The first ionization enthalpy of Na , Mg and Si , respectively, are: 496 , 737 and 786   kJ mo 1 - 1 .…

$$In the third period, $\mathrm{Al}$ comes after $\mathrm{Mg}$ and before $\mathrm{Si}$. Since in $\mathrm{Mg}$ ($1{\mathrm{s}}^2 2{\mathrm{s}}^2 2{\mathrm{p}}^6 3{\mathrm{s}}^2$ ), the last electron is to be lost from a fully filled $3\mathrm{s}$-orbital while in $\mathrm{Al}$ (${\mathrm{ls}}^2 2{\mathrm{s}}^2 2{\mathrm{p}}^6 3{\mathrm{s}}^2 3{\mathrm{p}}^1$) it is to be lost from a $3\mathrm{p}$-orbital, therefore, ${∆}_{\mathrm{i}}{\mathrm{H}}_1$ of $\mathrm{Al}$ is lower than that of $\mathrm{Mg}$ ($737 \mathrm{kJ} {\mathrm{mol}}^{-1}$). Further, nuclear charge increases in going from $\mathrm{Na}$ to $\mathrm{Al}$, therefore ${∆}_{\mathrm{i}}{\mathrm{H}}_1$ of $\mathrm{Al}$ is higher than that of $\mathrm{Na}$($496 \mathrm{kJ} {\mathrm{mol}}^{-1}$) but lower than that of $\mathrm{Mg}$ ($737 \mathrm{kJ} {\mathrm{mol}}^{-1}$). Therefore, the appropriate value of ionisation potential of  $\mathrm{Al}$ is $577 \mathrm{kJ} {\mathrm{mol}}^{-1}$.