ChemNoob
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VSEPR Help
Hey everyone,
Im currently working on a project for Grade 11 Chem and we have to make a 3D model of our molecule- mine is Kevlar (C14H10N2O2). I was wondering if
anyone would be able to help me figure out the shapes, or more verify that what I have is correct.
There is a picture of the molecular structure of Kevlar. So far, Ive determined that the benzene rings are all trigonal planar pieces, correct? Im not
too sure about the far left side though, where the nitrogen is bonded to C, H and technically another C when the polymer is attached to another one.
Would that shape be considered trigonal pyramidal since the N is being bonded to 3 atoms and there is one lone pair of electrons? I was also wondering
if anyone could help me out with the part in betwenn the two benzene rings, or point me in the right direction. Thanks in advance! 
[Edited on 29-5-2011 by ChemNoob]
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UKnowNotWatUDo
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My guess is that your teacher wants you to correctly identify the bond angles and configurations that are represented by this picture. The nitrogen
and carbon in between the two benzene rings is identical to the carbon on the right hand side and the nitrogen on the left hand side of the picture.
So the geometries will be exactly the same. So basically you are correct. However if you wish to hear a more "exact" analysis of the molecule, read
below. Otherwise good luck with your model.
Your are correct that the nitrogen would be trigonal pyramidal...at least sometimes. The bond between a nitrogen and a C=O, a carbonyl, is called an
amide. This means that there is a resonance structure where the nitrogen's lone pair attacks the carbon, pushing one of the C=O pairs onto the oxygen.
This creates a C=N bond and changes the C=O to a C-O. The nitrogen acquires a positive charge and the oxygen now has a negative charge. This causes
the hydrogen to jump from the nitrogen to the oxygen. The carbon will still have bond angles of 120 degrees but now so will the nitrogen. Actually the
nitrogen's long pair (where the hydrogen used to be) will cause the bond angle between the C=N and the C-N to be slightly less than 120 degrees, but
only slightly. So the actual structure of the molecule would be somewhere between these two configurations, being closer to one or the other or
exactly in between the two depending on where the equilibrium lies.
[Edited on 5/29/2011 by UKnowNotWatUDo]
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ChemNoob
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Alrighty, thanks a lot. Definately helped me out
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