1 and for ozone's 1 I could show a
diagram likeSimplified Bond Orders in Delocalized Systems
If there are problems understanding this new approach, please E-mail me at pk03@andrew.cmu.edu
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