A solid AB has NaCl type structure. If the radius of cation A is 100 pm, then the radius of the anion B will be

241 pm

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414 pm

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22.5 pm

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44.4 pm

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Solution

The correct option is A
241 pm

$\frac{r_{A^{+}}}{r_{B^{-}}} = 0.414; r_{B^{-}} = \frac{100}{0.414} = 241 p m$

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