Question

CsCl has bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance
in CsCI.
​

Answer 1

Answer:

Inter Ionic Distance = 346.4 pm

Explanation:

We Have :-

CsCl has BCC arrangement

Unit Cell Edge is 400 pm

To Find :-

Inter ionic distance in CsCl

Formula Used :-

Inter ionic distance in BCC = √3 / 2 * a

Solution :-

$Inter\ ionic\ distance=\frac{\sqrt{3} }{2} a\\\\= \frac{\sqrt{3} }{2} * 400\\\\= 1.732 * 200\\\\= 346.4 pm$