Question

$CsBr$ has $bcc$ structure with edge length $4.3\,?$ The shortest inter ionic distance in between $Cs^+$ and $Br^-$ is

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

Answer 1

Answer: (A)

3.72

Answer 2

For bcc structure, atomic radius, $r = \frac{\sqrt{3}}{4} a$ $= \frac{\sqrt{3}}{4} \times 4.3 = 1.86$ Since, r = half the distance between two nearest neighbouring atoms. $\therefore$ Shortest inter ionic distance = 2 $\times$ 1.86 = 3.72