Question

The distance between octahedral and tetrahedral voids in fcc lattice would be:
[A] $a\sqrt{3}$
[B] $\dfrac{a\sqrt{3}}{2}$
[C] $\dfrac{a}{\sqrt{3}}$
[D] $\dfrac{a\sqrt{3}}{4}$

Answer 1

Hint: To solve this remember that the octahedral voids lie at the body centre of the cube and the tetrahedral voids lie at one-fourth distance from the corners on the body-diagonal. Use the length of body-diagonal to find the locations of the two voids.

Complete step by step answer:
To answer this question, we have to understand the meaning of octahedral and tetrahedral voids.
The void surrounded by four spheres sitting at the corners of a regular tetrahedron is called a tetrahedral void. When two tetrahedral voids from two different layers are aligned, together they form an octahedral void. This kind of void is surrounded by 6 atoms.

Now, let us discuss the fcc lattice. In a face centred cubic lattice, there are atoms present on each corner of the cube and also in the face centre of each face. Pictorially, we can represent fcc lattice as-
The number of tetrahedral and octahedral voids in CCP unit cells is 8 and 4 respectively. The number of particles in ccp is 4. Hence, the number of tetrahedral and octahedral voids is 8 and 4 respectively.
Now, we have to find the distance between the octahedral and the tetrahedral voids.
Firstly let’s talk about the tetrahedral voids-
In fcc lattice, the tetrahedral voids are located on the body diagonal of the cube at one-fourth distance from the corner.
We know that body-diagonal length is $a\sqrt{3}$ , so the location of tetrahedral voids will be at $\dfrac{a\sqrt{3}}{4}$ where a is the edge length.
Now, the octahedral voids are located at the body centre of the cube and the edge centres. If we calculate for the body centre void, it will be equivalent to that of the each edge centre as the body centred void is in the exact centre of the cube.

So, if the octahedral void is present at the centre of the body diagonal then we can write that its location is half to that of the body-diagonal length i.e. $\dfrac{a\sqrt{3}}{2}$ .
Now, distance between the two voids will be $\dfrac{a\sqrt{3}}{2}-\dfrac{a\sqrt{3}}{4}=\dfrac{a\sqrt{3}}{4}$
So, the correct answer is “Option D”.

Note: Face Centered Cubic i.e. FCC and cubic close packed i.e. CCP, these are two different names for the same lattice. We can think of this cell as being made by inserting another atom into each face of the simple cubic lattice - hence the "face-centred cubic" name.
If we divide an FCC unit cell into 8 small cubes, then each small cube has 1 tetrahedral void at its own body centre. Thus, there are 8 tetrahedral voids in total in one unit cell.