Calculate the radius of an iridium atom, given that ir has an fcc crystal structure, a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.

Given dа ta:

Density of Iridium (Ir) = 22.4 g/cm3

Atomic mass of Ir = 192.2 g/mol

Therefore, volume of Ir atom/mol = 192.2 g. mol-1/22.4 g. cm-3 = 8.58 cm3/mol

1 mole of Ir has 6.023 * 10^23 Ir atoms

Hence, Volume/atom = 8.58 / 6.023 * 10^23 = 1.425*10^-23 cm3

Now, a Face centered cubic cell has 4 atoms/cell

Volume of a single unit cell = 4 * 1.425*10^-23 cm3 = 5.7 * 10^-23 cm3

Volume (V) of a cube is related to the length of its side (a) as:

V = a^3

a = (V) ^1/3 = (5.7 * 10^-23) ^1/3 = 3.91*10^-8 cm

For an FCC cell, the length of the side (a) is related to the atomic radius (R) as:

a = 2.83 R

R = a/2.83 = 3.91*10^-8/2.83 = 1.38 * 10^-8 cm