$$Al\left( {3{s^2}{p^1}} \right)$$   and $$Mg\left( {3{s^2}} \right).$$   Lower energy is required to remove $$3{p^1}$$ electron than $$3{s^1}$$ electron ( penetrating effect is $$s > p > d > f\,).$$
Secondly $$Mg$$  has stable electronic configuration than $$Al$$

Metalloids are shown by the zig-zag boxes and the elements present on the right side of the boundary are nonmetals. Metals are present on the left side of the periodic table.

On moving along the period, ionization enthalpy increases.
In second period, the order of ionization enthalpy should be as follows $$:F > O > N$$
But $$N$$  has half-filled structure, therefore, it is more stable than $$O.$$  That’s why its ionization enthalpy is higher than $$O.$$  Thus, the correct order of $$IE$$  is $$F > N > O.$$

Group 15 includes $$N, P$$  as non-metals, $$As$$  and $$Sb$$  as metalloids and $$Bi$$  is a metal.

Penetration effect order is $$s > p > d > f.$$

Hydrogen have one proton and one electron, when it ionise, i.e. it lose one electron, then only proton is left in the nucleus, so $${H^ + }$$  ion is formed during ionisation which is also called proton.
\[\underset{\left( \begin{smallmatrix} {{e}^{-}}=1 \\ p=1 \end{smallmatrix} \right)}{\mathop{H}}\,\to \underset{\text{Proton}}{\mathop{{{H}^{+}}}}\,+{{e}^{-}}\]