[Ksc]

Hello!

I have a question regarding polyatomic MO-diagrams like NH3, SF6, CH4 and so on.. what is a bit unclear for me is the filling in of electrons. At the examples i've seen they only take one electron from each of the polyatomic orbitals, i.e 3 electrons from H3, 6 from F6 and 4 from H4. That is perfectly normal for the hydrogen given that it only has 1 valence electron.. but for F6 I would assume that it'd be 6*7 =42 valence electrons. Could anyone give me a heads up?

[Hunter2]

You have to write down the electronically structure of fluorine.

It is [He] 2 s² 2 p⁵

You can see from the occupation of the p-orbitals only one is left what has space to get an electron to form flouride.

[Ksc]

Ok, thanks. But how do you know that we "want" to form fluoride?

[Hunter2]

I don't understand your question. Fluorine like to get the 8 electrons on the outer shell, so it will grab a electron from other elements to get it fulfilled. And the result is a fluoride.

[Ksc]

Ok, i'm not sure if I understand it.. It's a different method compared to diatomic molecules? Let's say CO for example - there you can just count all the valence electrons(10 combined).

[unsu]

The molecule of SF₆ has 48 valence electrons. Instead of making full MO diagram for all valence electrons, we can construct a partial MO diagram. For six fluorine atoms, we can only consider those p-orbitals that are pointed directly to the sulfur atom (2p_z). These orbitals are called radial orbitals. To form six bonds, you consider 6 valence electrons of the sulfur and one electron from each fluorine atom that occupies 2p_z orbital. So we have total of 12 electrons participating in the formation of S-F bonds. The other (paired) fluorine electrons in 2p_x and 2p_y orbitals are not pointed directly to the S atom and are not considered in the partial MO diagram approach. They do not contribute significantly to the formation of the six S-F bonds.

If you want to construct full MO diagram as for diatomic molecules, then yes, you need to use all 7 valence electrons from each fluorine, derive the reducible representations for Г_2s, Г_2px, Г_2py, Г_2pz, find their symmetries, etc. etc. Generally, this if very difficult and not really necessary.

This is how I understand it but I would like someone else to comment on this. Please also review Section 5.7 "Molecular orbital theory: learning how to use the theory objectively" in Housecroft/Sharpe Inorganic Chemistry textbook, you could find some answers there.

[Ksc]

Thanks alot!!  that cleared things up quite a bit. I'm using the book Inorganic Chemistry by Shriver & Atkins and I found it pretty vague on that subject. But, what about Sulfurs 2px and 2py orbitals?

[unsu]

In the octahedral point group the sulfur's 3s orbital possesses A_1g symmetry and the three 3p orbitals (x, y and z) is a degenerate set of T_1u symmetry.

The ligand group orbitals (LGO) constructed from 2p_z orbitals of fluorines have A_1g, E_g and T_1u symmetries. So, there are six ligand group orbitals.

Ok, 3s orbital of sulfur (a_1g) interacts with a_1g LGO making one bonding and one antibonding orbital.

Three 3p orbitals of sulfur (t_1u) interact with three t_1u ligand group orbitals.
"t" means triply degenerate (3 orbitals, same energy). This produces three bonding and three antibonding orbitals.

The LGO with E_g symmetry do not match the symmetry of the AO of the sulfur atom, so they form the non-bonding MOs.

So, now we have the following MO: one a_1g, three t_1u, two e_g, three t_1u* and one a_1g*

Now, count the electrons: 6 valence electrons from S and one 2pz electron from each F = 12 total electrons. They will occupy  a_1g, t_1u and e_g molecular orbitals.

You can consider the unused fluorine's 2s, 2p(x) and 2p(y) as lone pairs (non-bonding). 2s orbital is just too far in energy from the sulfur AOs. And 2p(x) and 2p(y) they either do not match the symmetry of sulfur AOs or, for t(1u) set, their overlap with the sulfur AO is inefficient (due to their positions).