Question

Iridium crystallizes in a face-centered cubic unit cell that has an edge length of 3.811 Å.

(a) Calculate the atomic radius of an iridium atom. (b) Calculate the density of iridium metal.

Answer 1

Solution :-

Crystal is FCC

Edge length is 3.811 A

Using endge length lets calculate the diagonal length

Diagonal length = squre root (2) * edge length

                              = square root (2) * 3.811 A

                              = 5.39 A

Now lets calculate the radius of the metal

r= diagonal length / 4

= 5.39 A/ 4

= 1.347 A

Now we can calculate density using the molar mass and volume we can calculate the density

In the FCC cell 4 atoms are present

So

Density = mass / volume

               = (4 atoms * 192.22 g per mol / 6.022*10^23 atoms) / (3.811*10^-8 cm)^3

               = 23.06 g/cm3

So the radius of the iridium metal is 1.347 A

And density = 23.06 g/cm3