Question

$KCl$ crystallises in the same type of lattice as does $NaCl$ . Given that $r_{N{a}^{+}}+/r_{C{l}^{-}}=0.55$ 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)
$\frac{r_{K^{+}}+r_{C{l}^{-}}}{r_{N{a}^{+}}+r_{C{l}^{-}}}=\frac{1.74}{1.55}=1.1226$