Palladium crystallizes in a face-centered cubic unit cell. Its density is 12.0 g/cm^3 at 27

Calculate the atomic radius of Pd.



A. 138 pm

B. 1.95 x 10^-8 nm

C. 1.95 x 10^-8 cm

D. 154 pm

E. 0.109 nm
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/cm



V = 5.89 x 10^-23 cm



For the FCC configuration, 4r = a*2



a = 4r /2, also V (cell) = a



a = 5.89 x 10^-23 cm



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 10^10 pm



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



r = 138 pm, answer A