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Topic: cyclopentadienyl anion and orbitals  (Read 4711 times)

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Offline stb05

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cyclopentadienyl anion and orbitals
« on: June 16, 2016, 12:59:31 PM »
Hello, can someone help me? I dont understand what happens with the orbitals in a cyclopentadienyl anion in his picture. im trying to find something about it know for 2 days.
Can someone explain me what happens with the orbitals and why they change +/- directions.

And how this is done, and how i can figure out this "change of the orbitals" of every molecule.

And why is in 2 of the forms the one orbital not seen? Where is it and in which direction does it stand.

Thanks
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Offline Enthalpy

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Re: cyclopentadienyl anion and orbitals
« Reply #1 on: June 19, 2016, 02:04:18 PM »
These represent molecular orbitals of the same molecule. Some of these molecular orbitals are occupied, others may be empty (and in an excited state, filling wouldn't necessarily favour the lowest energy).

The molecular orbital is represented as the orbitals of the constituting atoms that sum with varied orientations. This is a simplification. But it's the same story as bonding and antibonding between two atoms, just more general.

You could start with ethylene and butadiene, later go to cyclic molecules and ions.

Offline orgopete

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Re: cyclopentadienyl anion and orbitals
« Reply #2 on: June 19, 2016, 04:25:27 PM »
I'm not too strong on this stuff either, but on the left is the lowest energy state, the next is the next higher energy state. I think it is represented as it is because the missing one could be either way, and because of symmetry, you end up at the same level. Etc.
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Offline Babcock_Hall

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Re: cyclopentadienyl anion and orbitals
« Reply #3 on: June 20, 2016, 10:52:31 AM »
I would not say that the orbitals change direction but rather that they change algebraic sign.  In between where the sign changes is called a node, a place of zero probability of finding an electron.  Nodes are indicated with a dashed line.  I am not very knowledgeable about MO theory, but my understanding is that the more nodes in an MO, the higher its energy, all else held equal.

Offline Corribus

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Re: cyclopentadienyl anion and orbitals
« Reply #4 on: June 20, 2016, 11:16:41 AM »
They're not even atomic orbitals at this point. The drawing of p-orbitals on the carbon skeleton represents the relative contribution of the atomic orbital on each respective carbon center to the (globally delocalized) molecular orbital for a particular energy eigenvalue. In this case only the sign/phase but not the magnitude (with the exception of the nodes) of the respective p-orbital  is represented. This is somewhat misleading, but it's important to realize that the p-orbital drawings are just for bookkeeping: again, the electronic π structure does not involve pz-orbitals any longer - each state has a single molecular orbital that is constructed from linear combinations of atomic p-orbitals.

It is true that in general the number of nodes scales with the energy of the molecular orbital, as this also correlates to the amount of "antibonding" character. It becomes a little complicated here because of the degeneracy of the high level energy states; the nodes (dashed lines) here are somewhat arbitrarily places. Note that because of the symmetry of the molecule you can effectively rotate the nodes anywhere you want. When a node passes through a carbon center, there is no p-orbital drawn because the p-atomic orbital on this carbon does not contribute to the molecular orbital (it's contribution is zero). This isn't a binary proposition, though - it's not zero or one. A better way to draw it would be to scale the size of the drawn p-orbital with the relative magnitude of the contribution, but this becomes even more confusing to people who don't really know what they're looking at.

The plus-minus just refers to the phase of the atomic-orbital contributions with respect to each other. If you consider a simpler molecule like ethene, there are two carbon atomic p orbitals that combine to form two molecular orbitals. Each p orbital has two lobes of opposite phase (which we call arbitrarily + and -). When the two carbons are brought together the respective p orbitals can match phases (+ lobes interacting with + lobes and - lobes interacting with - lobes) or match opposite phases. There is some mathematical formulation to this but it's basically the equivalence of constructive and destructive interference in wave interaction. Anyway, in the constructive combination you get a bonding orbital with no nodes and with the destructive combination you get an antibonding (higher energy) orbital with one node between the two carbon centers where there is no electron density. The node also marks a point where the phases change (the + and - of the molecular orbital switch positions).

See here for an example figure: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-6-1.html

What happens in larger molecules like the one you are asking about is basically the same thing, except you have more possible combinations (and usually fractional contributions) because you have more p-atomic orbitals involved.
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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