Question

Volume occupied by single $CsCl$ ion pair in a crystal is $7.014\times {10}^{-23}{cm}^3$. The smallest $Cs-Cs$ internuclear distance is equal to length of the side of the cube corresponding to volume of one $CsCl$ ion pair. The smallest $Cs$ to $Cs$ internuclear distance is nearly :

• A. $4.3\mathring{A}$
• B. $4.5\mathring{A}$
• C. $4.4\mathring{A}$
• D. $4\mathring{A}$

Answer 1

The correct option is D $4\mathring{A}$

$CsCl$ crystal has one $CsCl$ formula unit per unit cell. The volume of the unit cell is $7.014\times {10}^{-23}c{m}^3$
The smallest $Cs$ to $Cs$ inter-nuclear distance corresponds to the edge length of the unit cell.

According to the given condition,

Edge length,

$\left(a\right)={\text{Volume of one Cs-Cl ion pair}}^{\frac{1}{3}}$

It is nearly $(7.014\times {10}^{-23}c{m}^3{)}^{1/3}=4.124\times {10}^{-8}cm\simeq 4\times {10}^{-8}cm=4A^o$

Hence, option D is correct.