JEE Main · 2009 · 12 Apr 2009 · Offline · Classification of Elements and Periodicity

Ionic size grows down a group and shrinks as positive charge rises, and for ions with the same number of electrons the one with the larger nuclear charge is smaller. Thus $\mathrm{N}{\mathrm{a}}^{+}$ is larger than $\mathrm{L}{\mathrm{i}}^{+}$, while among the two-electron pair $\mathrm{L}{\mathrm{i}}^{+}$ exceeds $\mathrm{B}{\mathrm{e}}^{2+}$ and among the ten-electron pair $\mathrm{N}{\mathrm{a}}^{+}$ exceeds $\mathrm{M}{\mathrm{g}}^{2+}$. Putting these together, the correct order is $\mathrm{N}{\mathrm{a}}^{+}>\mathrm{L}{\mathrm{i}}^{+}>\mathrm{M}{\mathrm{g}}^{2+}>\mathrm{B}{\mathrm{e}}^{2+}$.