Question

Palladium crystallizes in a face-centered cubic unit cell. Its density is 12.0 g/cm^3 at 27 degrees C. Calculate the atomic radius of Pd.

Answer 1

Density = mass / volume

There are 4 atoms to the unit cell in a fcc.
(4*atomic mass /6.02 x 10^23) = mass of unit cell

(4*106.42 g/mol) / (6.02 x 10^23) = 7.07 x 10^-22 g

Density = mass/volume
? V = mass / density

V = 7.07 x 10^-22 g / 12.0 g/cm³

V = 5.89 x 10^-23 cm³

For the FCC configuration, 4r = a*v2

a = 4r /v2, also V (cell) = a³

a³ = 5.89 x 10^-23 cm³

a = 3.89 x 10^-8 cm

a = 4r /v2 , where r = atomic radius

3.89 x 10^-8 cm = 4r / v2

r = 1.38 x 10-8 cm

1 cm = 1 x 10^10 pm

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

r = 138 pm

Answer 2

124 because of the emelu of the eme is so eme