Question

The unit cell cube length for $LiCl$ ( $NaCl$ type structure) is $5.14\mathring{A} $ . Assuming anion cation contact, calculate the ionic radius for chloride ion.

• A. $1.815$
• B. $3.63$
• C. $2.75$
• D. $5.14$

Answer 1

Answer: (A)

In $NaCl$ type of arrangement anion occupies corner of unit cell and and cation in octahedral void at edge center.
Length of unit cell $\left(a\right)=2r^{-}+2r^{+}$
Surface diagonal is,
$4r^{-}=\sqrt{2}a$ (given $\alpha =5.14\mathring{A} $ )
$r^{-}=\sqrt{2}a/4=\frac{1.414 \times 5.14}{4}=1.815\mathring{A} $