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 very close to:

Options

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

Answer

A

Explanation

From the figure, it can be seen that the cation `A^(+)` occupies octahedral void formed by `X^(-)` anions. The radius ratio for an octahedral void is `r_(A^(+))//r_(X)=0.414`. Now, given that the radius of anion `X^(-)` is 250 pm. So the radius of `A^(+)` is `r_(A^(+))=0.414xx250=103.50 cong 104` pm.