In face centred cubic (fcc) crystal lattice, edge length of the unit cell is $400\text{pm}$. Find the diameter fo the greatest sphere which can be fitted into the interstitial void without distortion of lattice.

The correct option is B $117.2pm$
In fcc unit cell, spheres touches along the face diagonal such that
$a\sqrt{2}=4r$
where,
a is the unit cell edge length
r is the radius of the sphere
$\therefore r=\frac{a\sqrt{2}}{4}=\frac{400\times \sqrt{2}}{4}$
$r=141.4\text{pm}$
In fcc there are two voids - smaller tetrahedral and larger octahedral. Octahedral voids are present at the centre of edges and at the body centre of fcc. The edge length of fcc contains two spheres at corner and one octahedral void in edge centre which are in contact with each other.
The formula for octahedral void is
$2(r+R)=a$
$2R=a-2r$
$2R=400-2\times 141.4$
$=117.2\text{pm}$
$\text{Diameter of greatest sphere}=117.2\text{pm}$