WBJEE 2026 Physics and Chemistry Paper

Concept:Hybridisation depends on the steric number, that is, the total number of bond pairs and lone pairs around the central atom.Explanation:For $\mathrm{XeF}_{6}$, the central Xe atom has $8$ valence electrons and forms $6$ bonds with F atoms.Steric number = $\frac{8+6}{2} = 7$.Thus it has $6$ bond pairs and $1$ lone pair, so hybridisation is $\mathrm{sp}^{3}d^{3}$, not $\mathrm{sp}^{3}d^{2}$.For $\mathrm{BrF}_{6}^{+}$, Br has $7$ valence electrons, and the $+1$ charge reduces the electron count.Steric number = $\frac{7+6-1}{2} = 6$, so it is $\mathrm{sp}^{3}d^{2}$ hybridised.For $\mathrm{IF}_{5}$, I has $7$ valence electrons and forms $5$ bonds.Steric number = $\frac{7+5}{2} = 6$, so it is $\mathrm{sp}^{3}d^{2}$ hybridised.For $\mathrm{XeF}_{4}$, Xe has $8$ valence electrons and forms $4$ bonds.Steric number = $\frac{8+4}{2} = 6$, so it is $\mathrm{sp}^{3}d^{2}$ hybridised.Hence, only $\mathrm{XeF}_{6}$ does not show $\mathrm{sp}^{3}d^{2}$ hybridisation.Answer:Option A: $\mathrm{XeF}_{6}$