Section 9.6 and delocalized pi-bonding systems

[Zellmer, Robert]

I've received questions concerning the number of electrons in the pi-bonding
system for molecules which are planar and for which you can draw resonance
structures.  This is discussed in Section 9.6.  They discuss benzene and how
to determine there are 6 electrons in the delocalized pi-bonding system.
Sample Exercise 9.7 does this for nitrate, NO3^- (which would apply to SO3
and CO3^2- since they all have 24 valence electrons, are trigonal planar, with
sp^2 hybridized atoms, all of them, and have 3 resonance structures conforming
to the octet rule).  I did some examples of this during the review session on
Sunday so you can watch that video.

Exam Prep question 9.14 asks you to do this for C2O4^2-.  Draw out the
molecule with just sigma bonds drawn in.  All the atoms are sp^2 hybridized.
The O atoms use two of their sp^2 hybrid orbitals for lone pairs and one
sp^2 hybrid orbital to overlap with an sp^2 hybrid orbital on the C atom
they're attached to form sigma bonds between the C and O atoms.  There
are a total of 34 valence electrons.  Each C is bonded to the other C
and two O atoms.  That means there's a sigma bond between the two C
atoms and sigma bonds from each C atom to two O atoms.  That's a total
of 5 sigma bonds accounting for 10 electrons.  Each O atom has 4 electrons
(2 lone pairs on each O atom in sp^2 hybrid orbitals).  There are 4 O atoms.
That's a total of 16 electrons.  That's a total of 26 of the 34 valence electrons
involved in sigma bonds and the lone pairs on the O atoms.  That leaves
8 electrons in the delocalized pi-bonding system that encompasses all
6 atoms in the molecule.

I hope this makes sense.  It's hard to describe in words w/o pictures.
Watch the review video (at the Lecture module in Carmen) for examples.
If I remember correctly, I went over EP 9.14.

Dr. Zellmer

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