Question

Which of the following statements is/are correct about tetrahedral voids (TVs) in an fcc unit cell ?

• A. Number of TVs per atom in fcc unit cell is $2$.
• B. Number of TVs per unit cell is $8$.
• C. Number of TVs is twice the number of atoms in the fcc unit cell.
• D. Number of TVs is equal to the number of atoms in the fcc unit cell.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Number of TVs per atom in fcc unit cell is $2$.
B Number of TVs per unit cell is $8$.
C Number of TVs is twice the number of atoms in the fcc unit cell.

Tetrahedral interstices : We have seen that in hexagonal close packing (hcp) and cubic close packing (ccp), each sphere of second layer touches with three spheres of first layer. Thus, they leave a small space in between which is known as tetrahedral site or interstices or the vacant space between $4$ touching spheres is called as tetrahedral void. Since a sphere touches three spheres in the below layer and three spheres in the above layer, there are two tetrahedral sites associated with one sphere. It may by noted that a tetrahedral site does not mean that the site is tetrahedral in geometry but it means that this site is surrounded by four spheres and the centres of these four spheres lie at the apices of a regular tetrahedron.
In FCC, one corner and its face centres form a tetrahedral void,
In FCC, two tetrahedral voids are obtained along one cube diagonal. So, in FCC, $8$ tetrahedral voids are present.
In FCC, total number of atoms $=4$.
In FCC, total number of tetrahedral voids $=8$.
So, we can say that in 3D close packing, $2$ tetrahedral voids are attached with one atom.