# Benzene - MO states [closed]

When constructing the different combined molecular orbital combinations (from the original 6pi orbitals) we normally say that p orbitals "combine" to produce the different molecular orbitals and antibonding orbitals etc.? I'll use just benzene as en example:

1. For the lowest energy molecular orbital (pi1, is it accurate to say it's because the electrons in particular are able to take "advantage" of the p orbitals of the carbon atoms and then can occupy the p orbitals equally, delocalize and share the p orbitals? (like in this manner they the p orbitals are "combining")? Because when I first saw that orbitals are combining, I thought it's inherently due to the electrons actually resulting in those orbitals.

2. Then how will the next two electrons "cause" to create a node in the next highest energy molecular orbital (pi 2 and pi 3)? I know that some of the p orbitals may be out of phase to create a node. But how does this conceptually happen with the electrons? Maybe the two electrons move, and one of the p orbitals, go out of phase when they realize the "pi 1" is full of electrons.

Sorry for the long worded question, yea I'm just conceptually looking for how these MOs form.

• If the p orbitals can overlap one another it is possible for the electrons to lower their overall energy by forming molecular orbitals. The delocalisation lowers the energy; think of a particle in a box, the bigger the box the lower the energy. All the orbitals are formed at the same time, the benzene symmetry determines where the nodes are going to be. The lowest energy has most delocalisation, the next lowest only half as much etc. The six pi electrons fill the lowest 3 orbitals with three remaining unfilled. Overall the energy is lowered. Look at the Huckel model for a simple calculation. – porphyrin Jun 23 at 8:02

Let's stay with your example of benzene, $$\ce{C6H6}$$. There are six carbon atoms bound in cyclic fashion with each other, and maintain each one bond to a hydrogen atom, too. This is explained conceptually that for each carbon atom the outmost $$s$$ and three $$p$$ atom orbitals were hybridized to yield three $$sp^2$$ of equal energy, leaving one $$p_z$$ orbital «unaffected».

If you draw the molecular structure of benzene as flat, regular hexagon, the $$p_z$$ orbital may be assumed to point orthogonally out of the plane of drawing. Joining the six carbon atoms will create the benzene molecule. For $$n$$ (integer number) of atomic orbitals $$p_z$$, there will be $$n$$ molecular orbitals, and because they are not of rotational symmetric around the C-C bonds, they are called $$\pi$$-bonds. In case of benzene, starting with 6 $$p_z$$ atomic orbitals, there will be 6 $$\pi$$ molecular orbitals ($$\pi_1, \pi_2, \pi_3; \pi^*_4, \pi^*_5, \pi^*_6$$). Compared to the energy level of the atomic orbitals, the energy levels of the molecular orbitals either are lower, or higher and contribute to the molecular energy with either bonding or anti-bonding character. (Antibonding $$\pi$$ orbitals have this asterisk, too.)

You recall Hund's rules and Pauli principle; specicially:

• if there are multiple orbitals of equal energy level, each one will fill first with one electron only
• at maximum, one orbital may accomodate two electrons at maximum

An useful tool to sort these molecular $$\pi$$-orbitals is the Musulin Frost cycle. Draw the benzene as a regular hexagon with a tip to the bottom. Either keep in mind, or mark the corners as atoms, too:

(altered from doi 10.1063/1.1698970)

As you see, energy wise, $$\pi_2$$ and $$\pi_3$$ are degenerate but (as $$\pi_1$$ as well) below the level of the atom orbitals we started (the sketch symbolized this by 0). The antibonding $$\pi$$ orbitals are higher than this level. Again, there are two ($$\pi^*_4, \pi^*_5$$) which are degenerate because of the symmetry in benzene. And you may fill the orbitals one by one, schematically by the added blue lines. This representation equally helps you to identify HOMO and LUMO levels accessed, e.g. by UV-Vis spectroscopy.

As shown in the example below, it equally allows you to understand the energetic difference between cyclopropyl cation and anion, too:

As for the name (cycle): It may be easier to draw a cycle around the (cyclic) molecule in question with a centre at an energy level = 0 as in the example below and to quantify the energy differences, too:

(loc. cit.)

• Thank you so much for the quick response! So in particular to how (like the process in which) the orbitals form, do the pi orbitals form sequentially or at the same time (pi 1, then pi 2, then pi 3, etc.). I imagine the electrons will be "in motion" from the Pz orbitals, forming the combined pi orbitals. So maybe 2 electrons form a pi2 orbital, but then since pi1 is of lower energy, they'd form actually go to the pi1 configuration, then pi 2 and pi 3 would form I imagine around the same time. – Matt Jun 24 at 15:03