charge in the gaps like a back row spectator at the World Cup seeing the game by peering through the gaps left by the guys in front?
If an atom has, say, 9 protons, and if the orbitals were concentric separated shells (they're not, they overlap), then the "inner" 2+6 electrons would still let the 9th feel one positive charge.
"instantaneously having particular location" is not what orbitals do. Orbitals are stationary, that is time-independant, and I say more crudely: immobile (they can have an orbital momentum though). But by writing one single wavefunction for all particles, quantum mechanics permits a similar situation. With a Psi(R1, R2) not being Psi(R1)*Psi(R2), you can have some simple distribution to find one electron in a zone (say, with spherical symmetry) and have a smaller probability to find a second electron near the first one - all that independently of time. A mental picture of this is difficult.
Then, you must factor in that electrons are fermions, so that Psi(R1, R2) = -Psi(R2, R1). This constraints the function Psi, but leaves many possible functions, and the electrostatic repulsion among electrons guides the choice of the function.
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"instantaneously having particular location" is not what orbitals do. Orbitals are stationary, that is time-independant, and I say more crudely: immobile (they can have an orbital momentum though). But by writing one single wavefunction for all particles, quantum mechanics permits a similar situation. With a Psi(R1, R2) not being Psi(R1)*Psi(R2), you can have some simple distribution to find one electron in a zone (say, with spherical symmetry) and have a smaller probability to find a second electron near the first one - all that independently of time. A mental picture of this is difficult.
Then, you must factor in that electrons are fermions, so that Psi(R1, R2) = -Psi(R2, R1). This constraints the function Psi, but leaves many possible functions, and the electrostatic repulsion among electrons guides the choice of the function.
Can we say that the electrons in inner orbitals, to any extent that they stand between in a interactive sense the nucleus and outer electrons, would resolve their pattern or position before those in further out shells, and that if there is a effect on the positions or patterns of electrons between shells if both equally could occupy one pattern or position the electrons in the inner shells would take priority?
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Priority to inner shells: I don't know how accurate this is.
Deeper orbitals are smaller, their bonding energy bigger, and so is the repulsion among electrons. That would suggest that outer electrons influence them less.
I'd also expect a full level to be stiffer than a partially filled one.
Though, orbitals in one atom overlap very much. It's even necessary in order to be orthogonal, in the sense <Psi1|Psi2>=0: take for instance the 1s which is positive everywhere, all others are positive and negative both where 1s has a significant density.
With the orbitals occupying much the same volume, the mental image of concentric shells can't be accurate, hence my caution.
To some extent, numerical simulations of atoms and crystals do suppose that higher orbitals influence lower ones little - finite elements for any significant number of electrons must be too hard for computers.
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The thought process that led me to think about this idea of some kind of priority in instantaneous characteristics associated with electrons in inner shells, whether those characteristics include spatial distance from a nucleus or not, taking your point, is that a sequence of determination of characteristics of electrons from inner to outer, outer being constrained in determined characteristics by the values just calculated for inner, would simplify any attempt at programming, as compared to electrons being equally determinative of the values of the characteristics of each other electron.
But I also take your point about the thought that x electrons exhibit a pattern of characteristics and x plus 1 electrons exhibit another pattern, and it may be to mask an opportunity to see a solution to "why" to only allow thinking in terms of concentric layered spatial zones of increasing distance from a central point.
Part of me though sees that as we are talking attraction and repulsion that a term in the function of attraction and repulsion must be a vector of the experience of that attraction and repulsion and connection and spatial proximity seems evident everywhere as a vector of such experience.
It's equally conceivable though that the arrival of another electron influences others from the outside then deeper as initially the change in spatial proximity is at the outer areas of the atomic structure as the new electron arrives?
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Let me see if I can get this right. If you draw the spin states for the MO’s of butadiene, they alternate, +-+-. This gives the lowest energy form, no nodes, in colors, all pink. The next higher energy level would be +--+ and a node between C2 and C3. If drawn with colors, this would half pink and half blue for this Psi state. Post two could be as stated by Corribus. This would be equivalent to the LUMO of ethylene, no pi bond, anti-bonding. If colors, pink and blue.
Post four shows the colors. In this case, the HOMO of butadiene and the LUMO of ethylene. This is not showing the possible spin states, just the psi states. For bonds to form, the colors must match.
I have contributed to this confusion in the past by referring to the spin states with plus and minus signs and then used plus and minus signs for the psi-states. I don't know if this is exactly what the poster is asking about, but it looks as though it might be. The symmetry rules for electrocyclic reactions seem to work quite well in predicting the resulting stereochemistry. One should know this. Because spin states can have a mirror image, they are not used, or it may be additional reasons.
