Question

A solid AB has CsCl type structure. Edge length of the unit cell is $4.04\mathring{A}$. The distance (in $\mathring{A}$) of closest approach between $A^{+}$ and $B^{-}$ is :

• A. $4.0$
• B. $6.4$
• C. $3.5$
• D. $7.0$

Answer 1

Answer: (C)

$3.5$

In BCC lattice structure, $\sqrt{3}a=2\left(r_{+}{+r}_{-}\right)$

$\therefore \left(r_{+}{+r}_{-}\right)=\frac{\sqrt{3}\times 4.04}{2}=3.5\mathring{A}$