Question

An element crystallizes in fcc lattice having edge length 400 pm. The maximum diameter of an atom which can be placed in interstitial site without distorting the structure is (in pm):

Answer 1

In fcc, there are octahedral voids and tetrahedral voids.

For tetrahedral voids, $\frac{r}{R}=0.225$

For octahedral voids, $\frac{r}{R}=0.414$

Where,
$r=$ radius of the smaller ion

$R=$ radius of the larger ion

For the maximum diameter of the atom in the interstitial site, octahedral voids must be taken because they can accommodate a larger atom.

$\frac{r}{R}=0.414$

For FCC,

$R=\frac{a}{2\sqrt{2}}=\frac{400}{2\sqrt{2}}$
$\frac{r}{R}=0.414$

$\Rightarrow r=0.414\times \frac{400}{2\sqrt{2}}=58.55$ pm

Diameter is $2r=2\times 58.55=117.1$ pm