JEE Main · 2025 · 23 Jan 2025 · Shift 1 · p-Block Elements

Nitrogen's lone pair is smaller and more available for bonding than phosphorus's, so $\mathrm{N}{\mathrm{H}}_3$ has a higher proton affinity than $\mathrm{P}{\mathrm{H}}_3$ — true.

Phosphorus, with accessible 3d orbitals, can expand its octet to form species such as $\mathrm{P}{\mathrm{F}}_5$, whereas nitrogen (period 2, no d-orbitals) cannot exceed a covalency of 4; so $\mathrm{P}{\mathrm{F}}_3$ exists but $\mathrm{N}{\mathrm{F}}_5$ does not — true.

$\mathrm{N}{\mathrm{O}}_2$ is an odd-electron, paramagnetic radical, and pairing its unpaired electron with that of another $\mathrm{N}{\mathrm{O}}_2$ molecule readily gives the dimer ${\mathrm{N}}_2{\mathrm{O}}_4$ — true.

Sulfur in $\mathrm{S}{\mathrm{O}}_2$ is in the $+4$ oxidation state, intermediate between $+6$ (as in $\mathrm{S}{\mathrm{O}}_3$, ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$) and lower states such as $0$ or $-2$. $\mathrm{S}{\mathrm{O}}_2$ can therefore be oxidised, acting as a reducing agent, or reduced, acting as an oxidising agent, depending on the other reactant — the claim that it acts only as an oxidising agent is false.

The incorrect statement is that $\mathrm{S}{\mathrm{O}}_2$ can act as an oxidising agent but not as a reducing agent.