Question

If 'b' is Van Der Waal constant of argon then the molecular diameter of Argon molecule can be represented by the expression________.

• A. ${\left[\frac{3b}{2\pi NA}\right]}^{1/3}$
• B. ${\left[\frac{3b}{2\pi NA}\right]}^{3/2}$
• C. ${\left[\frac{3b}{2\pi NA}\right]}^{1/2}$
• D. ${\left[\frac{3b}{\mathit{16}\pi NA}\right]}^{1/3}$

Answer 1

The correct option is A

${\left[\frac{3b}{2\pi NA}\right]}^{1/3}$

Explanation for correct option:-

Option (A) ${\left[\frac{3b}{2\pi NA}\right]}^{1/3}$

Step 1: Finding the molecular orbital

We know that, Van der Waal's constant b is:

$$\begin{gathered} \mathrm{b}=4\times \mathrm{volume}\mathrm{occuipied}\mathrm{by}\mathrm{the}\mathrm{molecules}\mathrm{in}1\mathrm{mole}\mathrm{of}\mathrm{gas} \\ \mathrm{b}=4\times {\mathrm{N}}_{\mathrm{A}}\times \left(\frac{4{\mathrm{\pi r}}^3}{3}\right) \end{gathered}$$

From here, we can calculate the diameter of the orbital:

$$\begin{gathered} \mathrm{b}=4\times {\mathrm{N}}_{\mathrm{A}}\times \left(\frac{4{\mathrm{\pi r}}^3}{3}\right) \\ \Rightarrow {\mathrm{r}}^3=\frac{3\mathrm{b}}{4\times {\mathrm{N}}_{\mathrm{A}}\times 4\times \mathrm{\pi }} \\ \Rightarrow \mathrm{r}={\left(\frac{3\mathrm{b}}{4\times {\mathrm{N}}_{\mathrm{A}}\times 4\times \mathrm{\pi }}\right)}^{\frac{1}{3}} \\ \Rightarrow 2\mathrm{r}=2{\left(\frac{3\mathrm{b}}{4\times {\mathrm{N}}_{\mathrm{A}}\times 4\times \mathrm{\pi }}\right)}^{\frac{1}{3}} \\ \Rightarrow \mathrm{d}={\left(\frac{3\mathrm{b}\times 8}{4\times {\mathrm{N}}_{\mathrm{A}}\times 4\times \mathrm{\pi }}\right)}^{\frac{1}{3}}[\because 2\mathrm{r}=\mathrm{diameter}] \\ \Rightarrow \mathrm{d}={\left(\frac{3\mathrm{b}}{2\times {\mathrm{N}}_{\mathrm{A}}\times \mathrm{\pi }}\right)}^{\frac{1}{3}} \end{gathered}$$

Hence it is the correct option

Therefore, molecular diameter of Argon molecule can be represented by the expression option (A) ${\left[\frac{3b}{2\pi NA}\right]}^{1/3}$