[Winga]

I don't understand why the pi3 & pi5* orbital are illustrated like that.
I mean why the 2 p-orbitals are disappeared (2 nodes are formed) in each of these 2 MOs.

[GCT]

It probably means that they are non-bonding orbitals, although I'm not quite sure, I'lll have to read up on this

[Garneck]

I believe it means that the orbitals can be oriented in both ways and they will have the same energy. Try drawing it, you will see that you will get the same orientation.

[Winga]

I believe it means that the orbitals can be oriented in both ways and they will have the same energy. Try drawing it, you will see that you will get the same orientation.

I see.

Can I say that the pi2 and pi3 orbitals are degenerate because both of them have same no. of constructive interferences (4) and same no. of destructive interferences (2)?

[movies]

Can I say that the pi2 and pi3 orbitals are degenerate because both of them have same no. of constructive interferences (4) and same no. of destructive interferences (2)?

Well, no because they don't have the same number, at least the way they are drawn here.

The disappearance of the p-orbitals in 3 and 5 are, as you said, because there are nodal planes through those carbons.  You can come to this conclusion by visuallizing a p-orbital at one of those carbons.  On one side there will be a bonding interaction but on the other side there will be an anti-bonding interaction.  The two effects cancel each other out to give a node.  You can't call them non-bonding because the "bonding", "non-bonding", and "anti-bonding" terms are reserved for molecular orbitals and can't be applied to atomic orbitals.

[Winga]

On one side there will be a bonding interaction but on the other side there will be an anti-bonding interaction.  The two effects cancel each other out to give a node.

e.g. the pi2 MO, just consider the p-orbital on the front left, it has bonding interaction with the left p-orbital and anti-bonding interaction with the right p-orbital.
Why there is no node?

[Mitch]

pi1 has 0 nodes

pi2 and pi3 have 1 node and are degenerate

pi4 and pi 5 have 2 nodes and are degenerate

pi6 has 3 nodes

I know my Aromaticity well, ask whatever you want.

[Winga]

pi1 has 0 nodes

pi2 and pi3 have 1 node and are degenerate

pi4 and pi 5 have 2 nodes and are degenerate

pi6 has 3 nodes

I know my Aromaticity well, ask whatever you want.

I only understand pi1.

I wonder what do the empty spaces want to represent.

[Winga]

pi1 has 0 nodes

pi2 and pi3 have 1 node and are degenerate

pi4 and pi 5 have 2 nodes and are degenerate

pi6 has 3 nodes

I know my Aromaticity well, ask whatever you want.

In pi2, I find that it has 1 nodal plane, but I think it should have 2 nodes.
In pi4, I find that it has 2 nodal planes, but I think it should have 4 nodes.
In pi6, I find that it has 3 nodal planes, but I think it should have 6 nodes.
What did I mix up?

[Mitch]

In pi2, I find that it has 1 nodal plane, but I think it should have 2 nodes.
In pi4, I find that it has 2 nodal planes, but I think it should have 4 nodes.
In pi6, I find that it has 3 nodal planes, but I think it should have 6 nodes.
What did I mix up?

A node is a plane that bisects the molecule. If the plane bisects through an atom the p-orbital will have zero as its coefficient.

Your second image is misleading and I have deleted it.

[Winga]

I still don't understand why there are nodes passing through the two carbons of each case. How come?

[Mitch]

Because, if you do the math it'll pop out. The resulting wavefunction is mutually orthogonal to wavefunction 2 and constitutes a valid representation of E₁, that came from performing the projection operator on D_6h using p-orbitals as your basis set..

Now, who ever told you a node couldn't pass through atoms in the first place? Remember all of this is based on quantum mechanics and group theory, so don't assume you can make sense from it. If you continue with Chemistry, you will be able to come back to these basic ideas and fully understand what is going on.

[movies]

I think that you are counting the node that passes through the entire ring, perpendicular to the p-orbitals.  Don't count that one, it will just confuse you more.  It's the same in all of them anyway.

Only count nodal planes that are perpendicular to the benzene ring.

[Winga]

For pi3 orbital, I can accept it if one pair of p-orbitals on one side are turned upside down, so that all the 4 p-orbitals are parallel to each other. Then, the rest of the 2 p-orbitals are antiparallel to those p-orbitals, therefore, the nodes are formed at these 2 carbons.

[movies]

For pi3 orbital, I can accept it if one pair of p-orbitals on one side are turned upside down, so that all the 4 p-orbitals are parallel to each other. Then, the rest of the 2 p-orbitals are antiparallel to those p-orbitals, therefore, the nodes are formed at these 2 carbons.

What you have described is the pi4 orbital.  There is one additional nodal plane with that description because there is a node between each of the mismatched adjacent p-orbitals.

The way that the p-orbitals in pi3 are shown is correct.  The orientation of the p-orbitals that have been omitted are irrelevant because you get the same result either way.