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CsBr has bcc stucture with edge length 4...

`CsBr` has bcc stucture with edge length `4.3` pm .The shortest interionic distance in between Cs and Br is

A

3.72 pm

B

1.86 pm

C

7.44 pm

D

4.3 pm

Text Solution

Verified by Experts

The correct Answer is:
A

`r_c + r_a = (sqrt3a)/(4)`
` r = (sqrt3 xx 4.3)/(4) = 1.86 `pm
shortest inert ionic distance = 2r
`1.86 xx 2 = 3.72 `pm
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