# Huckel's rule for Aromaticity-what is n?

 P: 10 True, n is just any integer, but it has to signify sumthing! i mean, come on, we're talking science, you can't just use any 'n' straightway without challenging its credibility. Yes, it's true that there are hardly few books which mention what 'n' is. I don't think it's there even in Morrison and Boyd. To answer the question, it's fundamentally based on the MOT: aromatic systems have 4n+2 electrons, where n is the number of pairs of degenarate bonding orbitals. Consider Benzene as an example. we concern ourselves only with the pi-orbital system. Benzene has six atomic p-orbitals, which give six pi molecular orbitals (MO's): three bonding orbitals,say $$\psi$$1, $$\psi$$2, $$\psi$$3, and three antibonding, say, $$\psi$$4, $$\psi$$5, $$\psi$$6. the 6 p-electrons arrange themselves in the 3 bonding orbitals. $$\psi$$1 has no node, while $$\psi$$2 & $$\psi$$3 have one node each. Furthermore, The energy level of the orbitals increases with increasing number of nodes. Thus, $$\psi$$1 is at a lower energy level than $$\psi$$2 and $$\psi$$3, which share the same energy level, having one node each. $$\psi$$2 and $$\psi$$3 are said to be degenerate. Benzene, thus has one pair of degenerate bonding orbitals (i.e, n=1). For higher aromatic systems, the number of pairs of degenerate bonding orbitals increases. Napthlene has 10 atomic p-orbitals, thus, 10 MO's. The 5 bonding orbitals contain 2 pairs of degenerate orbitals and along with $$\psi$$1. $$\psi$$1 can contain 2 electrons, while each degenerate pair has a capacity of 4 electrons. Thus, the rule: 4n+2, which is the configuration having all pi-bonding orbitals completely filled, associated with extra stability.