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

We begin by recalling that the hybridisation of the central atom in any species is decided by the total number of σ-bonds plus lone pairs around that atom. A convenient expression for the steric number (SN) is

$$\text{SN}=\frac12\left(V+M-C+A\right),$$

where

$$V=\text{valence electrons of the central atom},$$

$$M=\text{number of monovalent atoms attached},$$

$$C=\text{charge on the cation (if any)},$$

$$A=\text{charge on the anion (if any)}.$$

The hybridisation is then

$$ \begin{aligned} \text{SN}=2 &\;\Rightarrow\; sp,\\ \text{SN}=3 &\;\Rightarrow\; sp^2,\\ \text{SN}=4 &\;\Rightarrow\; sp^3,\\ \text{SN}=5 &\;\Rightarrow\; sp^3d,\\ \text{SN}=6 &\;\Rightarrow\; sp^3d^2. \end{aligned} $$

Now we analyse every option one by one, always determining the steric number before and after the reaction.

Option A: $$\underline{H_3PO_2}\;\overset{\text{disproportionation}}{\rightarrow}\;PH_3+H_3PO_3$$

• In $$H_3PO_2$$ the phosphorus forms two $$P\!-\!H$$ $$\sigma$$-bonds, one $$P\!-\!OH$$ $$\sigma$$-bond and contains one lone pair: SN $$=3\sigma+1\text{ LP}=4\Rightarrow sp^3.$$

• In $$PH_3$$ the phosphorus has three $$P\!-\!H$$ bonds and one lone pair: SN $$=4\Rightarrow sp^3.$$

• In $$H_3PO_3$$ the phosphorus has three $$P\!-\!OH$$ bonds and one lone pair: SN $$=4\Rightarrow sp^3.$$

Because the hybridisation remains $$sp^3$$ throughout, the underlined atom’s hybridisation is unchanged.

Option B: $$H_2SO_4+NaCl\overset{420\,\text{K}}{\rightarrow}\;NaHSO_4+HCl$$

• Sulphur in both $$H_2SO_4$$ and $$NaHSO_4$$ is present in the tetrahedral $$\text{SO}_4^{2-}$$ skeleton. Each has four σ-bonds and no lone pair on sulphur, so SN $$=4\Rightarrow sp^3.$$

Hence the hybridisation of sulphur does not change.

Option C: $$\underline{NH_3}\overset{H^+}{\rightarrow}NH_4^+$$

• In $$NH_3$$ we have three $$N\!-\!H$$ bonds and one lone pair: SN $$=4\Rightarrow sp^3.$$

• In $$NH_4^+$$ we have four $$N\!-\!H$$ bonds and no lone pair: SN $$=4\Rightarrow sp^3.$$

So the nitrogen retains $$sp^3$$ hybridisation; no change occurs.

Option D: $$\underline{Xe}F_4+SbF_5\;\longrightarrow\;XeF_3^++SbF_6^-$$

• In $$XeF_4$$ xenon is surrounded by four $$Xe\!-\!F$$ σ-bonds and two lone pairs: SN $$=4\sigma+2\text{ LP}=6\Rightarrow sp^3d^2.$$ (This corresponds to the square-planar shape after accounting for lone-pair positions.)

• In $$XeF_3^+$$ xenon carries three $$Xe\!-\!F$$ σ-bonds and now only two lone pairs because of the positive charge: SN $$=3\sigma+2\text{ LP}=5\Rightarrow sp^3d.$$ The geometry becomes T-shaped.

Thus the steric number drops from 6 to 5 and the hybridisation changes from $$sp^3d^2$$ to $$sp^3d.$$

Among all the options, only in Option D does the underlined atom experience a change in hybridisation.

Hence, the correct answer is Option 4.