Question

The ionic radius of $Cl^{-}$ ion is $1.81\, \mathring A $. The interionic distances of $NaCl$ and $NaF$ are $2.79 \, \mathring A $ and $2.31\, \mathring A $ respectively. The ionic radius of $F^{-}$ ion will be

• A. 0.98 $\mathring A $
• B. 0.80 $\mathring A $
• C. 1.33 $\mathring A $
• D. 2.29 $\mathring A $

Answer 1

Answer: (C)

$d_{ NaCl }=r_{ Na }++r_{ Cl }^- $

$ 2.79=r_{ Na^+ }+1.81 $

or $ r_{ Na ^{+}}=2.79-1.81=0.98 \,\mathring{A}$

$d_{ NaF }=r_{ Na ^{+}}+r_{ F ^{-}}, \text {i.e., } 2.31=0.98+r_{ F ^{-}} $

or $ r_{ F ^{-}}=2.31-0.98=1.33 \,\mathring{A}$