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A solid $A^+B^-$ has an NaCl type closest packed structure.If the radius of the anion is 250pm,what is the radius of the cation?

$\begin{array}{1 1}103.5pm\\104.5pm\\106.5pm\\203.5pm\end{array} $

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Answer : 103.5pm
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
$r_c=(0.414)r_a=(0.414)(250pm)=103.5$pm
answered Jul 16, 2014 by sreemathi.v
 

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