Question

CsBr has bcc structure with edge length 4.3. The shortest interionic distance in between Cs⁺ and BrÆ 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 = $\frac{\sqrt{3}}{4}$ × 4.3 = 1.86

we know that, r = half the distance between two nearest neighbouring atoms.

\ shortest interionic distance = 2 × 1.86 = 3.72