A solid AB has NaCl type structure. If the radios of the 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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225 pm

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44.4 pm

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Solution

The correct option is D 241 pm
Some concept can be applied to ionic solids in general.
$r_{N a} = 0.102 n m$
$r_{C l} = 0.181 n m$
$r_{N a} / r_{C l} = 0.564$
$∴$ Cation $({N a}^{+})$ prefer octahedral sites

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Similar questions

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

B^{-}

250

B^{-}

100

A^{+}

100

A B

N a C l

100 p m

B^{-}

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