Question

$CsBr$ has a BCC structure where the edge length is 4.3. The shortest inter ionic distance between $C{s}^{+}$ and $B{r}^{-}$ is:

• A. 3.72
• B. 1.86
• C. 7.44
• D. 4.3

Answer 1

The correct option is A

3.72

In BCC, the ions touch along the body diagonal. So, the distance between two ions is equal to half the body diagonal length.

$=\frac{1}{2}$ of body diagonal $=\frac{1}{2}\times \sqrt{3}a$

$=\frac{\sqrt{3}}{2}\times 4.3=3.72\stackrel{\circ }{A}$

Protip: You dont even have to do this calculation, look at the options and it will be evident right away. The number has to be smaller than 4.3 but greater than 3/4th its value as $\frac{\sqrt{3}}{2}$ = 0.866