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Charge density calculations of resonance hybrids of arenium ion

  1. Apr 10, 2012 #1
    In the diagram attached, the 3 resonance structures of arenium ion bonded to H electrophile are shown.I labelled the carbon atoms in one of the structures for convenience.

    From the diagram, the A and B carbons, and D and E carbons, have a double bond between them for two of the resonance structures; whereas the B and C carbons, and C and D carbons, have a double bond between them for only one of the resonance structures. From this can we safely assume that in the resonance hybrid, the bond between the A and B carbons, and D and E carbons have a high percentage of double bond character; compared to the bond between the B and C carbons, and C and D carbons, which have a higher percentage of single bond character?

    Also since the bond between A and B carbons (also between C and D carbons) are double bonds for 2 of the 3 resonance structures can we say that in the resonance hybrid, the charge between the A and B carbons is (2/3)(2) (as the single bond remains fixed and the 2 electrons of double bond become delocalized)? And that each of the carbon atoms has a charge of -(2/3).
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2012 #2
    help anyone?:confused:
     
  4. Apr 11, 2012 #3

    DrDu

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    I don't understand completely how you get the numbers in the last paragraph. You have a cation, so I would expect all charges on the carbons to be positive, not negative. What do you mean with charge "between" the A and B carbon?

    In general, you can make only qualitative statements from the consideration of possible valence bond structures, i.e. that some bonds are stronger than others or that there is probably more charge on one atom than on the other. For qualitative statements you would have to know how much each of the valence bond structures actually contributes and take into account also ionic VB structures with several charged atoms.
     
  5. Apr 11, 2012 #4
    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) (–[itex]\frac{2}{3}[/itex])= –([itex]\frac{4}{3}[/itex]), with each atom possessing –([itex]\frac{2}{3}[/itex]). Both atoms, A and B, lack –([itex]\frac{1}{3}[/itex]) charge from becoming neutral, giving each of them a net positive charge of +([itex]\frac{1}{3}[/itex]). Same argument goes for the other carbons with the net result that A, C and E carbons each gain a net positive charge of +([itex]\frac{1}{3}[/itex]), 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?
     
  6. Apr 12, 2012 #5

    DrDu

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    Yes, but isn't that a rather complicated way to obtain that atoms A, C, E have 1/3 of a positive charge when all resonance structures have equal weight? This can be directly read off from the structures.
     
  7. Apr 12, 2012 #6
    ....yeah..I kind of figured that out later........:biggrin:

    Thanks DrDu.
     
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