QJEEM20Ch889b: JEE Main 2020 Chemistry - Class 11 JEEM20Ch889b Solution: The reaction in which the hybridisation of the underlined atom…

To determine the option in which the hybridisation of the underlined atom is affected, we need to look at each reaction and predict the changes in hybridisation that might be occuring with the reactants and products.

Option A:

In the reaction $$\underset{―}{Xe}{F}_4+Sb{F}_5\to$$, xenon tetrafluoride (XeF₄) reacts with antimony pentafluoride (SbF₅). XeF₄ is a square planar molecule, and the central Xe atom is $sp^3d^2$ hybridised. The reaction with SbF₅ can lead to the formation of a complex where the hybridisation of Xe changes if it coordinates with SbF₅. Thus, the hybridisation of Xe may indeed be affected in this case.

$$\underset{―}{Xe}$$F4	+	SbF5	$$\to$$	[XeF3]+[SbF6]-
sp3d2				sp3d   sp3d2

Option B:

In the reaction $$H_2\underset{―}{S}{O}_4+NaCl\stackrel{420K}{⟶}$$, sulfuric acid (H₂SO₄) reacts with sodium chloride (NaCl). The underlined atom here is sulfur (S). In sulfuric acid, sulfur is hybridised as $sp^3$ . Upon heating, there is typically no change in the hybridisation of the sulfur atom within sulfuric acid as part of this reaction. Therefore, the hybridisation of S is likely not affected.

Option C:

In the reaction $$H_3\underset{―}{P}{O}_2\stackrel{Disproportionation}{⟶}$$, phosphorous acid (H₃PO₂) is undergoing a disproportionation reaction. The underlined atom is phosphorus (P). In H₃PO₂, the hybridisation of the phosphorus atom is $sp^3$ . A typical disproportionation reaction of phosphorous acid would result in phosphine (PH₃) and orthophosphoric acid (H₃PO₄). The phosphorus atom in PH₃ will have $sp^3$ hybridisation while in H₃PO₄ the phosphorus atom is $sp^3$ hybridised. Given this information, the hybridisation of phosphorus may or may not change, depending on the resulting state of the phosphorus atom in the products. It's not a clear-cut case without knowing the specific products.

Option D:

In the reaction $$\underset{―}{N}{H}_3\stackrel{H^{+}}{⟶}$$, ammonia (NH₃) is reacting with a proton (H⁺). The underlined atom is nitrogen (N). In NH₃, the nitrogen is $sp^3$ hybridised. When ammonia accepts a proton, it forms ammonium ion (NH₄⁺), which is also $sp^3$ hybridised. Therefore, there is no change in the hybridisation of nitrogen in this case.

Given this information, the option in which the hybridisation of the underlined atom is most clearly affected is Option A, where the reaction can lead to coordination changes to the Xe atom, potentially affecting its hybridisation.