Question

The arrangement of $ X^- $ ions around $ A^+ $ ion in
solid $AX$ is given in the figure (not drawn to scale). If the
radius of $ X^- $ is $250\, pm$, the radius of $ A^+ $ is

• A. 104pm
• B. 125pm
• C. 183pm
• D. 57pm

Answer 1

Answer: (A)

Given arrangement represents octahedral void and for this
$ \frac{r_+ (cation)}{r_- (anion)} = 0.414 $
$ \frac{r(A^+)}{r(X^-)} = 0.414 $
$ r(A^+) = 0.414 \times r(X^-) = 0.414 \times 250pm $
$ = 103.5 pm \approx 104 pm $