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Topic: Constructing MO diagrams  (Read 8284 times)

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Offline Donaldson Tan

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Constructing MO diagrams
« on: January 25, 2005, 05:38:57 AM »
My chemistry tutor tells me that the general chemistry my chemical engineering department covers in two terms is covered in equal depth by my school's chemistry department. Anyway, I am new to MO theory and been trying to construct MO diagrams. It draws to me that MO theory is based on experimental results and not some qualitative theory, so unlike valence bond theory.

In constructing MO diagrams, I am not too sure how the high the antibonding orbitals and how low the bonding orbitals should be. I am not too sure how these orbitals should be placed relative to each other. eg. in drawing the MO diagram for H2, there gotta be a reason better than the fact that H2 exists, therefore the magnitude of the relative height of the antibonding orbital to the hydrogen 1s is less than that for the bonding orbital.

Moreover, in constructing a heteronulear diatomic system. eg. CO, it's pretty eveident that the 2s and 2p of oxygen is lower than that of carbon because of oxygen's higher proton number. However, given the case where I have two different energy levels to construct a single a molecular orbital, where should the pi and pi* be placed relative to it's sigma and sigma*. Perhaps someone can educate me and provide a reason why it should be so. thank you in advance.
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Offline Mitch

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Re:Constructing MO diagrams
« Reply #1 on: January 25, 2005, 12:16:21 PM »
First off, Welcome to MO Theory. :)

MO Theory is based quanitatively just like VSPER is, the one advantage of MO theory is that it actually matches and predicts experimental results, VSEPR can not.

Antibonding orbitals will always be higher than bonding orbitals. I can't think of a single exception at the moment. This is why He2 does not exist; although, one summer my project was to make it  ::) .

Depending on the electronegativity of the atom it will depend whether the 2pi is lower than the 2sigma.
« Last Edit: January 27, 2005, 08:59:26 AM by Mitch »
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Offline Donaldson Tan

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Re:Constructing MO diagrams
« Reply #2 on: January 26, 2005, 05:23:08 AM »
I dont understand how 2pi will be lower than 2sigma due to electronegativity.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Demotivator

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Re:Constructing MO diagrams
« Reply #3 on: January 26, 2005, 11:37:55 AM »
The normal or generic order is:
sigma2s sigma2s* sigma2p sigma2pi sigma2pi* sigma2p*.

However, complications arise concerning the light second row elements like B2, C2, N2.  These elements have p and s atomic orbitals that are close enough in energy (due to smaller Zeff). As a result, the MOs that are formed are close enough in energy  so that pairs of the same symmetry interact and mix further to introduce some sp character. This extra mixing alters the energies.
For one pair, down for  sigma2s, up for sigma2p.
For the other pair, down sigma2s*, up sigma2pi*.
The net order (the sigma2p sigma2pi  flip):
sigma2s sigma2s* sigma2pi sigma2p sigma2pi* sigma2p*

Heavy diatomics like O2 and F2 mix less and have the normal pattern. Heterodiatomics like CO are more difficult to predict. CO happens to have enough sp mixing to get the non normal order.
« Last Edit: January 27, 2005, 08:59:41 AM by Demotivator »

pizza1512

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Re:Constructing MO diagrams
« Reply #4 on: January 27, 2005, 08:44:47 AM »
What is MO?...

 ???

Offline Mitch

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Re:Constructing MO diagrams
« Reply #5 on: January 27, 2005, 09:00:26 AM »
Molecular Orbital Theory.
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