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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 ........ |
Answer» Solution : `(r_A^+)/(r_(X^-)) = 0.414implies r_(A)^(+) + 0.414 XX 250` = 104 pm. |
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