Question

A solid $A^{+}$ and $B^{-}$ has $NaCl$ type closed packed structure. If the anion has a radius of $250pm$, what should be the ideal radius of the cation?
Can a cation $C^{+}$ having a radius of $180pm$ be slipped into the tetrahedral site of the crystal of $A^{+}{B}^{-}$, without disturbing the crystal ? Give reason for your answer.

• A. $153pm$, yes
• B. $56pm$, no
• C. $103pm$, yes
• D. $103pm$, no

Answer 1

The correct option is D $103pm$, no

In $N{a}^{+}C{I}^{-}$ crystal each $N{a}^{+}$ ion is surrounded by 6 $C{I}^{+}$ ions and vice versa. Thus $N{a}^{+}$ ion is placed in

octahedral hole.

The limiting radius ratio for octahedral site = $0.414$

or $\frac{A^{+}}{B^{-}}=\frac{r}{R}=0.414$

Given that radius of anion $\left(B^{-}\right)R=250$ pm

i.e. radius of cation $\left(A^{+}\right)r=0.414R=0.414\times 250$ pm

or $r=103.5$ pm

Thus ideal radius for cation $\left(A^{+}\right)$ is $r=103.5$ pm.

We know that $\left(r/R\right)$ for tetrahedral hole is $\mathrm{0.225.}$

$\therefore \frac{r}{R}=0.225$

or $r=0.225$ $R=0.225\times 250=56.25pm$

Thus ideal radius for cation is $56.25$ pm for tetrahedral hole. But the radius of $C^{+}$ is $180$ pm. It is much larger

then ideal radius i.e. $56.25$ pm. Therefore we can not slipped cation $C^{+}$ into the tetrahedral site.