February 17, 2020, 02:25:49 AM
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Topic: MO Cancellation/Symmetry Help?  (Read 281 times)

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Offline sciencenerd358

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MO Cancellation/Symmetry Help?
« on: September 07, 2019, 05:28:52 PM »
ORGO Help: Explanation for MO cancellation? Pretty confused as to which MO I would cancel in this situation. I specifically was stuck on MO #3 cancellation. I just do not understand why there needs to be symmetry and how to go about it when canceling specific MO’s and placing my nodes. My professor told me to see it as "folding a book," but I couldn't grasp onto the concept still. Thank you.

If you need more clarification lmk. ):

Offline spirochete

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Re: MO Cancellation/Symmetry Help?
« Reply #1 on: September 08, 2019, 02:40:24 PM »
You labeled the bonding orbitals slightly wrong. MO #3 is "non-bonding" and not bonding. It has zero bonding or anti-bonding interactions.

The orbitals always alternate between having planes of symmetry and 180 degree rotational symmetry (C2 axis). "Folding a book" seems to be another way of saying "has a plane of symmetry".

You cancel the MO's when you need to do so to give them symmetry. Placing the phases up or down can cause the orbital to have either a plane of symmetry or the rotational symmetry.

Offline hollytara

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Re: MO Cancellation/Symmetry Help?
« Reply #2 on: September 08, 2019, 10:40:54 PM »
When you generate the MOs from the AOs using LCAO, there are a number of rules that need to be followed. 

-each MO has to conform to the symmetry of the system.  With each symmetry element, the MO has to be either symmetric (symmetry operation leaves it the same) or antisymmetric (symmetry operation just changes all orbital signs)

- each MO has to be orthogonal to all other MOs.  This means that the net overlap has to go to zero - this is probably what you mean by "cancellation"

at higher levels - each AO has to contribute "1" orbital overall, and each MO has to be equal to "1" orbital overall.  At a qualitative level this isn't important. 

Why symmetry and orthogonality?  Mathematically, this is necessary for the equations to work.  Physically, these will be the lowest energy orbitals. 

For simple linear systems like you have here, the orbitals have increasing number of nodes as the energy increases.  The lowest energy orbital always has zero, then 1, then 2, and so on until there is a node between each pair of atoms (for N atoms, N-1 nodes).  These nodes have to be distributed symmetrically. 

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