`IE_(1)` of Na is less than that of Mg but `IE_(2)` of Na is higher than that of Mg- explain.

Explanation

`IE_(1)` of Na is less than that of .Mg. Reason:
Na- has electronic configuration [Ne] `3s^(1)`
Mg -has electronic configuration [Ne] `3s^(2)`
Mg has completely filled configuration so Mg has more `IE_(1)` than Na.
`IE_(2)` of Na is higher than that of Mg
`rarr Na underset(-e^(Θ))overset(IE_(1))rarr underset(("stable"))(Na^(+)) overset(IE_(2))rarr Na^(+2)`
`rarr Na^(+)` has stable inert gas configuration so `IE_(2)` of Na is very high
`Mg underset(-e^(-))overset(IE_(1))rarr Mg^(+) underset(-e^(-))overset(IE_(2))rarr underset(("stable"))(Mg^(+2))`
`rarr` By the lose of one electron from `Mg^(+)` ion forms `Mg^(+2)` ion which is more stable so low amount of energy is required.
`therefore IE_(2)` of Na is higher than Mg.