Question

CsBr crystal has bcc structure. It has an edge length of $4.3A^0$. The shortest interionic distance between $C{s}^{+}$ and $B{r}^{-}$ ions is?

• A. $1.86A^0$
• B. $2.86A^0$
• C. $3.72A^0$
• D. $4.72A^0$

Answer 1

Answer: (C)

$3.72A^0$

For BCC $r=\frac{\sqrt{3}}{2}a$

$r^{+}+r^{-}=\frac{\sqrt{3}}{2}a$

$r^{+}+r^{-}=\frac{\sqrt{3}\times 4.3}{2}$

$\therefore r^{+}+r^{-}=3.72\mathring{A}$