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

Reaction of H₂SO₄ and NaCl forms Na₂SO₄ and HCl. Hybridization of S in H₂SO₄ is sp³ and hybridization of S in Na₂SO₄ is also sp³.

 $\underset{{\mathrm{sp}}^3}{{\mathrm{H}}_2{\mathrm{SO}}_4}+2\mathrm{NaCl}\stackrel{420 \mathrm{K}}{\to }\underset{{\mathrm{sp}}^3}{{\mathrm{Na}}_2{\mathrm{SO}}_4}+2\mathrm{HCl}$

Ammonia forms ammonium ion. Hybridization of N in ammonia is sp³ and in ammonium ion is also sp³.

$\underset{{\mathrm{sp}}^3}{{\mathrm{NH}}_3}+{\mathrm{H}}^{+}\to \underset{{\mathrm{sp}}^3}{{\mathrm{NH}}_4^{+}}$

On disproportionation, hypo phosphorous acid (H₃PO₂) forms phosphoric acid (H₃PO₄).

(c) $\underset{{\mathrm{sp}}^3}{2{\mathrm{H}}_3{\mathrm{PO}}_2}\stackrel{\mathrm{Disproportionation} \mathrm{reaction}}{\to }\underset{{\mathrm{sp}}^3}{{\mathrm{H}}_3{\mathrm{PO}}_4}+\underset{{\mathrm{sp}}^3}{{\mathrm{PH}}_3}$

Reaction of Xenon fluoride with Antimony pentafluoride is:

(d) $\underset{{\mathrm{sp}}^3{\mathrm{d}}^2}{{\mathrm{XeF}}_4}+{\mathrm{SbF}}_5⟶\underset{{\mathrm{sp}}^3\mathrm{d}}{{\left[{\mathrm{XeF}}_3\right]}^{+}}{\left[{\mathrm{SbF}}_6\right]}^{-}$

In the above reaction, hybridization of Xe changes from sp³d² to sp³d.