Question

The unit cell cube length for $LiCl$ $\left(NaCl\text{like structure}\right)$ is $5.14\stackrel{o}{A}.$ Assuming anion-anion is in contact, the ionic radius for chloride ion will be-

Answer 1

$LiCl$ has the same structure as $NaCl$.
Thus, $C{l}^{-}$ anions form the fcc arrangement. These occupy the corners and face centres of the cubic unit cells.
$L{i}^{+}$ occupy all the octahedral voids in the unit cell.


Here the anion-anion is in contact
$\therefore$
$4r^{-}=a\sqrt{2}$
$a=5.14\stackrel{o}{A}$
$r^{-}=\frac{5.14\times 1.414}{4}$
$r^{-}=1.817\stackrel{o}{A}$