Question

In CsCl structure, caesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is 412 pm and chloride ion has a radius of 181 pm, calculate the radius (In pm) of caesium ion.
A. 35.24
B. 89.2
C. 134.4
D. 175.8

Answer 1

Find out the given compound caesium chloride forms which type of unit cell. It is given in the question that ions are in contact along the body diagonal of the unit cell so accordingly, identify the unit cell. We have been given dimensions of anion and we need to find the radius of cation that is, caesium.

Complete step by step answer:
From the question we can identify that CsCl forms a body centered cubic cell that is, BCC.
In a body-centred cubic lattice of CsCl, anionic radius of chloride is represented as ${{r}^{-}}$ which is 181pm and the length of the side of the unit cell represented as ‘a’ is 412pm.
We know the radius of BCC unit cell is given by, $a=\dfrac{4}{\sqrt{3}}r$ in general but here, caesium and chloride are two different atoms having different radius and so the formula needs to be modified. Therefore, $\sqrt{3}a=4r$.
Length of the diagonal of BCC cell is $\sqrt{3}a$ which has a radius of four atoms or ions, two cations and two anions. Therefore,
$\sqrt{3}a=2{{r}^{+}}+2{{r}^{-}}$ as radius = radius of cation + radius of anion

& \sqrt{3}\times 412=2\text{r}{{}^{+}}^{}+2{{\text{r}}^{-}} \\\ & \text{713}\text{.6}=2\text{r}{{}^{+}}+2\times 181 \\\ & 713.6=2\text{r}{{}^{+}}+362 \\\ & 2\text{r}{{}^{+}}=351.6 \\\ & \text{r}{{}^{+}}^{}=175.8 \end{aligned}$$ So, $\text{r}{{}^{+}}=175.8$ pm **So, the correct answer is “Option D”.** **Note:** Body centred lattice has lattice points at the eight corners of the unit cell plus an additional point at the centre of the cell. The simplest crystal structures are those in which there is only a single atom at each lattice point. In the bcc structure the spheres fill 68% of the volume. The number of atoms in a unit cell is two $\left( 8\times \dfrac{1}{8}+1=2 \right)$.