On this Wikipedia page, the electrical conductivity of various materials are given in the third column ($\sigma \text{ (S/m) at 20}^\circ \text{C}$). I am interested in the entry for Carbon (graphite):

  • $2$ to $3 \times 10^5 \text{ S/m} \perp \text{basal plane}$
  • $3.3 \times 10^2 \text{ S/m} \parallel \text{basal plane}$

Why is conductivity lower parallel to the plane than in the direction perpendicular to the plane?

I am surprised by the relative magnitudes of these $\sigma$ values because I thought that one of the amazing things about graphene -- an atomic layer of graphite -- was that its conductivity in the plane (i.e., parallel to the basal plane) is very high. I would expect a similar trend parallel to the basal plane in graphite.

Do you have any suggestions for reading that I can do to understand this?


1 Answer 1


Good catch - you are absolutely correct. This is a mistake in the Wikipedia page, as a quick check with the reference listed there confirms. The values are the wrong way around.

Changing it now...

Edited to add: this of course raises the question of why it should be this way around. A somewhat hand-wavy answer is that, just like in benzene, the $p_z$ orbitals overlap within each graphene sheet so that the electron contributed by each carbon is delocalised in this plane (the other three are bound up in the three $\sigma$ bonds). So in a sense this is conjugation, but over a whole 2D plane rather than around a ring or along a single chain of atoms. On the other hand, the graphene sheets are rather further apart from one another than the covalently bonded atoms within a sheet (as you might expect since the sheets are only weakly bound through van der Waals interactions), and so the overlap is greatly reduced and conduction is much more difficult.

A more complete answer would rely on a study of graphite's electronic band structure, which is an active research area and beyond the scope of this response!

  • $\begingroup$ Just as a follow-up, why is the electrical conductivity higher in the parallel direction? $\endgroup$
    – Andrew
    Jul 17, 2012 at 16:29
  • 1
    $\begingroup$ You assumed that already in Your question, why? $\endgroup$
    – Georg
    Jul 19, 2012 at 9:48

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