Classification of Elements Part 2

Concept:Ionization enthalpy generally increases across a period due to increasing effective nuclear charge ($Z_{\text{eff}}$), but exceptions occur due to stable electronic configurations (half-filled or fully-filled orbitals).Explanation:In the $3^{\text{rd}}$ period, the elements are $\mathrm{Na, Mg, Al, Si, P, S, Cl, Ar}$.From left to right, $Z_{\text{eff}}$ increases, so ionization enthalpy usually increases.However, $\mathrm{Al}$ ($1s^2 2s^2 2p^6 3s^2 3p^1$) has a lower first ionization enthalpy than $\mathrm{Mg}$ ($3s^2$) because the electron lost is from a $3p$ orbital, which is easier to remove.Among the given options, $\mathrm{Al}$ has the lowest value ($578$ kJ/mol).Then $\mathrm{Si}$ ($786$ kJ/mol) comes next.$\mathrm{S}$ ($1000$ kJ/mol) has a lower ionization enthalpy than $\mathrm{P}$ ($1011$ kJ/mol) because in $\mathrm{P}$ ($3p^3$) the $3p$ orbitals are half-filled, giving extra stability, while in $\mathrm{S}$ ($3p^4$) one orbital is doubly occupied, causing repulsion and easier removal of an electron.Finally, $\mathrm{Cl}$ ($1255$ kJ/mol) has the highest value due to the highest $Z_{\text{eff}}$ in this set.Thus the correct increasing order is: $\mathrm{Al} < \mathrm{Si} < \mathrm{S} < \mathrm{P} < \mathrm{Cl}$.Shortcut:Remember the order: $\mathrm{Al} < \mathrm{Si} < \mathrm{S} < \mathrm{P} < \mathrm{Cl}$ for the $3^{\text{rd}}$ period exceptions ($\mathrm{P}$ is higher than $\mathrm{S}$}$). < br> < br> < div class="hscrollenable"> < img src="https://examimages.github.io/cdn1/jee/mains/prev/2026/jee_mn_23_jan_2026_s1/jemain_23_jan_2026_s1_sol_58.png" /></div> < br> < br> < b>Answer:</b> < br>Option B:$\mathrm{Al} < \mathrm{Si} < \mathrm{S} < \mathrm{P} < \mathrm{Cl}$ is correct.