a.
The atomic radius of iridium is 1.355 Å
For a face-centered cubic unit cell, with edge length, a and atomic radius r,
we have a² + a² = (4r)²
2a² = 16r²
r² = a²/8
r = a/√8
Since a = 3.833 Å,
Substituting this into the equation, we have
r = a/√8
r = 3.833 Å/√8
r = 3.833 Å/2.828
r = 1.355 Å
So, the atomic radius of iridium is 1.355 Å
b.
The density of iridium metal is 22.67 g/cm³
To find the density of the iridium metal, we need to find the mass of atoms per unit cell.
So, for the face centered cubic iridium atom, there are 6 faces per unit cell × 1/2 atoms per face + 4 corners per unit cell × 1/4 atoms per corner = 3 atoms + 1 atom = 4 atoms per unit cell.
So, the mass of the atoms in the unit cell is m = number of atoms/cell × number of moles/atom × molar mass of Iridium
= 4 atoms/cell × 1 mol/6.022 × 10²³ atoms × 192.22 g/mol
= 768.88/6.022 × 10²³ g/cell
= 127.68 × 10⁻²³ g/cell
= 1.2768 × 10⁻²¹ g/cell
Next, we need to find the volume of the unit cell which is V = a³
= (3.833 Å)³
= (3.833 × 10⁻¹⁰ m)³
= 56.314 × 10⁻³⁰ m
= 5.6314 × 10⁻²⁹ m
So, the density of iridium = mass of unit cell/volume of unit cell
= m/V
= 1.2768 × 10⁻²¹ g/cell ÷ 5.6314 × 10⁻²⁹ m
= 0.2267 × 10⁸ g/m³
= 0.2267 × 10⁸ g/m³ × 10⁻⁶ m³/cm³
= 0.2267 × 10² g/cm³
= 22.67 g/cm³
So, the density of iridium metal is 22.67 g/cm³
Learn more about face-centered cubic unit cell here:
https://brainly.com/question/17192154
