A solid $A B$ has $N a C l$ structure. If radius of $A^{+}$ is $100 p m$ . What is the radius of $B^{-}$ ion?

120 p m

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136 t o 241.6 p m

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280 p m

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Unpredictable

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Solution

The correct option is B $136 t o 241.6 p m$
For $N a C l$ type compounds (Rock Salt)
$\frac{r^{+}}{r^{-}} \in [0.414, 0.732]$
Given, $r^{+} = 100 p m$
$\frac{r^{+}}{r^{-}} = 0.414. \Rightarrow r^{-} = 241.5$
$\frac{r^{+}}{r^{-}} = 0.732 \Rightarrow r^{=} 136.6 p m$
$r^{-} \in [136.6 p m, 241.6 p m]$
So radius of $Y^{-}$ lies between $136.6$ to $241.6 p m$

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