Published by:
CGP EDU Academic Team
Published on: August 13, 2026
CsBr has bcc structure with edge length 4.3. The shortest interionic distance in between Cs + and Br¯ is –
Text Solution
Verified by ExpertsThe correct answer is:
A
For bcc structure,
Atomic radius, r =
a =
× 4.3 = 1.86
we know that, r = half the distance between two nearest neighbouring atoms.
∴ shortest interionic distance = 2 × 1.86 = 3.72
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