We were required to find the distance between two graphene layers of a graphite structure. We were given the atomic radius of carbon as $r_c$ and the bond length was to be considered equal to double the radius. Carbon forms an HCP lattice and has carbons in alternate tetrahedral voids. Do we have to find the half of the height of the hcp lattice i.e. $2a\sqrt{\frac 2 3}$ ? Also how will we calculate the side of the unit cell? Will it be $2r_c$ = $\frac{a\sqrt{3}}{4}$ (distance of tetrahedral void from edge). But wasn't this for FCC? Is it right to use it here?


1 Answer 1


Carbon forms an HCP lattice and has carbons in alternate tetrahedral voids.

If you remove every second carbon for a graphite layer (they are not in tetrahedral voids, though), you do get a layer of an HCP lattice:

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The layers are oriented in an ABAB fashion, but the distance between layers is not the distance you expect from HCP. In an HCP lattice, all 12 neighbors of an atom have the same distance. In graphite, the distance between bonded carbons is 0.143 nm, between alternate carbons in a layer 0.247 nm, and the distance between layers is similar to the distance between base pairs in DNA, 0.34 nm. For the within-layer distances, covalent radii are relevant. For the between-layer distances, van der Waals radii are relevant.

If you just follow the logic of the question, though, the layer distance should be the height of a tetrahedron with edge length of 0.247 nm, the distance of the atoms supposedly forming an HCP lattice.


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