How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

(A) The electronic configuration of $Na$ is $[Ne] 3s^{1}$ and that of $Mg$ is $[Ne] 3s^{2}$.
The configuration of $Mg$ is more stable (being completely filled) than that of $Na$.
Therefore,the first ionization enthalpy of $Mg$ is higher than that of $Na$.
After the loss of one electron from $Na$,it acquires the stable noble gas configuration of $Ne$ $(1s^{2} 2s^{2} 2p^{6})$.
On the other hand,for $Mg$,the configuration after the first ionization becomes $[Ne] 3s^{1}$.
Thus,the electronic configuration of $Na^{+}$ is more stable than that of $Mg^{+}$,making the removal of the second electron from $Na^{+}$ much more difficult than from $Mg^{+}$.
$Na ([Ne] 3s^{1}) \longrightarrow Na^{+} ([Ne]) + e^{-}$
$Mg ([Ne] 3s^{2}) \longrightarrow Mg^{+} ([Ne] 3s^{1}) + e^{-}$
$IE_{1}(Na) < IE_{1}(Mg)$
$IE_{2}(Na) > IE_{2}(Mg)$