A crystalline substance, ${\mathrm{X}}^{+}{\mathrm{Y}}^{-}$ has a $\mathrm{NaCl}$ type close packed structure. If ${\mathrm{Y}}^{-}$  has a radius of 250 pm, what will be the ideal radius of cation, ${\mathrm{X}}^{+}$. Can a cation, ${\mathrm{A}}^{+}$ of radius 110 pm be slipped into the tetrahedral hole of the crystal of ${\mathrm{X}}^{+}{\mathrm{Y}}^{-}?$

In a $\mathrm{NaCl}$ crystal, six ${\mathrm{Cl}}^{-}$  are surrounded around one Na+ ion and vice-versa. Therefore, ${\mathrm{Na}}^{+}$  ion is placed in an octahedral hole or void. The limiting radius ratio of octahedral site is 0.414

$\begin{array}{ccc}\frac{(\mathrm{Radius} \mathrm{of} \mathrm{cation}) {\mathrm{r}}^{+}}{(\mathrm{Radius} \mathrm{of} \mathrm{anion}) {\mathrm{R}}^{-}}& =& 0.414\\ \frac{{\mathrm{r}}^{+}}{250 \mathrm{pm}}& =& 0.414\\ {\mathrm{r}}^{+}& =& 103.5 \mathrm{pm}\end{array}$

For a tetrahedral hole, limiting radius ratio is 0.225.

  $$\begin{gathered} \frac{{\mathrm{r}}^{+}}{{\mathrm{R}}^{-}}=0.225 \\ \frac{{\mathrm{r}}^{+}}{250\mathrm{pm}}=0.225 \\ {\mathrm{r}}^{+}=56.25 \mathrm{pm} \end{gathered}$$

The ideal radius for cation for tetrahedral hole is 56.25 pm. But the radius of cation, A⁺ to be slipped in the tetrahedral hole is 110 pm. Since 110 pm radius is much larger than 56.25 pm. Hence, A⁺ cation cannot be slipped into tetrahedral hole.