An element X which crystallises in face centred cubic structure. Its metallic radius is 145 pm, the length of the side of the unit cell is:

410.1 pm

The correct answer is Option (2) → 410.1 pm

To find the length of the side of the unit cell for an element \(X\) that crystallizes in a face-centered cubic (FCC) structure, we can use the relationship between the metallic radius and the edge length of the unit cell.

Relationship in FCC Structure

In an FCC lattice, the atoms at the corners and the face centers are arranged such that:

\(\text{Length of the face diagonal} = 4 \times r\)

where \(r\) is the metallic radius. The face diagonal of the cube can also be expressed in terms of the side length \(a\) of the cube:

\(\text{Face diagonal} = a\sqrt{2}\)

Setting the Equations Equal

Setting these two expressions for the face diagonal equal gives:

\(a\sqrt{2} = 4r\)

Solving for \(a\)

Rearranging the equation to find \(a\):

\(a = \frac{4r}{\sqrt{2}} = \frac{4 \times 145 \, \text{pm}}{\sqrt{2}}\)

Calculating \(a\):

\(a = \frac{580 \, \text{pm}}{1.414} \approx 410.1 \, \text{pm}\)

Conclusion

Rounding this to one decimal place, the length of the side of the unit cell is approximately: 410.1 pm