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#### oneat

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##### huckel's rule - how to explain
« on: January 15, 2012, 02:58:12 AM »

Can somebody explain me how to explain huckels rule?
Or maybe how these orbitals http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch11/ch11-4-1.html

are created ?
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#### Honclbrif

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##### Re: huckel's rule - how to explain
« Reply #1 on: January 15, 2012, 07:33:46 AM »

This really would work better if I had a piece of paper to draw on, but I'll do my best to explain via text. To draw an orbital diagram for a simple conjugated ring, start by drawing the ring (pentagon, hexagon, heptagon, etc...) with a corner facing down (if there's an even # of sides there will be a point at the top and bottom, odd = point at bottom, flat side on top)

Wherever there is a vertex, there's an orbital, so draw a horizontal line there. The lower half of the digram are the bonding orbitals, the upper half are antibonding.  Now, determine the number of electrons in the conjugated system and fill in the diagram as you usually would.

For benzene we get:

__

__     __     antibonding
--------------------
__     __     bonding

__

Filling it in with 6 conjugated electrons: /| = up arrow, \| = down arrow
__

__     __     antibonding
----------------------
bonding
/| \|    /| \|

/| \|

All of the electrons are paired up in bonding orbitals, so stability if maximized.
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#### oneat

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##### Re: huckel's rule - how to explain
« Reply #2 on: January 15, 2012, 09:18:49 PM »

Ok but this is a instruction. But why it has to be done like that?
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#### Yakimikku

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##### Re: huckel's rule - how to explain
« Reply #3 on: January 16, 2012, 09:07:51 AM »

The method described by Honclbrif is a shortcut for making MO diagrams for cyclic pi systems (and props to Honclbrif for doing a good job with just text!).

If you want understand why that shortcut works, you need to go deeper into MO theory. In summary, here's how it works: for 'n' number of p orbitals in a cycle, you make linear combinations of those p orbitals using the symmetry of the cycle. You end up with a set of 'n' orbitals representing the pi system. There is always one completely bonding MO (no nodes) from combining 'n' p orbitals all in phase. This MO is lowest in energy and remains at the bottom of the MO diagram. With the polygon shortcut, this orbital is the reason why one vertex of the polygon points at the bottom. For cycles with 'n' even, you have a completely anti-bonding mo where the p orbitals are in and out of phase (see the mo at the top of the diagram in your link). This is the highest in energy. Other MO-s that are generated happen to be two-fold degenerate. This means that there are two orbitals generated with the same energy, but one alone is not sufficient to describe the MO with that symmetry. So the shortcut works because the horizontal alignment of other vertices in the polygon account for degenerate MO-s!

So there you have it! I'm not sure if my explanation was good for you, so please describe what parts are confusing so the community can further aid you. Note that this simplification with polygons only works for these cyclic systems. It might be worth looking into Huckel MO theory in more general.

If you are truly interested in really understanding this deeply, I recommend "Symmetry and Spectroscopy" by Harris and Bertolucci. It's a great introductory text for the foundation of all this, and it's about \$12. Highly recommended. (http://www.amazon.com/Symmetry-Spectroscopy-Introduction-Vibrational-Electronic/dp/048666144X)
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