Question

Iron occurs as body-centred as well as face-centred cubic systems. If the effective radius of an atom of iron is 124 pm, calculate the density of iron in both the structures.

Answer 1

Given: For the body-centred cubic system,

`r = (a sqrt 3)/4` and Z = 2

Atomic mass of Fe = 55.8

∴ `a = (4 r)/sqrt 3`

= `(4 xx 124)/sqrt 3`

= `(4 xx 124)/1.73`

= `496/1.73`

= 286.7 pm

= 286.7 × 10⁻¹⁰ cm

Hence, the density

`rho = (Z xx M)/(a^3 xx N_A)`

= `(2 xx 55.8)/((286.7 xx 10^-10)^3 xx 6.022 xx 10^23)`

= `(111.6)/(23565848.3 xx 10^-30 xx 6.022 xx 10^23)`

= `(111.6)/(23.56 xx 10^-24 xx 6.022 xx 10^23)`

= `(111.6)/(141.87 xx 10-^1)`

= `(111.6)/(14.187)`

= 7.86 g cm⁻³

For the face-centred cubic system,

`r = (a sqrt 2)/4` and Z = 4

`a = (4 r)/sqrt 2`

= `(4 xx 124)/sqrt 2`

= `(4 xx 124)/1.414`

= 350.7 pm

= 350.7 × 10⁻¹⁰ cm

= 3.507 × 10⁻⁸ cm

Hence, the density

`rho = (Z xx M)/(a^3 xx N_A)`

ρ = `(4 xx 55.8)/((350.7 xx 10^-10)^3 xx 6.022 xx 10^23)`

= `(223.2)/(43132764.84 xx 10^-30 xx 6.022 xx 10^23)`

= `(223.2)/(43.13 xx 10^-24 xx 6.022 xx 10^23)`

= `(223.2)/(259.72 xx 10^-1)`

= `(223.2)/(25.972)`

ρ = 8.59 g cm⁻³