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

Rhodium has a FCC structure because the edge length of the unit cell is smaller than 4 times the atomic radius divided by the square root of 3.

Step-by-step explanation:

BCC stands for Body-Centered Cubic structure and FCC stands for Face-Centered Cubic structure. To determine which structure rhodium has, we need to compare the atomic radius of rhodium with the edge length of the unit cell for both structures.

For BCC, the diagonal distance of the body-centered cube is equal to 4 times the atomic radius. Therefore, the edge length of the unit cell for BCC is given by 4 times the atomic radius divided by √3.

For FCC, the face diagonal distance of the face-centered cube is equal to 2 times the atomic radius. Therefore, the edge length of the unit cell for FCC is given by 2 times the atomic radius.

By comparing the given atomic radius of 0.1345 nm to the calculated edge lengths, we can determine that rhodium has a FCC structure because 2 times the atomic radius (2 x 0.1345 nm) is smaller than 4 times the atomic radius divided by √3 (4 x 0.1345 nm / √3).

Answer 2

Rhodium has FCC structure.

Step-by-step explanation:

Formula used :

`\rho=(Z* M)/(N_(A)* 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* 10^(-8) cm`

Edge length = a

For FCC, a = 2.828 × r :

a =
`2.828* 1.345* 10^(-8) cm=3.80366* 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=(4* 102.91 g/mol)/(6.022* 10^(23) mol^(-1)* (3.80366* 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* 10^(-8) cm`

Edge length = a

For BCC, a = 2.309 × r :

a =
`2.828* 1.345* 10^(-8) cm=3.105605* 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=(2* 102.91 g/mol)/(6.022* 10^(23) mol^(-1)* (3.105605* 10^(-8)cm)^(3))`

`12.41 g/cm^3` ≠
`11.41 g/cm^3`

Rhodium has FCC structure.