Question

The number of carbon atoms in the unit cell of diamond is ________.
(A) $6$
(B) $1$
(C) $4$
(D) $8$

Answer 1

Carbon is $s{p^3}$ hybridised and thus, the atoms are arranged in tetrahedral shape in diamond. In a unit cell of diamond, the carbon atoms are arranged in the half of tetrahedral voids and in the fcc lattice.

Complete step by step answer:
The most common crystal structure of diamond is called diamond cubic. In this crystal structure, the atoms are arranged according to face centred cubic lattice and in the half of centre of tetrahedral voids.
For atoms arranged in face centred cubic lattice:
Number of atoms at the corners of unit cell $= 8$
Each corner atom is shared by eight unit cells. Thus, effective number of corner atoms $= 8 \times ({1 \mathord{\left/ {\vphantom {1 {8)}}} \right.} {8)}} = 1$
Number of atoms at the centre of faces of unit cell $= 6$
Each face centred atom is shared by two unit cells. Thus, effective number of face centred atoms $= 6 \times ({1 \mathord{\left/ {\vphantom {1 {2)}}} \right.} {2)}} = 3$
Total number of atoms arranged in fcc lattice $= 1 + 3 = 4$
For atoms arranged in centre of tetrahedral voids:
The number of tetrahedral voids is equal to double the number of effective atoms in the unit cell. Therefore, number of tetrahedral voids $= 2 \times 4 = 8$
Half of these voids are occupied by atoms. Therefore, effective number of atoms in tetrahedral voids $= 8 \times ({1 \mathord{\left/ {\vphantom {1 {2)}}} \right.} {2)}} = 4$
Thus, the total number of carbon atoms in the unit cell of diamond is $= 4 + 4 = 8$
Hence, the correct option is (D).

Note:
The number of octahedral voids is equal to the effective number of atoms in the unit cell. We should not confuse it with tetrahedral voids.
The atoms in the centre of tetrahedral or octahedral voids are completely inside the unit cell. Therefore, they are considered equal to the effective number of atoms.