Question

Rhodium has an atomic radius of 0.1345 nm and density of 12.41 gm/cm3 . Determine whether it has an FCC or BCC crystal structure.

Answer 1

FCC.

Step-by-step explanation:

Hello,

In this case, since the density is defined as:

`\rho =(n*M)/(Vc*N_A)`

Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:

`Vc_(BCC)=((4r)/(√(3) ) )^3=((4*0.1345x10^(-7)cm)/(√(3) ) )^3=2.997x10^(-23)cm^3\\\\Vc_(FCC)=(2√(2)r)^(3) =(2√(2) *0.1345x10^(-7)cm)^3=5.506x10^(-23)cm^3`

Hence, we compute the density for each crystal structure:

`\rho _(BCC)=(n_(BCC)*M)/(Vc_(BCC)*N_A)=(2*102.9g/mol)/(2.337x10^(-23)cm^3*6.022x10^(23)/mol) =14.62g/cm^3\\\\\rho _(FCC)=(n_(FCC)*M)/(Vc_(FCC)*N_A)=(4*102.9g/mol)/(5.506x10^(-23)cm^3*6.022x10^(23)/mol) =12.41g/cm^3`

Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.

Regards.