Based on VSEPR model, match the xenon compounds given in **List-I** with the corresponding geometries and the number of lone pairs on xenon given in **List-II** and choose the correct option....

The correct option is (B). Click to read the detailed step-by-step conceptual solution.

JEE Advanced 2024 Paper 1 Chemistry Question 15 Solution

According to the VSEPR theory and the calculation of Steric Number (SN = Number of sigma bonds + Number of lone pairs):

(P) $XeF_2$ : Xenon has 8 valence electrons. It forms 2 sigma bonds with Fluorine. Remaining 6 electrons form 3 lone pairs. $SN = 2 + 3 = 5$ . The electron geometry is Trigonal bipyramidal with 3 lone pairs. (P matches 5)
(Q) $XeF_4$ : Xenon has 8 valence electrons. It forms 4 sigma bonds with Fluorine. Remaining 4 electrons form 2 lone pairs. $SN = 4 + 2 = 6$ . The electron geometry is Octahedral with 2 lone pairs. (Q matches 3)
(R) $XeO_3$ : Xenon has 8 valence electrons. It forms 3 double bonds with Oxygen (3 sigma + 3 pi). Remaining 2 electrons form 1 lone pair. $SN = 3 + 1 = 4$ . The electron geometry is Tetrahedral with 1 lone pair. (R matches 2)
(S) $XeO_3F_2$ : Xenon has 8 valence electrons. It forms 3 double bonds with Oxygen (3 sigma) and 2 single bonds with Fluorine (2 sigma). All 8 valence electrons are used in bonding ( $3 \times 2 + 2 = 8$ ). Lone pairs = 0. $SN = 5 + 0 = 5$ . The electron geometry is Trigonal bipyramidal with 0 lone pairs. (S matches 4)

Matching results: P-5, Q-3, R-2, S-4. Correct Option: B