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The thought process that led me to think about this idea of some kind of priority in instantaneous characteristics associated with electrons in inner shells, whether those characteristics include spatial distance from a nucleus or not, taking your point, is that a sequence of determination of characteristics of electrons from inner to outer, outer being constrained in determined characteristics by the values just calculated for inner, would simplify any attempt at programming, as compared to electrons being equally determinative of the values of the characteristics of each other electron.
(Some more points please, hard to read :( )
This is, somehow, done in most calculations that use a split-valence basis. But you still need to say goodbye to the idea of different electrons. you can't distinguish them! Thinking in Quantum Mechanical Terms, which obviously fit atoms better than classical mechanics, there is no such thing as "one electron close to the nucleus than another". look at the Hamiltonian and Psi expressed as an (superposition of) slater-determinants.
Let me see if I can get this right. If you draw the spin states for the MO’s of butadiene, they alternate, +-+-. This gives the lowest energy form, no nodes, in colors, all pink. The next higher energy level would be +--+ and a node between C2 and C3. If drawn with colors, this would half pink and half blue for this Psi state. Post two could be as stated by Corribus. This would be equivalent to the LUMO of ethylene, no pi bond, anti-bonding. If colors, pink and blue.
Post four shows the colors. In this case, the HOMO of butadiene and the LUMO of ethylene. This is not showing the possible spin states, just the psi states. For bonds to form, the colors must match.
I have contributed to this confusion in the past by referring to the spin states with plus and minus signs and then used plus and minus signs for the psi-states. I don't know if this is exactly what the poster is asking about, but it looks as though it might be. The symmetry rules for electrocyclic reactions seem to work quite well in predicting the resulting stereochemistry. One should know this. Because spin states can have a mirror image, they are not used, or it may be additional reasons.
What?
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For butadiene, the lowest energy is ++++, the second lowest ++--, second highest +--+, and highest +-+-.
http://www.chem.ucla.edu/harding/butadieneorbitals.html
http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch10/ch10-6-2.html
which corresponds to the general idea that electrons have a lower kinetic energy when the sign or phase changes slowly over distance - in other words, the energy usually increases with the wave vector, similar to E=(h/2pi)2k2/2m (but not always, especially at the top of a crystal's energy band).
Spin has only indirect relations with that. We're speaking about electron pairs in ground state, that is with ++++ and ++-- occupied in butadiene. From time to time, I've read attempts to explain chemical bonds by magnetic moment attraction, and this is flawed. Spin permits two electron to occupy the same favourable orbital, but the magnetic energy is far too weak to explain the bond.
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so what has this to do with the previous discussion?
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For butadiene, the lowest energy is ++++, the second lowest ++--, second highest +--+, and highest +-+-.
http://www.chem.ucla.edu/harding/butadieneorbitals.html
Spin has only indirect relations with that. We're speaking about electron pairs in ground state, that is with ++++ and ++-- occupied in butadiene. From time to time, I've read attempts to explain chemical bonds by magnetic moment attraction, and this is flawed. Spin permits two electron to occupy the same favourable orbital, but the magnetic energy is far too weak to explain the bond.
I was really trying to explain how the poster may have been confused. I'm not sure this is helping. Electrons are given four designations, shell, subshell, magnetic, and spin. In order for a bond to form, the spins must be anti-parallel. If one is drawing the lowest energy form of butadiene, the all pink energy state, the pi-electron spins must all be anti-parallel, or I presume so. You may see this written with up and down arrows, but you also see it written with + and - signs in the orbitals. This is different than the psi designation, though I think it is not unrelated.
If this is in no way related to the poster's question, then I'm sorry for creating any confusion. Sometimes, this does come up. I believe the psi states correspond to the different possible bonding and spin states for the electrons.
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It is possible - only you can decide - that you mix up the spin with the sign of the orbitals.
The + and - tell the sign of the molecular orbital near the individual atoms. Each of these orbitals contains two electrons of opposed spin, so these signs are NOT the spin orientation.
In the lowest molecular orbital of butadiene, ++++ means that the pi orbital has identical signs at all four carbons, for instance plus above the molecule and minus below, if the carbon chain is horizontal. Within this ++++ orbital, you find two electrons, one up and the other down. These signs do NOT describe the spin.
Same for the other occupied orbital at rest, ++--, which contains one electron pair with up and down orientation, both being spread identically among the four carbons.
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In addition to the excited ethylene example, where two atoms keep together despite some electrons are not paired, you have molecules with some unpaired electrons in the ground state. One example is the normal oxygen molecule 02 (=triplet)
http://en.wikipedia.org/wiki/Triplet_oxygen
which has two more electrons than the binding molecular orbitals can contain, so these electrons go to antibinding ones, one electron to each orbital because both orbitals have the same energy and electrons repel an other.
So normal oxygen at rest is a case where
- some electrons must populate antibonding orbitals
- two electrons are not paired