Question

Palladium crystallizes in a face-centered cubic unit cell. Its density is 12.02 g/cm^3 .Calculate the atomic radius of Pd.

A) calculate the average mass of 1pd atom

B) calculate the number of pd atoms in FCC

C) use the density to calculate the volume

D) Determine the edge lenth using volume

E) determine the atomic radius using the pythagorean theorem .

Answer 1

A) Pd atomic mass is -106g/mol

That is for 1mol of atoms 6.023x10²³ atoms the atomic mass is -106g

For 1 Pd atom it is : 106g/mol6.023x10²³ mol

= 17.6x10^-23 g

B)Number of Pd atoms in FCC:

Each atom on the face is shared by two unit cells.

Each atom in the corner is shared by 8 unit cells.

Hence,the number of atoms in the FCC unit cell = 8x1/8 + 6x1/2 = 4atoms

C) Density = mass / volume

There are 4 atoms to the unit cell in a fcc.

(4xatomic mass /6.023 x 10²³) = mass of unit cell

(4x106.0 g/mol) / (6.023 x 10²³) = 7.07 x 10^-22 g

Density = mass/volume ∴ V = mass / density

V = 7.07 x 10^-22 g / 12.02 g/cm³

V = 5.89 x 10^-23 cm³

D) For the FCC configuration, 4r = √2a

a = 4r /√2,

also V (cell) = a³

a³ = 5.89 x 10^-23 cm³

a = 3.89 x 10^-8 cm

a = 4r /√2

where r = atomic radius

3.89 x 10^-8 cm = 4r / √2

r = 1.38 x 10^-8 cm

1 cm = 1 x 10¹⁰ pm

1.38 x 10^-8 cm x (1 x 10¹⁰ pm / 1 cm) = 138 pm

r = 138 pm