JEE Main 2025 Apr 3 Shift 1, Chemistry Q18: Valence Bond Theory And Hybridisation

${\text{PF}}_5$: 5 P–F sigma bonds, no lone pair on P. Steric number 5 $\Rightarrow$ $s{p}^{3}d$ (trigonal bipyramidal). A → II.

${\text{SF}}_6$: 6 sigma bonds, no lone pair. Steric number 6 $\Rightarrow$ $s{p}^3{d}^2$ (octahedral). B → III.

${\text{Ni(CO)}}_4$: Ni is in oxidation state 0. CO is a strong field ligand and pairs up the $3d$ electrons, giving the empty $4s$ and three $4p$ orbitals for bonding: $s{p}^3$ (tetrahedral). C → IV.

$[{\text{PtCl}}_4{]}^{2-}$: Pt(II) is a $5d^8$ ion. In a strong field of ${\text{Cl}}^{-}$ (acting as borderline / strong here due to the 5d nature of Pt), it adopts a square planar geometry using $ds{p}^2$ hybridisation. D → I.

Matching: A-II, B-III, C-IV, D-I.