True everywhere except here

The inert-pair effect is a real, well-taught trend — heavier p-block elements do get more reluctant to use all their valence electrons in bonding — and it's tempting to apply it flatly and conclude antimony, like the heavier bismuth, simply can't reach +5. It can: $S{b}_2{O}_5$ is a known, real compound. The inert-pair effect only becomes strong enough to rule out the higher oxidation state one element further down the group, at bismuth. Statement II over-extends a genuine trend one element too far.

Statement I is genuinely true, and the instinct that carries into Statement II is a familiar one from elsewhere in the syllabus: zymogens like pepsinogen get switched on by acid. Applying that same acid-activates-it intuition to the Bt protoxin is the trap — inside an insect's gut, which runs alkaline rather than acidic, that's precisely what flips the inactive protoxin into its active, toxic form. The general "acid activates it" pattern doesn't hold in this specific gut.

Glucose has a free aldehyde group, and the textbook rule for aldehydes is that they react with reagents like these — reduce Tollens', add HCN, add NH${}_2$OH, give Schiff's test, form a bisulphite adduct with NaHSO${}_3$. Correctly remembering that glucose is the exception to part of that rule is exactly what makes this option tempting: it catches the famous exception, Schiff's reagent (glucose failing that test is the classic proof its aldehyde is "hidden" as a cyclic hemiacetal), then pairs it with HCN — extending the same "the ring blocks it" logic one reagent too far. HCN addition still goes through, via the small equilibrium amount of open-chain aldehyde always present, so the real second exception is NaHSO${}_3$, not HCN — the rule breaks for exactly two of these five reagents, and this option catches only one of the right two.

This is the order you get by trusting the general rule that ionisation enthalpy climbs across a period — and across most of Period 2, it does. Beryllium is the exception this question is built around: its $2s^2$ configuration is a filled subshell, unusually hard to break into, so Be actually sits above boron, whose lone $2p$ electron is comparatively easy to remove. The safe, general trend gives exactly this — wrong — monotonic order; the real order dips at boron.

A and C are both genuinely force-requiring scenarios, so the instinct that "moving something away from a strong magnet always takes force" feels well-supported by two out of three. D is built to break that instinct: a sheet that is both non-conducting and non-polar has no magnetic attraction to fight (it isn't magnetic) and no eddy currents to oppose its motion (it doesn't conduct), so nothing resists it being moved away at all. The general "near a magnet, moving things costs force" intuition needs a magnetic or conducting property to actually apply, and D has neither.

Statement II is a faithful summary of Gause's principle — the inferior competitor is eliminated when resources are limiting — and its correctness lends Statement I credibility by association. But Statement I quietly changes the condition the principle depends on: competitive exclusion is about two species competing for the same limited resource, not different ones. Species after different resources can coexist indefinitely; it's the word "different" doing the damage here, not any error in the rest of the sentence.

Echinoderms are the one phylum on this list NCERT teaches with a twist built in: bilateral as larvae, radial as adults — a fact memorable enough that "echinoderms and bilateral symmetry go together" tends to stick. The question specifically asks about adults, which is exactly the scope where that memorable exception stops applying — adult echinoderms are radially symmetric, same as ctenophores and coelenterates. The bilateral-symmetry fact about echinoderms is true, it's just true of the wrong life stage for this question; the actual answer, Hemichordata, carries none of that larva/adult complication — it's simply, unambiguously bilateral throughout.