The reaction in which the hybridisation of the underlined atom is affected is

🧠 Check whether the hybridization of the underlined atom changes before and after the reaction.

🗺️ Analyse each option:

(a) $\underline{\mathrm{H_3PO_2}}$ → disproportionation products: P is sp³ in $\mathrm{H_3PO_2}$ (4 electron pairs). Disproportionation gives $\mathrm{H_3PO_3}$ and $\mathrm{PH_3}$, where P remains sp³ in both. No change ✗

(b) $\mathrm{H_2\underline{S}O_4 + NaCl}$: S in $\mathrm{H_2SO_4}$ is sp³ (4 bonds, no LP). Product $\mathrm{NaHSO_4}$ also has S in sp³. No change ✗

(c) $\mathrm{NH_3 + H^+}$: N in $\mathrm{NH_3}$ is sp³ (3 bonds + 1 LP); in $\mathrm{NH_4^+}$ N is sp³ (4 bonds, 0 LP). Hybridization remains sp³. No change ✗

(d) $\underline{\mathrm{Xe}}\mathrm{F_4 + SbF_5}$:

• Xe in $\mathrm{XeF_4}$: 4 bonds + 2 LP → 6 electron pairs → sp³d²
• Reaction: $\mathrm{XeF_4 + SbF_5 \rightarrow [XeF_3]^+[SbF_6]^-}$ (SbF₅ abstracts F⁻)
• Xe in $\mathrm{[XeF_3]^+}$: 3 bonds + 2 LP → 5 electron pairs → sp³d (T-shaped)
• Hybridization changes: sp³d² → sp³d ✓

$\boxed{\text{Answer: } \mathrm{XeF_4 + SbF_5}}$