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.

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Their formal charges would differ, correct?

What is the consequence of the differing charges?


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:

Free radicals

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

Triple Bonds

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

'Other Stuff'


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."

  • $\begingroup$ Makes sense. Thank you! I didn't think of the possibility of there existing a lone pair and neither did my teacher it appears. $\endgroup$ – Dissenter May 29 '14 at 2:55
  • $\begingroup$ You're welcome! Please note though, those are not lone pairs - they are free radicals (unpaired electrons). A lone pair would be stable but would result in a net charge. The free radical has no net charge, but would be extremely unstable. I was just showing it so that you could see that the formal charge would be zero in all cases. $\endgroup$ – thomij May 29 '14 at 2:57
  • $\begingroup$ Oops. Good point! $\endgroup$ – Dissenter May 29 '14 at 3:00
  • $\begingroup$ While the formal charge of carbon in benzene is zero, the actual charge is much closer to minus one. Unfortunately I cannot access the linked article, but I think that the concept of a triple bond applies to the edges of graphene only. How do they resolve the bonding picture in this case? Overlap must be terrible and $\ce{H4C6}$ is already highly reactive and unstable. How do they keep the structure stable? No Air? Matrix? $\endgroup$ – Martin - マーチン May 29 '14 at 9:18
  • $\begingroup$ Yes, in the article I linked the samples were under vacuum. The benzene ring is meant to represent the edge of graphene sheet - the triple bond would only be on the edge bonds. If I get some time today I'll change the drawing to make it more obvious. Of course, since these are resonance structures, it would really be something between a double and a triple bond along the edges. You make a good point about formal vs. actual charge - something that we often forget when looking at Lewis structures. $\endgroup$ – thomij May 29 '14 at 12:53

Yes, the formal charge at such atoms would be different. Such atoms, typically found at the at the edge of some crystalline or regular solid, are known as "defects" in the solid. Because they have an incomplete outer shell and lack a stable bonding configuration (bonding vacancies), they usually exist as a radical, anion or cation. They define an area of extremely high reactivity on the solid's surface and mark the spot where reactions often begin.

Edit: references to charges at defect sites

this link covers inorganic systems, see p.5 for some nice pictures.

this link covers graphene systems, from the "Conclusions" section; "We have performed scanning gate microscopy on graphene and observed such charge inhomogeneity and puddles due to extrinsic doping from possible sources such as metal contacts, graphene edges and surface residues."

  • $\begingroup$ For cations and anions the formal charge changes. For radicals it usually does not. However, the actual charge (in terms of electron density) may differ a lot. $\endgroup$ – Martin - マーチン May 29 '14 at 9:20

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