Palladium crystallizes in a face-centered cubic unit cell.Its density is 12.0 g/cm³ at 27°C.Calculate the atomic radius of Pd. A)154 pm B)138 pm C)1.95 × 10^-8 nm D)0.109 nm E)1.95 × 10^-8 cm

Question

Quizplus · Accepted Answer

Given, density of palladium (ρ) = 12.0 g/cm^3
The unit cell of FCC has 4 atoms
Let the edge length of the unit cell be 'a' and the atomic radius of Pd be 'r'
The volume of a unit cell of FCC = a^3/4
The mass of 4 atoms = 4 × atomic mass of Pd = 4 × 106.42 g (atomic mass of Pd)
Density (ρ) = mass/volume
=> 12.0 = 4 × 106.42/(a^3/4)
=> a = 3.89 Å
The face diagonal of a FCC unit cell = 4r√2/3
For a unit cell of Pd, the face diagonal = a√2
=> a√2 = 4r√2/3
=> r = a/2√2 = 0.138 nm ≈ 138 pm

Palladium Crystallizes in a Face-Centered Cubic Unit Cell