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A solid $A^{+}B^{-}$ has an $NaCl$ type closet packed structure . If the radius of the anion is $250\;pm$ What is the radius of the cation ?

$\begin{array}{1 1} 103.5\;pm \\ 7 \;pm \\ 8.4\;pm \\ 9.5\;pm\end{array} $

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Answer :$ 103.5\;pm$
The structure of NaCl involves face- centred cubic arrangement of $Cl^{-}$ ions and its octahedral holes are occupied by $Na^{+}$ ions. Now for a closest - packed structure $r_c/r_a=0.414$ ,Hence
$r_c =(0.414)r_a=(0.414)(250\;pm)=103.5\;pm$
answered Aug 8, 2014 by meena.p

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