# How are the electrons of benzene actually configurated?

I found this diagram:

According to it, it is not true that the six electrons of carbons in the pi-orbitals above the ring are all identical.

But, how can it be the case, seeing that this molecule should have an order-6 rotational symmetry?

• The electrons are indistinguishable, you cannot look at the system and say "this electron is in the HOMO whereas that one is not" – orthocresol May 14 '16 at 16:32
• But I sure can count the number of electrons in the non-bonding pi-orbital? – Kenny Lau May 14 '16 at 16:32
• Yes but you can't say which electrons they are. There are two of them in each orbital but they are not individually distinguishable. – bon May 14 '16 at 16:34
• I don't think I ever said they are? – Kenny Lau May 14 '16 at 16:37
• The $C_6$ rotation can alter the phase of the molecule's electronic wavefunction, which is what is depicted in the above diagram, but it does not affect the electron density $|\psi|^2$ which is the quantity that must have invariance under $C_6$. In this case the wavefunction itself should also actually be invariant under the $C_6$ because it is a closed shell molecule (symmetry $A_{1g}$) – orthocresol May 14 '16 at 16:42

So, as I told you in chat what you have here is the orbitals of something which looks like, but which is not, benzene. It is the orbitals of one the all its two mesomeric form, it is so a static view.

Now let me show you the orbitals which are in the benzene. I only draw with the p-orbitals, I won't do the hybridation with the hydrogen it will be too difficult for me two draw and it will not be more effective.

How to draw a molecular diagram when the molecule has a lot of atoms?

First draw the diagram like if in the molecule there is just its squeleton (atoms which interact to create the shape of the molecule, for example with benzene only for the carbon atoms) and do like if these atoms were hydrogen atoms.

Second draw the correct orbitals with the same symetry of what you got with hydrogen just before.

You are obviously not obliged to do it but it is easier to see which orbitals in the fragments have the same symetry or not.

First step Here for the benzene I will draw the MO's diagram of benzene with $\ce{H_6}$ first, with a molecule which is a rectangle of hydrogen atoms and the other fragment is a longer dihydrogen molecule than the original one. After seeing which of them have the good symetries I get this

Second step

Here we want the same but with the p-orbital of the carbons so keeping the same symetries or antisymetries you get this

And then here you have the correct orbitals.

Be careful

The color on the orbitals represent the sign of the wave function which has no meaning in the reality. So you can make the black and the white anywhere you want if you obviously respect the molecule symetries. In this example I draw exactly the same orbital.

I say that because the orbitals I drawn for benzene are not exactly in the same color logic of the MO's diagram of $\ce{H_6}$ but as you can see each one have the same symetries or antisymetries.

On the MO's diagram you show they show the interaction of the p-orbitals in one of the mesomeric structure, like this

And then the symetries are not the same as in the benzene but at one of its mesomeric form which is a static view like this

In fact you need to imagine the same but with an interaction of all p-orbitals and the get a ring orbital for $\psi_1$, and the it explain the real shape of the benzene like this

Remember that orbital diagrams are done to think about what you observe in the nature and not to predict it. Well you can use quantum theory to predict things but it is an other subject.

I hope it can help !

• Some questions: 1. The s orbitals don't react, right? 2. The first picture is not really an MO diagram, but the possible configurations, right? 3. How do you know they are the actual mesomeric forms? Sure, they look symmetrical, but... – Kenny Lau May 15 '16 at 0:59
• Some answers: 1. The s orbitals of which atoms? Orbitals of hydrogens obviously interact if not the hydrogens will not make bond with carbons... 2.The first picture really is a MO's diagram, just it's not one of a real molecule. 3. There are because of the experiment you did. As I told you "diagrams are done to think about what you observe in the nature not to predict it" so if you want to make the MO's diagram of benzene you must need to know it is an aromatic coumpound. :-) – ParaH2 May 15 '16 at 13:24
• How do you go from, you know, "it is aromatic", to the pictures there? – Kenny Lau May 15 '16 at 13:27
• en.wikipedia.org/wiki/Aromaticity you should read about that. It is a complex subject. – ParaH2 May 15 '16 at 13:31
• Your depiction of the orbitals is almost the same as in the question, with some leeway they are both qualitatively in the realm of correct. (Even for a schematic representation, they miss certain key features.) It does, however not answer the question why these reproduce the molecular symmetry. And the part about mesomeric structures is completely wrong. An MO scheme cannot really show such things, and at the same time adhere to symmetry, which the OP's clearly does. The part that you have to know benzene is aromatic to construct the diagram is also completely false. – Martin - マーチン Sep 14 '18 at 17:48