JEE Advance - Chemistry (2013 - Paper 1 Offline - No. 14)
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
104 pm
125 pm
183 pm
57 pm
Explanation
From the figure, it can be seen that the cation A+ occupies octahedral void formed by the anion X$$-$$. The radius ratio for an octahedral void is rA+ / rX$$-$$ = 0.414. Now, given that the radius of anion X$$-$$ is 250 pm, so the radius of A+ is
rA+ = 0.414 $$\times$$ 250 = 103.50 $$ \simeq $$ 104 pm
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