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Topic: Molecular orbitals of butadiene  (Read 315 times)

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Offline bananizer2222

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Molecular orbitals of butadiene
« on: December 30, 2021, 11:38:38 AM »
Hi everyone!
I have some trouble understanding π-bonding with p-orbitals and the different molecular orbitals of butadiene.

First here are some things I think I understand:
1. p-orbitals have both a positive and a negative phase, that have to do with wavefunction, so orbitals can interfere constructively or destructively
2. constructive interference increases electron density between nuclei making bonding orbitals, destructive interference creates antibonding orbitals, or a node
3. electron density at nodes is zero, there is a horizontal node on the plane of the molecule, vertical nodes increase with the number of molecular orbitals
4. larger molecular orbitals mean lower energy, the higher energy an electron has the shorter its wavelength is, and this also correlates with the nodes
5. the C1-C2 and C3-C4 bonds are longer then the π-bonds of ethylene, and the C2-C3 bond is shorter then a regular σ-bond
6. the molecule is planar, and there are rotational barriers between C1-C2, C3-C4 and also C2-C3
7. the energy of an electron is determined by the orbital it occupies, so the energy differrence between π1 and π2 is a result of the number p-orbitals interacting in-phase and out-of-phase
8. the number of molecular orbitals equals the number of atomic orbitals
Please correct me or add to these in any way you feel like!

What causes my mindlock is that if 4 p-orbitals all housing 4 individual electrons all interacted in-phase with eachother (and this is possible because the original p-orbitals are large enough to overlap), forming the largest, lowest energy molecular orbital π1, which is full, housing 2 electrons;
than have can they interact again, this time with a node, and two of the p-orbitals with flipped phases in π2?
Also can electrons cross through the nodes? Is there 1 electron on both sides of the molecule in π1? Are the electrons on opposite ends also on opposite sides in π2?
Do the 4 p-orbitals donate less and less of themselves while creating the MOs making them smaller and higher energy? If they donated equal parts that would make them the same energy right? If that is the case, does that mean they can donate their different phase lobes wherever they want?

Am I missing something, or just trying to think more into it?

Offline Babcock_Hall

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Re: Molecular orbitals of butadiene
« Reply #1 on: December 30, 2021, 04:04:55 PM »
I gave what you wrote a quick read.  Here are my thoughts:

2.  A node is a place of zero electron density, and it is also a place where the wavefunction changes sign.  I would not directly connect it with anti bonding orbitals.
3.  As the number of nodes increases, the energy of the orbital increases.
4.  The molecular orbitals of butadiene cover all four carbon atoms; therefore, I don't see one as being larger or smaller than another.
7.  In molecular orbital theory, there are no longer any p orbitals.  In some sense they were used up in creating the molecular orbitals.

Quantum mechanics is full of paradoxes; the electron has no trouble crossing over the node as I understand it.

Although molecular orbitals are created out of atomic orbitals, the amounts that each AO contributes are not necessarily equal to one another.

Offline Corribus

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Re: Molecular orbitals of butadiene
« Reply #2 on: December 30, 2021, 04:11:32 PM »
What causes my mindlock is that if 4 p-orbitals all housing 4 individual electrons all interacted in-phase with eachother (and this is possible because the original p-orbitals are large enough to overlap), forming the largest, lowest energy molecular orbital π1, which is full, housing 2 electrons;
than have can they interact again, this time with a node, and two of the p-orbitals with flipped phases in π2?
See your point number 8. If you start with four atomic orbitals, then they combine to form four molecular orbitals. As a first approximation, linear combinations are taken, which gives rise to the various configurations and nodes.

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Also can electrons cross through the nodes? Is there 1 electron on both sides of the molecule in π1? Are the electrons on opposite ends also on opposite sides in π2?
This is classical physics thinking. In quantum mechanics we can't specify where the electrons are exactly, only where we expect to find them on average.

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Do the 4 p-orbitals donate less and less of themselves while creating the MOs making them smaller and higher energy? If they donated equal parts that would make them the same energy right? If that is the case, does that mean they can donate their different phase lobes wherever they want?
Not exactly sure what you are asking here, but there are ways to estimate the relative contribution each atomic orbital donates to the molecular orbital. When summed over all the molecular orbitals, the atomic orbitals contribute equally, but any specific molecular orbital will have different contributions from each atomic orbital. Calculating these takes some effort and comes down a lot to symmetry. Sounds like it is a bit beyond your current level. But the implication is that some nuclear centers have more electron density than others, which impacts reactivity.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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