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)138 pm B)1.95 * 10^-8 nm C)1.95 * 10^-8 cm D)154 pm E)0.109 nm

Question

Quizplus · Accepted Answer

Given data:
Density of palladium at 27°C = 12.0 g/cm³
We know that,

Density (ρ) = (Z × M)/NA × a³

Where, 
Z = 4 (for FCC unit cell)
M = molar mass of Pd = 106.4 g/mol
NA = Avogadro number = 6.022 × 10²³ mol⁻¹
a = edge length of the unit cell

Putting values in the above equation, we get:

a = [(4 × 106.4)/(12.0 × 6.022 × 10²³)]^(1/3) 
a = 3.890 Å

Now, we have to find the atomic radius, which can be calculated as half of the edge length of the unit cell.

Atomic radius of Pd = a/2
Atomic radius of Pd = 3.890/2 = 1.945 Å = 194.5 pm ≈ 138 pm

Therefore, the correct answer is A) 138 pm.

Palladium Crystallizes in a Face-Centered Cubic Unit Cell