Q12.119CP - Chemistry: Molecular Nature Of Matter And Change

Question

Barium is the largest nonradioactive alkaline earth metal. It has a body-centred cubic unit cell and a density of  $\mathbf{3}\mathbf{.}\mathbf{62}\mathbf{g}/{\mathbf{cm}}^3$ . What is the atomic radius of barium? (Volume of a sphere: $\mathbf{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi r}}^3$ .)

Step-by-Step Solution

Answer

The atomic radius is the distance from the nucleus to the atom and has the value  $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{-\mathbf{8}}$ m.

Step 1: Definition

Density can be defined as the ratio of mass to volume. It is inversely proportional to the volume of the unit cell as the mass of the atom is constant.

 

 $\mathbf{\rho }\mathbf{=}\frac{Z\times M}{\mathbf{Na}\mathbf{\times }{\mathbf{a}}^3}$

 

 

For BCC, the value of $\text{Z = 2}$   is the number of atoms in contact in a particular structure.

Step 2: The atomic radius of barium.

Given,

Density of the unit cell, ρ=  ${\text{3.62 g/cm}}^{\text{3}}$

Avogadro number, N_a = $\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{23}$  atoms/molecules

Molar mass of atom or molecule, m = 137 g/mole

 

As,

 $\begin{array}{c}\mathrm{\rho }=\frac{\mathrm{Z}\times \mathrm{M}}{\mathrm{Na}\times {\mathrm{a}}^3}\\ 3.62=\frac{2\times 137}{6.02\times {10}^{23}\times {\mathrm{a}}^3}\end{array}$

 

Hence,

 

  $\begin{array}{c}{\mathrm{a}}^3=\frac{2\times 137}{6.02\times {10}^{23}\times 3.62}\\ \mathrm{a}=(125\times 10^{-24}{)}^{\frac{1}{3}}\\ \mathrm{a}=5\times {10}^{-8}\end{array}$

The atomic radius of the body-centered unit cell is  $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{-\mathbf{8}}$ m.