Chemistry - PERIODIC TABLE Question with Solution | TestHub

In the third period, Al comes after Mg and before Si. For Mg ($1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^2$), the last electron is lost from a fully filled $3\mathrm{s}$-orbital. For Al ($1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^1$), the last electron is lost from a $3\mathrm{p}$-orbital. Therefore, the first ionization enthalpy (${\Delta }_{\mathrm{i}}{\mathrm{H}}_1$) of Al is lower than that of Mg ($737\mathrm{kJ}{\mathrm{mol}}^{-1}$).

 

Furthermore, the nuclear charge increases from Na to Al. Thus, the ${\Delta }_{\mathrm{i}}{\mathrm{H}}_1$ of Al is higher than that of Na ($496\mathrm{kJ}{\mathrm{mol}}^{-1}$) but lower than that of Mg ($737\mathrm{kJ}{\mathrm{mol}}^{-1}$). Therefore, the appropriate value for the ionization potential of Al is $577\mathrm{kJ}{\mathrm{mol}}^{-1}$.