Rhodium has a face-centered cubic structure and has a density of 12.4 g/cm3. What is its atomic radius? (Note: for a face-centered cubic structure, )

Explanation:

For face centered cubic structure , no of atoms in a cell will be 4 .

so n = 4

density = n x atomic mass / a³ x N

N is avogrado no

12.4 = n x 103 / (a³ x 6.02 x 10²³)  

a³ = 4 x 103 / (12.4 x 6.02 x 10²³ )

= 5.52 x 10⁻²³

= 55.2 x 10⁻²⁴

a = 3.8 x 10⁻⁸ cm

if atomic radius is r

a = 2√2 x r

r = a / 2√2

= 3.8 x 10⁻⁸ /2√2 cm

= 1.34 x 10⁻⁸ cm

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8