Question

The set(s) representing the correct order of ionic radius is/are:

• A. $L{i}^{+}>B{e}^{2+}>N{a}^{+}>M{g}^{2+}$
• B. $N{a}^{+}>L{i}^{+}>M{g}^{2+}>B{e}^{2+}$
• C. $L{i}^{+}>N{a}^{+}>M{g}^{2+}>B{e}^{2+}$
• D. $M{g}^{2+}>B{e}^{2+}>L{i}^{+}>N{a}^{+}$

Answer 1

The correct option is B $N{a}^{+}>L{i}^{+}>M{g}^{2+}>B{e}^{2+}$

More the positive charge, smaller will be the ionic radius.

Both $M{g}^{2+}$ and $B{e}^{2+}$ have the same $2+$ charge but down the group, the size increases.
Therefore, the radius, $R_{M{g}^{2+}}>R_{B{e}^{2+}}$. Similarly, $R_{N{a}^{+}}>R_{L{i}^{+}}$

Therefore, $N{a}^{+}>L{i}^{+}>M{g}^{2+}>B{e}^{2+}$.
i.e.,option (b) is correct.