What is the total number of atoms in the lattice structure of diamonds?

Diamond is a covalent crystal in which each C-atom is sp³ hybridized. Thus, each C-atom is covalently bonded to four other C-atoms tetrahedrally. These tetrahedral are linked together into a 3D giant molecule. The structure of diamond is similar to that of ZnS. In diamond, we have C-atoms in place of Zn²⁺ and S^2- ions.

Thus, diamond has face-centered cubic structure in which C-atoms are present at the corners as well as face centers and one half of tetrahedral voids. As there are 8 tetrahedral voids in FCC.

$\text{ Contribution of }\mathrm{C}\text{ -atoms at}\frac{1}{2}\text{ of tetrahderal voids}=8\times \frac{1}{2}=4$

$\text{ Contribution of }\mathrm{C}\text{ -atoms at the corners }=8\times \frac{1}{8}=1$

$\text{ Contribution of C-atoms at face centers }=6\times \frac{1}{2}=3$

$\text{Total number of }\mathrm{C}\text{ -atoms in the unit cell }=1+3+4=8$