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

Explanation

The electronic configurations of `Na` and `Mg` are
`Na:1s^(2)2s^(2)2p^(6)3s^(1) Mg:1s^(2)2s^(2)2p^(6)3s^(2)`
Thus, the first electron in both the cases has to be removed from the `3s`-orbital but the nuclear charge of `Na(+11)` is lower than that of `Mg(+12)`, therefore, the `IE_(1)` of sodium is lower than that of magensium.
After the loss of first electron, the electronic configuration of `Na^(o+)` is `1s^(2)2s^(2)2p^(6)`. Here, the electron is to be removed from inert(neon) gas configuration which is very group stable and hence removal of second electron from sodium is very difficult.
However, in case of magesium, after the loss of first electron, the electronic configuration of `Mg^(o+)` is `1s^(2)2s^(2)2p^(6)3s^(1)`. Here, the electron is to be removed from a `3s` orbital which is much easier than to remove an electron from inert gas configuration. Therefore, the `IE_(2)` of sodium is higher that that magnesium.