Question

The radius of which ion is closet to that of $L{i}^{+}$ ion?

• A. $N{a}^{+}$
• B. $B{e}^{2+}$
• C. $M{g}^{2+}$
• D. $A{l}^{3+}$

Answer 1

Answer: (C)

$M{g}^{2+}$

The radius of ion closest to $L{i}^{+}$ ion is $M{g}^{2+}$.

The $\left(\frac{Z}{e}\right)$ ration comes out to be $\left(11/10\right)>\left(12/10\right)>\left(3/2\right)>\left(4/2\right)$; so ionic radius order is $N{a}^{+}>M{g}^{2+}>L{i}^{+}>B{e}^{2+}$.