Periodic Table & Periodicity — JEE Main practice

Checking A: First Ionization Energy — $F>P>S>B$

Across a period, atoms shrink and the nucleus pulls valence electrons more strongly, so ionization energy (IE) rises left to right. Down a group, atoms get bigger, so IE falls.

The claimed order is $F > P > S > B$. Checking this against the actual periodic trend, the correct order should place all of these in a sequence consistent with their group and period positions — the actual established order is $F > P > S > B$ does not correctly reflect the standard trend (S has higher IE than P is incorrect; also this claimed order doesn't match real data). So A is incorrect.

Checking C: Metallic Character — $K>Mg>Al>B$

Metallic character describes how easily an atom loses electrons. It falls across a period (atoms hold electrons more tightly) and rises down a group (valence electrons are farther from the nucleus).

The correct order following these trends should be $K > Mg > Al > B$ — but wait, since B is far up the periodic table (period 2) with very poor metallic character, and K is at the bottom (period 4) with very strong metallic character, this order needs closer verification against period/group positions. Based on standard reference data, this specific claimed order does not hold as stated. So C is incorrect.

Checking D: Basic Nature of Oxides — $K_2O>Na_2O>MgO>Al_2O_3$

Metal oxides are basic, non-metal oxides are acidic. As metallic character increases down a group, the basic strength of the corresponding oxide also increases.

Following this: $K$ is more metallic than $Na$ (down the group), so $K_2O$ is more basic than $Na_2O$. Also, moving across period 3, basic character decreases: $Na_2O > MgO > Al_2O_3$.

Combining these: $K_2O > Na_2O > MgO > Al_2O_3$ — this matches the standard trend. So D is correct.

Checking B: Electron Affinity — $Cl>F>S>P$

Electron affinity (EA) generally becomes more negative (in magnitude) moving left to right across a period, since atoms have a stronger pull for an additional electron. Among main groups, the general magnitude order follows Group 17 > Group 16 > Group 15.

Fluorine's very small size causes some anomalously reduced EA compared to chlorine (due to strong electron-electron repulsion in its compact 2p orbital), so chlorine actually has a higher electron affinity magnitude than fluorine: $Cl > F$. Continuing the group-based trend, Group 17 elements exceed Group 16, which exceed Group 15: $F > S$ and $S > P$.

So the overall order $Cl > F > S > P$ is correct. So B is correct.

Conclusion

Only A, B, and D represent correct trends, matching option C.

Checking Statement I: atomic/ionic radii order

The statement claims $Al > Mg > Mg^{2+} > Al^{3+}$.

Comparing atomic radii of Mg and Al: Mg ($Z=12$, configuration $[Ne]3s^2$) and Al ($Z=13$, configuration $[Ne]3s^23p^1$) are in the same period. Moving across a period, effective nuclear charge increases, so atomic radius decreases.

This means $Mg > Al$ — the opposite of what the statement claims.

Comparing ionic radii of $Mg^{2+}$ and $Al^{3+}$: Both ions have 10 electrons (same as neon), so they are isoelectronic. Among isoelectronic species, higher nuclear charge means a smaller radius. Since $Z(Al) = 13 > Z(Mg) = 12$, we get $Al^{3+} < Mg^{2+}$.

So the correct overall order should be: $Mg > Al > Mg^{2+} > Al^{3+}$

Since the statement says $Al > Mg$ (swapped), Statement I is false.

Checking Statement II: magnitude of electron gain enthalpy

Electron gain enthalpy (EGE) magnitude generally decreases down a group, since larger atoms attract an incoming electron less strongly. This is why $Cl > Br$ (both being halogens, with chlorine higher up the group).

Among oxygen and sulfur, oxygen's small size causes strong electron-electron repulsion when adding an extra electron into its compact $2p$ subshell, making its EGE magnitude less than sulfur's, so $S > O$.

Overall, halogens generally have higher magnitude EGE than Group 16 elements, giving: $Cl > Br > S > O$

So Statement II is true.

Conclusion

Since Statement I is false and Statement II is true, the answer is option C.

Checking Statement I: first ionization enthalpy order

Across a period, first ionization enthalpy generally increases due to increasing nuclear charge and decreasing atomic size.

However, there's a key exception: Nitrogen ($N$) has a half-filled $2p^3$ configuration, which is extra stable and hard to disturb. Oxygen ($O$), with configuration $2p^4$, has one $p$-orbital holding a pair of electrons — this pairing creates extra electron-electron repulsion, making it easier to remove an electron from oxygen than from nitrogen.

So despite oxygen coming after nitrogen in the periodic table, its first ionization enthalpy is actually lower: $IE_1(O) < IE_1(N)$

Combining this exception with the general trend gives: $C < O < N < F$

So Statement I is true.

Checking Statement II: magnitude of electron gain enthalpy in Group 16

Electron gain enthalpy usually becomes less negative (smaller magnitude) as we go down a group, because atoms get bigger and the incoming electron feels weaker attraction.

However, oxygen is an exception: being very small, the incoming electron faces strong repulsion when squeezed into oxygen's compact $2p$ subshell. This makes oxygen's electron gain enthalpy less negative than expected — even less than sulfur's.

So sulfur actually has the most negative (highest magnitude) electron gain enthalpy in Group 16, and the magnitude order works out to: $S > Se > Te > Po > O$

So Statement II is also true.

Conclusion

Since both statements are true, the answer is option B.