Question

Molybdenum forms body-centred cubic crystals and at 20°C the density is 10.3 g/cm³. Calculate the distance between the centres of the nearest molybdenum atoms.

Answer 1

In a body-centred cubic system, the atoms are situated at the corners of the cube in addition to an atom at the body-centre. Therefore, the distance between the nearest neighbour

d = 2 r

= `(sqrt 3 a)/4`

= `(sqrt 3 a)/2`

Moreover, for a bcc system, Z = 2

The atomic mass of molybdenum (M) = 95.94

∵ `rho = (Z xx M)/(a^3 xx N_A)`

∵ `a = ((Z xx M)/(rho xx N_A))^(1//3)`

∵ `a = ((2 xx 95.94)/(10.3 xx 6.022 xx 10^23))^(1//3)`

∵ `a = ((191.88)/(62.026 xx 10^23))^(1//3)`

= `(3.093 xx 10^23)^(1/3)`

= 3.14 × 10⁻⁸ cm

= 3.14 Å

= 3.14 × 100 pm   ...(1 Å = 100 pm)

= 314 pm

Hence, the distance between the nearest molybdenum atoms `d = (sqrt 3 a)/2`

= `(sqrt 3 xx 3.14)/2`

= `(1.73 xx 3.14)/2`

= 2.72 Å 

= 2.72 × 100 pm   ...(1 Å = 100 pm)

= 272 pm