Question

In the CsCl structure, $\frac{r^{-}}{r^{+}}=\mathrm{1.37.}$ The side length of the unit cell in terms of $r^{-}$, will be:

• A. $a=2r^{-}$
• B. $a=r^{-}$
• C. $a=1.414r^{-}$
• D. $a=4r^{-}$

Answer 1

The correct option is A $a=2r^{-}$

In $CsCl$ structure, $C{l}^{-}$ form simple cubic lattice. Here, the $C{l}^{-}$ occupies the corners of the cubic unit cell.
$C{s}^{+}$ ion occupies cubical void present in the body centre.

A cube of edge length "a" has a body diagonal of $\sqrt{3}\times a$.

In a BCC structure cations and anions are touching each other along the body diagonal such that
Body diagonal $=2(r^{+}+r^{-})=\sqrt{3}a$
Given,
$r^{+}=\frac{r^{-}}{1.37}$
Substituting the $r^{+}$ term in above equation gives,
$2(\frac{r^{-}}{1.37}+r^{-})=1.732a$
$2(0.732r^{-}+r^{-})=1.732a$
$2r^{-}(0.73+1)=1.732a$
$2r^{-}\times 1.73=1.732a$
$2r^{-}=a.$
Thus, option (a) is correct.