Question

$\mathrm{CsBr}$ has bcc structure with edge length $4.3\text{Å}$. What is the shortest inter-ionic distance between ${\mathrm{Cs}}^{+}$and ${\mathrm{Br}}^{-}$?

Answer 1

Step 1: Relation between edge length and radius in bcc

$r=\frac{\sqrt{3}}{4}a$

Here, $r$ is half the distance between the two nearest neighboring atoms., and $a$ is the edge length.

Step 2: Calculating the radius of the sphere

$$\begin{gathered} r=\frac{\sqrt{3}}{4}\times 4.3\text{Å} \\ r=1.86\text{Å} \end{gathered}$$

Step 3: Calculating the shortest interionic distance

Since, $r$ is half the distance between the two nearest neighboring atoms. So, the shortest interionic distance is as follows:-

Shortest interionic distance = $2\times 1.86\text{Å}=3.72\text{Å}$

Hence, the shortest interionic distance is $3.72\text{Å}$.