JEE Main 2023 — Classification of Elements and Periodicity in Properties Question with Solution

Mg have $2$ electrons in valance shell and after donation of $2$electrons it converts into ${\mathrm{Mg}}^{2+}$ ion and attains inert gas configuration, now to ionize one more electron from inert gas configuration is very difficult, hence it is easy to ionize $\mathrm{Mg}$ to${\mathrm{Mg}}^{2+}$.  ${\mathrm{IE}}_2$ is always greater than ${\mathrm{IE}}_1$ because on removal of ${\mathrm{e}}^{-}$, size decreases and effective nuclear charge increases.So Assertion is correct.

${\mathrm{Mg}}^{2+}$ is a smaller ion than ${\mathrm{Mg}}^{+}$ and has a greater charge density. This means that ${\mathrm{Mg}}^{2+}$ has a stronger attraction to its electrons than ${\mathrm{Mg}}^{+}$ and requires more energy to remove an electron from it. So reason is also correct.