Question

A solid $AB$ has $NaCl$ structure. If radius of $A^{+}$ is $100pm$. What is the radius of $B^{-}$ ion?

• A. $120pm$
• B. $136to241.6pm$
• C. $280pm$
• D. Unpredictable

Answer 1

The correct option is B $136to241.6pm$

For $NaCl$ type compounds (Rock Salt)

$\frac{r^{+}}{r^{-}}\in [0.414,0.732]$

Given, $r^{+}=100pm$

$\frac{r^{+}}{r^{-}}=\mathrm{0.414.}\Rightarrow r^{-}=241.5$

$\frac{r^{+}}{r^{-}}=0.732\Rightarrow r^{=}136.6pm$

$r^{-}\in [136.6pm,241.6pm]$

So radius of $Y^{-}$ lies between $136.6$ to $241.6pm$