Question

In face centered cubic (fcc) lattice, edge length is 400pm. Find the diameter of the greatest sphere which can be fit into the interstitial void without distortion of lattice.

Answer 1

Hint: In order to find the diameter (R) of the greatest sphere which can be fit into the interstitial void, we could use the equation $a=2(r+R)$. We are given the edge length (a) and we could find the value of r by using the equation for radius of an atom in fcc lattice.

Complete step by step answer:
  - Let’s start with the face centered cubic or fcc lattice. They are the arrangement of atoms in crystals in such a way that the atomic centers are disposed in space, so that one atom is located at the center of each face and one at each of the corners of the cube. The fcc unit cell is the simplest repeating unit in a cubic close packed structure.
  - A we know the fcc lattice structures provide a framework for octahedral holes or voids. Also, there are four octahedral voids in a face centered cubic lattice with a relationship of one octahedral void per anion.
  - We are given the edge length of the fcc crystal and we are asked to find the diameter of the greatest sphere which can be fit into the interstitial void. The formula for octahedral void can be given as follows
\[a=2(r+R)\]
  Where a is the edge length (400 pm)
r is the radius of the lattice sphere
R is the radius of void sphere
The formula for finding the r is given below
\[r=\dfrac{a\sqrt{2}}{4}\]
\[r=\dfrac{400\sqrt{2}}{4}\]
  \[r=141.4 \]pm.

By substituting the value of r in the equation to find the radius of the lattice sphere we get,
\[a=2(r+R)\]
\[2R=a-2r\]
\[R=\dfrac{a-2r}{2}\]
\[R=\dfrac{400-2(141.44)}{2}\]
 \[R=58.56pm\]
 Since we are asked to find the diameter, multiply the radius by two and we will have the value as 117.12 pm.
Thus the diameter of the greatest sphere which can be fit into the interstitial void in an fcc crystal lattice is 117.12 pm.

Note: It should be noted that in addition to the octahedral voids, there is also another category of voids and tetrahedral voids. The tetrahedral void is surrounded by four atomic spheres and hence the coordination number for the tetrahedral void is four whereas the octahedral void is surrounded by six atomic spheres and has a coordination number of six.