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

Explanation

The E.C. Of Na is `1s^(2) 2s^(2) 2p^(6) 3s^(1)` and E.C. Of Mg is `1s^(2) 2s^(2) 2p^(6) 3s^(2)` . In both the cases first electron is to be removed from the 3s orbital but nuclear charge of Mg (+2) is hgiher than that of Na (+11). Further more 3s- orbital in Mg is completely filled (more stable ) while in Na it is only half- filled (less stable ) . Therefore `Delta_(i) H` of Na is lower than that of Mg.
`Na (1 s^(2) 2s^(2) 2p^(6) 3s^(1)) overset(Delta_(i)H_(1))(to) Na^(+) (1s^(2) 2s^(2) 2p^(6)) overset(Delta_(i)H_(2))(to) Na^(2+) (1s^(2) 2s^(2) 2p^(5))`
Neon gas configuration
(more stable )
`Mg (1s^(2) 2s^(2) 2p^(6) 3s^(2)) overset(Delta_(i) H_(1))(to) Mg^(+) (1s^(2) 2s^(2) 2p^(6) 3s^(1)) overset(Delta_(i)H_(2))(to) Mg^(2+) (1s^(2) 2s^(2) 2p^(6))`
Neon gas configuration
(more stable)
After the removal of the the first electron `Na^(+)` acquired the more stable neon gas configuration . Thus for the second electron from Na is to be removed from the stable noble (i.e., neon ) gas configuration but the loss of second electron from Mg gives the more stable neon gas configuration .Thus `Delta_(i) H_(2)` of Na is more than `Delta_(i) H_(2)` of Mg.