Problem 14 - Entrance Test

The structure of diamond is a face-centered cubic (FCC) lattice with carbon atoms at the lattice points, and additionally, carbon atoms in half of the tetrahedral voids. Each unit cell of diamond has 8 carbon atoms (4 from FCC lattice points and 4 from half of the 8 tetrahedral voids). In the diamond structure, each carbon atom is tetrahedrally bonded to four other carbon atoms. The shortest C-C bond length is between a carbon atom at a lattice point and a carbon atom in an adjacent tetrahedral void. Let 'a' be the edge length of the unit cell. In an FCC lattice, there are 8 tetrahedral voids located at (1/4, 1/4, 1/4) and equivalent positions relative to the corners. If a carbon atom is at a corner (0,0,0), a nearby tetrahedral void (1/4, 1/4, 1/4) would contain another carbon atom. The distance between these two atoms (which is the C-C bond length) is the body diagonal of a small cube of side a/4. The distance 'd' between an FCC lattice point atom (e.g., at (0,0,0)) and an atom in a tetrahedral void (e.g., at (a/4, a/4, a/4)) is given by the formula for the body diagonal of a cube with side length x = a/4: d = x√3 = (a/4)√3. Given that the C-C bond length (d) = 154 pm. So, 154 pm = (a/4)√3 Rearranging to solve for 'a': a = (154 pm × 4) / √3 a = 616 pm / 1.732 a = 355.65 pm. This value is very close to 356 pm. Let's verify with the relation for the radius if needed. The radius of a carbon atom (r) = 154/2 = 77 pm (though not directly used here for bond length calculation). So, the edge length of the unit cell is approximately 356 pm.