p-Block Elements (Groups 15-18): NEET previous year questions

Both nitrogen and oxygen are period-2 elements lacking accessible d-orbitals, so nitrogen cannot form a d$\pi$-p$\pi$ bond with oxygen, making statement D incorrect. Nitrogen does form p$\pi$-p$\pi$ multiple bonds (as in $\mathrm{N}_{2}$), the heavier elements P and As have vacant d-orbitals allowing d$\pi$-d$\pi$ overlap with transition metals, and P, As, Sb show catenation, so those statements are correct.

In ClF$_{3}$ the central Cl has five electron domains (3 bond pairs + 2 lone pairs) and sp$^{3}$d hybridisation, giving a trigonal bipyramidal electron arrangement. The two lone pairs occupy equatorial positions, leaving a T-shaped molecular geometry. Hence ClF$_{3}$ is T-shaped with two lone pairs on the Cl atom.

Why the other options are wrong:

- (2) ClF$_{3}$ has two, not three, lone pairs on chlorine.
- (3) Trigonal pyramidal (like NH$_{3}$) has 3 bond pairs + 1 lone pair, not the case here.
- (4) Planar trigonal (like BF$_{3}$) has no lone pairs on the central atom.

In both reactions $\mathrm{MnO}_{2}$ acts as the OXIDISING agent on the halide, so manganese itself goes from +4 down to +2 - it is REDUCED, not oxidised. Statement-I therefore states the opposite of what happens and is wrong, while Statement-II describes the same reduction correctly for the iodide. This is the classic laboratory route to chlorine, $\mathrm{MnO}_{2} + 2\mathrm{NaCl} + 2\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{MnSO}_{4} + \mathrm{Na}_{2}\mathrm{SO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + \mathrm{Cl}_{2}$. NCERT Class 12, The p-Block Elements (Group 17, preparation of chlorine).

Peroxodisulphuric acid ($\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}$) contains the peroxide linkage (-O-O-), which is not present in the other oxoacids listed. Chapter: p-Block Elements (Groups 15-18).