A element X with atomic mass 60 g/mol has a density of `6.23"g cm"^(-3)`. If the edge length of the unit cell is 400 pm, identify the type of the cubic unit cell. Calculate the radius of the atoms of the element.

Explanation

Step-I. Identifying the type of the element
We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))" or "Z=(rhoxxa^(3)xxN_(0)xx10^(-30))/(M)`
According to available data,
Edge length (a) = 400 pm
Atomic mass of the element (M) = 60 g `mol^(-1)`
Density of unit cell `(rho)=6.23"g cm"^(-3)`
Avogadro's number `(N_(0))=6.022xx10^(23)mol^(-1)`
`Z=((6.23"g cm"^(-3))xx(400)^(3)xx(6.022xx10^(23)mol^(-1))xx(10^(-30)cm^(3)))/((60g mol^(-1)))=4`
The element X belongs to face centred cubic (fcc) unit cell.
Step II. Calculation of the radius of radius of the atom of the element
For fcc, radius (r) = `(a)/(2sqrt2)=(("400 pm"))/(2xxsqrt2)=141.5"pm"`