Types of Voids in FCC:

In a face-centered cubic (FCC) structure, there are two types of empty spaces, called voids: tetrahedral voids and octahedral voids.

Ratio for Void Sizes:

For a tetrahedral void: $$\frac{r}{R}=0.225$$

For an octahedral void: $$\frac{r}{R}=0.414$$

Here, $r$ is the radius of the smaller atom that goes into the empty site (the interstitial atom), and $R$ is the radius of the atom making up the structure.

Choosing the Correct Void:

The largest atom that can fit into the FCC structure without changing its shape fits in the octahedral void, because this void is bigger than the tetrahedral one. So, we use the octahedral void ratio: $$\frac{r}{R}=0.414$$

Finding the Radius ($R$) of the Main Atom:

In FCC, the relation between the radius $R$ and the edge length $a$ is: $$R=\frac{a}{2\sqrt{2}}$$

The question says $a = 400$ pm.

Calculate $R$:

$$R=\frac{400}{2\sqrt{2}}$$

Find $r$ (radius of the maximum atom that fits in an octahedral site):

Substitute the value of $R$ into the octahedral void ratio:

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

Calculate Diameter:

The diameter of the atom is twice the radius:

Diameter = $2r = 2 \times 58.55 = 117.1~\text{pm}$

Final Answer:

The largest atom that can fit into the empty site in an FCC lattice (without changing the structure) has a diameter of 117.1 pm.

Note:

Only the octahedral void is considered for calculating the maximum possible diameter, because it is the largest void in FCC structures.

JEE Advanced Subjects