Thursday, December 14, 2017

Formal Charge of Terminal Atoms


Would the formal charge of terminal atoms differ from the FC of the non-terminal atoms?


Say you had a sheet of graphite; inside the network covalent solid all the carbons are bound to one another, but the terminal atoms are obviously missing some C-C bonds since they are terminal.



enter image description here


Their formal charges would differ, correct?


What is the consequence of the differing charges?



Answer



The terminal carbons may or may not be missing some bonds. In any case, unless there is a charge on the surface/edge, the formal charge will always be zero.


Let's look at a single graphene ring to make it easier to see:



A graphene ring with single electrons on each carbon


In this first case, there are no atoms attached, and it is pure carbon. However, for the molecule to be electrically neutral, there must be a total of 6*4 valence electrons: 24. If we only have alternating single and double bonds, that gives us 18, so the other 6 have to go somewhere. In this structure I have drawn them as free radicals (although it is extremely unlikely they would be for long). In this case, you can see that the formal charge on each carbon is still zero:


# valence electrons - [$\frac{1}{2}$ # bonding electrons + #non-bonding electrons] = formal charge



4 - ($\frac{1}{2}$*6 + 1) = 0



A graphene ring with triple bonds


The second image is more likely if we don't allow any other elements to attach. Since the alternating double bonds are are really resonance structures formed by the overlap of p-orbitals, those lone electrons can do the same and form another $\pi$-bond, resulting in a triple bond. This idea is support by this article, which shows that under vacuum, they have observed that graphene sheets can form triple bonds along the edge. Examining each carbon, we again see that the formal charge is zero:


4 - ($\frac{1}{2}$8 + 0) = 0



Benzene


The third picture is what we commonly assume happens - something (in this case hydrogen) is attached. This would again give a total of four bonds, resulting in a formal charge of zero for each carbon.


4 - ($\frac{1}{2}$8 + 0) = 0


So the answer to your first question:




Their formal charges would differ, correct?



Is most likely "no" unless the surface is charged.


However, your second question:



What is the consequence of the differing charges?



is a good one, considering that regardless of formal charge, the terminal atoms will obviously have a different character.


How exactly it will change depends on the specifics, but in general, we can say that they are "more reactive."



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