Question

In diamond, each carbon atom is bonded to four other carbon atoms tetrahedrally. Alternate tetrahedral voids are occupied by carbon atoms. The number of carbon atoms per unit cell is:
(a)4
(b)6
(c)8
(d)12

Answer 1

Hint: In diamond, carbon atoms occupy face-centered cubic lattice, it has eight corners of the unit cell and also at the centers of each face of the unit cell. A unit cell is the smallest representation of a crystal.

Complete answer:
Face centered cubic unit lattices have eight lattice points at each corner of the face centered crystal and there are additional lattice points at the middle of every face of the cube. This unit cell contains four atoms:
($8$ Corners of a given atom $ \times $ $\dfrac{1}{8}$ of the given atom’s unit cell) $ + $($6$ Faces $ \times \dfrac{1}{2}$ contribution) $ = 4$ atoms.
So, for the diamond it has FCC structure and carbon is placed at the tetrahedral voids. Therefore there will be $4$ carbon atoms. Carbon atoms are present in one half the tetrahedral voids. There are $8$ tetrahedral voids in face centered cubic unit structure. So, the remaining carbon atoms will be $8 \times \dfrac{1}{2} = 4$ carbon atoms.
Thus, the total carbon atoms present are $4 + 4 = 8$ atoms.

Note:
Each carbon atom in a diamond is covalently bonded to four other carbons in a tetrahedron. These tetrahedrons together form a three-dimensional network of six-membered carbon rings. This stable network of covalent bonds is the reason that diamond is a very strong substance. It has a high melting point because very strong carbon-carbon covalent bonds have to be broken. It is very hard because of the strong intermolecular force of attraction between the carbon atoms.