Question

$ KCl $ crystallises in the same type of lattice as does $ NaCl $ . Given that $ {{r}_{Na^+}}+/{{r}_{C{{l}^{-}}}}=0.55 $ and $ {{r}_{{{K}^{+}}}}/{{r}_{C{{l}^{-}}}}=0.74 $ . Calculate the ratio of the side of the unit cell for $ KCl $ to that of $ NaCl $ .

• A. 1.123
• B. 0.0891
• C. 1.414
• D. 0.414
• E. 1.732

Answer 1

Answer: (A)

Given $ {{r}_{N{{a}^{+}}}}/{{r}_{C{{l}^{-}}}}=0.55 $
$ {{r}_{{{K}^{+}}}}/{{r}_{C{{l}^{-}}}}=0.74 $
$ \frac{{{r}_{KCl}}}{{{r}_{NaCl}}}=? $
$ \frac{{{r}_{N{{a}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=0.55 $
$ \frac{{{r}_{N{{a}^{+}}}}}{{{r}_{C{{l}^{-}}}}}+1=0.55+1 $
$ \frac{r_{Na}^{+}+{{r}_{C{{l}^{-}}}}}{{{r}_{C{{l}^{-}}}}}=1.55 $ .. (i)
$ \frac{{{r}_{{{K}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=0.74 $
$ \frac{{{r}_{{{K}^{+}}}}}{{{r}_{C{{l}^{-}}}}}+1=0.74+1 $
$ \frac{{{r}_{{{K}^{+}}}}+{{r}_{C{{l}^{-}}}}}{{{r}_{C{{l}^{-}}}}}=1.74 $ .. (ii)
Eqs (ii)/(i)
$ \frac{{{r}_{{{K}^{+}}}}+{{r}_{C{{l}^{-}}}}}{{{r}_{N{{a}^{+}}}}+{{r}_{C{{l}^{-}}}}}=\frac{1.74}{1.55}=1.1226 $