NEET Chemistry The Solid State Questions Solved


CsBr has a (bcc) arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl [CBSE 1993]

(1) 346.4 pm

(2) 643 pm

(3) 66.31 pm

(4) 431.5 pm

(1) The (bcc) structure of CsBr is given in figure

The body diagonal AD=a3, where a is the length of edge of unit cell

On the basis of figure

AD=2(rCs++rCl)

a3=2(rCs++rCl) or (rCs++rCl)=a32=400×32=200×1.732=346.4pm  

Difficulty Level:

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