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Q.
A solid $A^+ B^-$ has NaCl close packed structure.The radius of the cation when the radius of the anion is 250 pm is
The Solid State
Solution:
Since the solid $A^+ B^-$ has $NaCl$ type close packed structure, it belongs to a system with coordination number $6$. In these cases the ratio of the cation to the anion radii is given by
$\frac{r^+}{r^-} \ge 0.414$
Now $r^- = 250 \,pm$
$\therefore r^+ = 250 \times 0.414 $
$ = 103.5\,pm$