JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

From the given reaction conditions: Xe(g) + 2 F _2 (g) [ 1 : 5 ratio ] 873 K, 7 bar XeF _4 Thus, P is XeF _4. XeF _4 + O _2 F _2 143 K XeF _6 + O _2 Thus, Q is XeF _6. XeF _6 + 3 H _2 O complete hydrolysis XeO _3 + 6 HF Thus, R is XeO _3. Evaluating the given options: (A) P ( XeF _4) has 8 valence electrons on the central Xe atom. It forms 4 single bonds with F, leaving 4 electrons which form 2 lone pairs. This statement is correct. (B) Q ( XeF _6) has 6 bond pairs and 1 lone pair on the central Xe atom. Its geometry is distorted octahedral, not perfect octahedral. This statement is incorrect. (C) Q ( XeF _6) is a strong fluorinating agent. This statement is correct. (D) R ( XeO _3) has 3 double bonds and 1 lone pair on the central Xe atom. Its molecular structure is trigonal pyramidal. This statement is correct. Answer: P has two lone pairs of electrons on the central atom.; Q can act as a fluorinating agent.; The molecular structure of R is trigonal pyramidal.