Molecular Orbital Help

[ecobeco]

I ran into a question which asks if you were able to reduce ethyne to a di-anion, would it be diamagnetic or paramagnetic?

 

My guess is paramagnetic as it would be unpaired electrons being a di-anion... how could I draw the MO diagram to find this out? I don't really understand this stuff...

[collegedude22]

Basically you're letting the protons on either end of the molecule dissociate, so you're left with C2(2-). Each carbon has the same valence orbitals (2s, 2px, 2py, 2pz). So if you were to draw an MO diagram, you'd draw an energy line for 2s and 3 lines of higher energy for the 2p orbitals on each side of your diagram. Your two 2s orbitals will have a sigma bonding interaction of lower energy and a sigma antibonding interaction of higher energy. Each carbon atom's 2pz orbital will interact with the other in a sigma bond of lower energy and a sigma antibond of higher energy as well (this is all above the 2s sigma interactions in energy, though. When I say 'higher' or 'lower' I mean relative to the orbitals' original energies). Finally, your two 2px orbitals will interact in a pi bond (of lower energy) and a pi antibond (of higher energy), and your two 2py orbitals will also interact in a pi bond and pi antibond. Though the 2px and 2py orbitals start at the same energy as the 2pz orbitals, when they form a pi bond, the energy isn't as low a the 2pz sigma bond, and when they form a pi antibond, the energy isn't as high as the 2pz sigma antibond. They're sort of nested in between.

 

You should have, in order of lowest to highest energy, molecular orbitals of:

 

1. sigma (from the two 2s orbitals)

2. sigma star (from the two 2s orbitals)

3. sigma (from the two 2z orbitals)

4. 2 pi (1 each from the interaction of both 2px and then both 2py)

5. 2 pi star (1 each from the antibonding of both 2px and then both 2py)

6. sigma star (from the antibonding of the two 2z orbitals)

 

Now count your electrons:

2 carbons x (4e-/carbon) + 2e- (dianion charge) = 10 electrons

 

Start filling:

 

1. sigma (2)

2. sigma star (2)

3. sigma (2)

4. 2 pi (4)

5. 2 pi star

6. sigma star

 

So there are no unpaired electrons!

[ecobeco]

Ahh okay makes a lot more sense now, but shouldn't the pi bonds be forming between the Py and pz orbitals?? And the sigma between the px?

[collegedude22]

No, the z axis is always taken to be the axis with the highest rotational symmetry, so it is the axis that connects the nuclei of the two carbon atoms, so the other two 2px and 2py orbitals lie in the horizontal plane (if z is taken to be vertical).

 

If you want to get more in depth, the ethyne dianion molecule has D∞h point group symmetry, under which the z-axis (on which lies the 2pz orbital) transforms separately from the x-axis and y-axis set. Basically this just further confirms that the 2px and 2py are the two pi bonds and the 2pz is the sigma bond.

[Sara92]

Thank you so much! This helped heaps