Question

CsBr crystal has bcc structure. It has an edge length of $4.3\stackrel{\circ }{A}$. The shortest interionic distance between $C{s}^{+}$ and $B{r}^{-}$ ions is

• A. $1.86\stackrel{\circ }{A}$
• B. $3.72\stackrel{\circ }{A}$
• C. $4.30\stackrel{\circ }{A}$
• D. $7.44\stackrel{\circ }{A}$

Answer 1

The correct option is B

$3.72\stackrel{\circ }{A}$

$\frac{1}{2}$ (Body Diagonal)
$\Rightarrow 2(r_c+r_a)=\sqrt{3}a\Rightarrow r_c+r_a=\frac{\sqrt{3}a}{2}=\frac{\sqrt{3}\times 4.3}{2}=3.72\stackrel{\circ }{A}$