Question

Iridium crystallizes in a face-centered cubic unit cell that has an edge length of 3.833 Å. The atom in the center of the face is in contact with the corner atoms, as shown in the drawing.

Part A Calculate the atomic radius of an iridium atom. Express your answer using four significant figures.

Part B Calculate the density of iridium metal. (Figure 1) Express your answer using four significant figures.

Answer 1

Given that Iridium crystallizes in a face-centered cubic unit cell.

A) Given that edge length a = 3.833 Å

For FCC, atomic radius = a /(8)^1/2 = 3.833 Å / (8)^1/2 =1.355 Å

Therefore,  atomic radius of Iridium atom = 1.355 Å

B)

Given that edge length a = 3.833 Å= 3.833 x 10^-8 cm

atomic mass of Iridium M = 192.2 g mol^-1

  no of units in FCC unit cell = 4

avagadro number = 6.023 x 10²³ mol^-1

density = [M x no of units in BCC unit cell] /[ a³ x avogadro number]

= [ 192.2 g mol^-1 x 4 ] / [ (3.833 x 10^-8 cm)³ x 6.023 x 10²³ mol^-1]

= 22.6 g/cm³

  Therefore, density of iridium metal =  22.6 g/cm³