Question

A solid $A^{+}{B}^{-}$ has NaCl close packed structure.The radius of the cation when the radius of the anion is 250 pm is

• A. 103.5 pm
• B. 207 pm
• C. 69 pm
• D. 276 pm

Answer 1

Answer: (A)

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 $r^{-}=250pm$
$\therefore r^{+}=250\times 0.414$
$=103.5pm$