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-

• A. $1.817\stackrel{o}{A}$
• B. $2.8\stackrel{o}{A}$
• C. $3.8\stackrel{o}{A}$
• D. $4.815\stackrel{o}{A}$

Answer 1

The correct option is A $1.817\stackrel{o}{A}$

$LiCl$ has the same structure like $NaCl$
Thus, $C{l}^{-}$ anions forms the fcc arrangement. It occupies the corners and face centres of the cubic unit cells.
$N{a}^{+}$ occupies all the octahedral voids in the unit cell.
Generally in $NaCl$ crystal lattice, the $C{l}^{-}$ ion in corner of cubic unit cell does not touches the $C{l}^{-}$ ion in face centre of the unit cell.
Thus,
$4r^{-}\ne a\sqrt{2}$
But, here the anion-anion is in contact
$\therefore$
$4r^{-}=a\sqrt{2}$
$a=5.14\stackrel{o}{A}$
$r^{-}=\frac{5.14\times 1.414}{4}$
$r^{-}=1.817\stackrel{o}{A}$