# Charge density calculations of resonance hybrids of arenium ion

by tasnim rahman
Tags: arenium, calculations, charge, density, hybrids, resonance
 P: 70 Charge density calculations of resonance hybrids of arenium ion For the last paragraph, I was trying to calculate the negative charge possessed by each carbon atom, if we assume that each resonance structure contributes equally and that the 4 pi-bond electrons are delocalized over the structure; based on the resonance structures. From the diagram, there is a pi-bond between the A and B carbon atoms for two of the three resonance structures, indicating that the bond between A and B in the real molecule has a high degree of double bond character. For calculation of the charge possessed by A and B, due to the delocalization of the pi-bond electrons; two electrons form a pi-bond between A and B for two out of three resonances, therefore the average negative charge possessed by the two atoms, through their bond (other than the single bond), in the real molecule, is (2) (–$\frac{2}{3}$)= –($\frac{4}{3}$), with each atom possessing –($\frac{2}{3}$). Both atoms, A and B, lack –($\frac{1}{3}$) charge from becoming neutral, giving each of them a net positive charge of +($\frac{1}{3}$). Same argument goes for the other carbons with the net result that A, C and E carbons each gain a net positive charge of +($\frac{1}{3}$), and the B, D and F carbon atoms remain neutral (Note: B and D carbon also possess some charge density from bonding with C carbon). This also causes delocalization of the positive charge. Right?
 Quote by tasnim rahman From the diagram, there is a pi-bond between the A and B carbon atoms for two of the three resonance structures, indicating that the bond between A and B in the real molecule has a high degree of double bond character. For calculation of the charge possessed by A and B, due to the delocalization of the pi-bond electrons; two electrons form a pi-bond between A and B for two out of three resonances, therefore the average negative charge possessed by the two atoms, through their bond (other than the single bond), in the real molecule, is (2) (–$\frac{2}{3}$)= –($\frac{4}{3}$), with each atom possessing –($\frac{2}{3}$). Both atoms, A and B, lack –($\frac{1}{3}$) charge from becoming neutral, giving each of them a net positive charge of +($\frac{1}{3}$). Same argument goes for the other carbons with the net result that A, C and E carbons each gain a net positive charge of +($\frac{1}{3}$), and the B, D and F carbon atoms remain neutral (Note: B and D carbon also possess some charge density from bonding with C carbon). This also causes delocalization of the positive charge. Right?