Question

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

• A. 346.4 pm
• B. 643 pm
• C. 66.31 pm
• D. 431.5 pm

Answer 1

Answer: (A)

346.4 pm

Answer 2

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

The body diagonal $AD = a\sqrt{3}$, where a is the length of edge of unit cell

On the basis of figure

$AD = 2(r_{Cs +} + r_{Cl^{-}})$

$a\sqrt{3} = 2(r_{Cs^{+}} + r_{Cl^{-}})$

or$(r_{Cs^{+}} + r_{Cl^{-}}) = \frac{a\sqrt{3}}{2} = 400 \times \frac{\sqrt{3}}{2} = 200 \times 1.732 = 346.4pm$