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The arrangement of X^(-) ions around A^(...

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

104 pm

B

125 pm

C

183 pm

D

57 pm

Text Solution

Verified by Experts

The correct Answer is:
A
Cation `A^(+)` occupies octahedral void of arrangement of `X^(-)` anions
`(r_(A^+))/(r_(X^-)) = 0.414 implies (r_(A^+))/(250) = 0.414`.
or `r_(A^+) = 0.414 xx 250 = 104 " pm" `.
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