Calculate Radius of Cation in Octahedral Solid Lattice

Based on the given figure, the $A^+$ ion is situated in an octahedral void surrounded by six $X^-$ ions (coordination number = $6$).

For a stable octahedral arrangement where the cation $A^+$ fits into the octahedral void formed by the anions $X^-$, the limiting radius ratio is given by: $\frac{r_{A^+}}{r_{X^-}} = \sqrt{2} - 1 \approx 0.414$

Given: $r_{X^-} = 250\text{ pm}$

Substituting the value into the radius ratio relation: $r_{A^+} = 0.414 \times r_{X^-}$ $r_{A^+} = 0.414 \times 250\text{ pm} = 103.5\text{ pm} \approx 104\text{ pm}$

Therefore, the radius of the $A^+$ ion is $104\text{ pm}$, which corresponds to Option (A).