Simplified Bond Orders in Delocalized Systems

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In an attempt to re-explain the bond orders that emerge from delocalized molecular orbitals (MOs) such as we visited in the cases of ozone and "1,3 butadiene" (H2C=CH-CH=CH2), the following scheme is offerred:

The bond order per electron for the lowest energy MO was approximated in the case of 1,3 butadiene by saying that one electron was equally distributed among the three CC links. That approximation gives one-third electron per link or a bond order, for that MO, of one-sixth due to one electron. Two electrons in that MO, of course, would give a bond order of one-third for each link. In the case of ozone, one electron equally distributed between the two oxygen-oxygen links gives one-half electron per link or a bond order of one-quarter at each link. Again, that's for just one electron. If I invent an expression for these partial bond orders in delocalized MOs due to one electron -- let's call it partial MO bond order per electron -- then for 1,3 butadiene's 1 and for ozone's 1 I could show a diagram like

^---- +0.166------^--- +0.166-----^--- +0.166-----^      and       ^--- +0.25-----^--- +0.25-----^

respectively. (The little "--^--" above represents an atom's location.) These correspond (more or less) to the lecture slides below and the lowest energy delocalized MO in each case.

To get the bond order for each link, you need multiply the above partial MO bond order per electron by the number of electrons in the MO. What I will do, instead of indicating which regions are bonding, which are antibonding and which are nonbonding is to give a diagram of all the so-called partial MO bond orders per electron. Thus, ozone with its three delocalized MOs, the middle one of which was nonbonding and the upper one of which was antibonding everywhere, would appear as

^--- -0.25-----^---- -0.25----^     for 3*

^---- 0.00-----^---- 0.00-----^     for 2nb

^--- +0.25----^--- +0.25----^      for 1

An electron configuration for the pi-electrons that would be (1 )2(2)2 and each oxygen-^xygen pi-bond order would be (2)*(+0.25) + (2)*(0.00) = 0.5. When the underlying sigma bond is added in the total oxygen bond order is 1 + 0.5 = 1.5 as in lecture. If we had an excited state of ozone corresponding to an electron configuration written as (1 )2(2)1(3 *)1 then each oxygen-oxygen bond order would be (2)*(+0.25) + (1)*(0.00) + (1)(-0.25) = 0.25 and the total oxygen-oxygen bond orders would both be 1.25, as in lecture.

If that's relatively clear, the cases for 1,3 butadiene bond orders may be extracted from the following diagram for its partial MO bond orders per electron.

^---- -0.166------^--- -0.166-----^--- -0.166-----^      for 4*

^---- -0.25 -------^--- +0.50 -----^--- -0.25 -----^      for 3*

^---- +0.25 ------^--- +0.00 -----^--- +0.25 -----^      for 2

^---- +0.166-----^--- +0.166-----^--- +0.166----^      for 1     

For the unexcited molecule, the pi MOs corresponding to the ground state configuration (1 )2(2)2 would give bond orders (2 electrons per MO here) in the central bond of (2)(+0.166) + (2)(0.00) = 0.33. Adding the sigma bond gives a total in the center of 1.33. For the outer bonds (identical) we get (2)(+0.166) + (2)(0.25) = 0.83. Including the sigma gives 1.83 as obtained in lecture. Hence, if I give a diagram like those just illustrated with the partial MO bond order per electron you should be able to figure out bond orders for any configuration in them. I think it's easier to follow than the more arbitrary description where I would indicate what regions of a particular MO were bonding, antibonding, or nonbonding. Note that positive values (in red) corresonding to bonding situations, negative to antibonding, and zero to nonbonding. To test yourself, see what you get for (1 )2(2)1(3 )1. (The answer is below.)






Sol'n: For the central bond, (2)(+0.166) + (1)(0.00) + (1)(0.50) + 1 (for the sigma) = 1.83 and for the outer bonds, (2)(+0.166) + (1)(+0.25) + (1)(-0.25) + 1 = 1.33