Question

Rhodium has an atomic radius of 0.1345 nm and a density 12.41g/cm^3. De-termine if it comes in FCC or BCC structure.

Answer 1

Answer:

Rhodium has FCC structure.

Explanation:

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

1) If it FCC cubic lattice

Number of atom in unit cell of FCC (Z) = 4

Atomic radius of Rh= 0.1345 nm = $1.345\times 10^{-8} cm$

Edge length = a

For FCC,  a = 2.828 × r :

a = $2.828\times 1.345\times 10^{-8} cm=3.80366\times 10^{-8}cm$

Density of Rh= $12.41 g/cm^3$

Atomic mass of Rh(M) = 102.91 g/mol

On substituting all the given values , we will get the value of 'a'.

$12.41 g/cm^3=\frac{4\times 102.91 g/mol}{6.022\times 10^{23} mol^{-1}\times (3.80366\times 10^{-8}cm)^{3}}$

$12.41 g/cm^3\approx 12.4214 g/cm^3$

2) If it BCC cubic lattice

Number of atom in unit cell of BCC (Z) = 2

Atomic radius of Rh= 0.1345 nm = $1.345\times 10^{-8} cm$

Edge length = a

For BCC,  a = 2.309 × r :

a = $2.828\times 1.345\times 10^{-8} cm=3.105605\times 10^{-8}cm$

Density of Rh= $12.41 g/cm^3$

Atomic mass of Rh(M) = 102.91 g/mol

On substituting all the given values , we will get the value of 'a'.

$12.41 g/cm^3=\frac{2\times 102.91 g/mol}{6.022\times 10^{23} mol^{-1}\times (3.105605\times 10^{-8}cm)^{3}}$

$12.41 g/cm^3$ ≠ $11.41 g/cm^3$

Rhodium has FCC structure.