JEE Main 2025 Apr 2 Shift 2, Chemistry Q4: Valence Bond Theory And Hybridisation

Checking each species against all three criteria:

$\mathrm{P}{\mathrm{F}}_5$ : $s{p}^{3}d$ hybridised, axial and equatorial P-F bond lengths differ, but the central P has no lone pair. Fails (c).

$\mathrm{Xe}{\mathrm{F}}_4$ : $s{p}^3{d}^2$ hybridised. Fails (a).

$\mathrm{S}{\mathrm{F}}_4$ : $s{p}^{3}d$ hybridised (see-saw geometry), axial S-F bonds longer than equatorial, and one lone pair on S. Satisfies (a), (b) and (c).

$\mathrm{Xe}{\mathrm{F}}_2$ : $s{p}^{3}d$ hybridised and has three lone pairs on Xe, but the two axial Xe-F bonds are identical. Fails (b).

$\therefore$ The required molecule is $\mathrm{S}{\mathrm{F}}_4$.