JEE Main · 2026 · 05 Apr 2026 · Shift 1 · p-Block Elements

Nitrogen in the intermediate oxidation states from $+1$ to $+4$ is unstable in acid and disproportionates, as in $3\mathrm{HN}{\mathrm{O}}_2\to \mathrm{HN}{\mathrm{O}}_3+2\mathrm{NO}+{\mathrm{H}}_2\mathrm{O}$, so statement A is correct. Lacking $d$ orbitals in its valence shell, nitrogen forms $p\pi \text{–}p\pi$ and not $d\pi \text{–}p\pi$ bonds, so statement B is wrong. The $\mathrm{N}-\mathrm{N}$ single bond is weakened by lone-pair repulsion at the short internuclear distance and is actually weaker than the $\mathrm{P}-\mathrm{P}$ single bond, so statement C is wrong. Being a low-mass gas, nitrogen has the lowest, not the highest, density in its group, so statement D is wrong. With only one $s$ and three $p$ valence orbitals, nitrogen can form at most four bonds, so statement E is correct. The correct statements are A and E.