Benzene, C6H6, consists of six member ring of sp2 hybridized p orbital. How many (pi)_2p bonding, anti bonding, and nonbonding molecular orbitals exist for benzene?
This may not help much but it's all I could find.
http://en.wikipedia.org/wiki/Benzene#mediaviewer/File:Benzene_Representations.svg
To determine the number of π (pi) molecular orbitals for benzene, we need to consider its molecular structure.
Benzene consists of a six-membered ring of carbon atoms, each with one hydrogen atom attached. The carbon atoms in benzene are sp2 hybridized, meaning that each carbon atom has three sp2 hybrid orbitals and one unhybridized p orbital.
The six p orbitals from the carbon atoms in the ring overlap sideways, forming a ring of delocalized π electrons above and below the plane of the molecule. This delocalization gives rise to π molecular orbitals.
In benzene, there are a total of six carbon atoms, and each carbon contributes one unhybridized p orbital. Therefore, there are six p orbitals available for π bonding and antibonding molecular orbitals.
To determine the number of π bonding (π2p) and antibonding (π*2p) molecular orbitals, we use the formula:
Number of π bonding (π2p) molecular orbitals = Number of p orbitals / 2
Number of π antibonding (π*2p) molecular orbitals = Number of p orbitals / 2
For benzene, the number of π bonding and antibonding molecular orbitals would be:
Number of π bonding (π2p) molecular orbitals = 6 p orbitals / 2 = 3
Number of π antibonding (π*2p) molecular orbitals = 6 p orbitals / 2 = 3
So, there are three π bonding (π2p) molecular orbitals and three π antibonding (π*2p) molecular orbitals in benzene.
It's important to note that the remaining p orbitals that are not involved in bonding are called nonbonding molecular orbitals. In this case, since all six p orbitals are involved in the formation of π bonds, there are no nonbonding molecular orbitals for benzene.