Question

$CsBr$ has similar type structure as of $CsCl$ with edge length of $6\text{pm}$. The shortest inter ionic distance in between $C{s}^{+}$ and $B{r}^{-}$ is:

• A. $5.196\text{pm}$
• B. $3.464\text{pm}$
• C. $2.598\text{pm}$
• D. $1.299\text{pm}$

Answer 1

The correct option is A $5.196\text{pm}$

As $CsBr$ has similar structure as of $CsCl$, then the $C{s}^{+}$ is present at the body centre and $B{r}^{-}$ ion is present at the corner of the lattice.
In $CsBr$, the the edge length is $\left(a\right)=6\text{pm}$
$\therefore \sqrt{3}\times a=2(r_{+}+r_{-})\Rightarrow r_{+}+r_{-}=1.732\times 6\times \frac{1}{2}=5.196\text{pm}$