QJEEM09Ch4c99: JEE Main 2009 Chemistry - Class 11 JEEM09Ch4c99 Solution: The set representing the correct order of ionic radius is :

In periodic table $$Li,Na,Mg,Be$$ are present in following ways :-

Group  1	Group  2
Li	Be
Na	Mg

Note :

$$(1)$$ In periodic table from left to right the size of element decreases.

$$(2)$$ In periodic table, from top to bottom size of element increases.

So, $$B{e}^{+2}$$ will be the smallest in size.

Among $$L{i}^{+}$$ and $$N{a}^{+},$$ size of $$N{a}^{+}>L{i}^{+}$$ and among $$B{e}^{+2}$$ and $$M{g}^{+2},$$ size of $$M{g}^{+2}>B{e}^{+2}$$.

Note : Size of $$L{i}^{+}>M{g}^{+2},$$ this is a exception case.

So, the correct order will be, $$N{a}^{+}>L{i}^{+}>M{g}^{+2}>B{e}^{+2}$$