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

Solution

Chemistry ma’am · 3 steps

Count σ bonds plus lone pairs on the atom before and after; only xenon ends up with a different total.

1. The rule. Hybridisation follows the steric number: σ bonds plus lone pairs on that atom. If the total stays the same, the hybridisation stays the same, even when the shape changes.
2. Fluoride transfer. SbF₅ is a strong fluoride acceptor, so XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]⁻. In [XeF₃]⁺, Xe has 8 - 1 = 7 valence electrons. Three go into Xe–F bonds and four are left as 2 lone pairs, so the steric number falls from 6 to 5.

   Option	Before	After	Changed?
   (a) Xe	XeF₄: 4 bonds + 2 lone pairs, sp³d²	[XeF₃]⁺: 3 bonds + 2 lone pairs, sp³d	Yes
   (b) S	H₂SO₄: 4 σ bonds, sp³	HSO₄⁻: 4 σ bonds, sp³	No
   (c) P	H₃PO₂: 4 σ bonds, sp³	PH₃: 3 bonds + 1 lone pair, sp³. The P oxyacid formed: 4 σ bonds, sp³	No
   (d) N	NH₃: 3 bonds + 1 lone pair, sp³	NH₄⁺: 4 bonds, sp³	No

3. So. Only in (a) does the steric number change, from 6 to 5. Xe goes from sp³d² (square planar) to sp³d (T-shaped).

The idea to keep. Hybridisation changes only when the steric number changes. A lone pair turning into a bond pair keeps the count the same. Losing a whole ligand, as in fluoride transfer, changes it.

Answer from the NTA final answer key. Solution by PrepWiser.