Match List-I with List-II
List-I List-II
A. XeO3 (I) sp3d; linear
B. XeF2 (II) sp3; pyramidal
C. XeO

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Question

Match List-I with List-II
List-I List-II
A. XeO3 (I) sp3d; linear
B. XeF2 (II) sp3; pyramidal
C. XeOF4 (III) sp3d3; distorted octahedral
D. XeF6 (IV) sp3d2; square pyramidal
Choose the correct answer from the options given below :
(1) A-IV, B-II, C-I, D-III
(2) A-II, B-I, C-IV, D-III
(3) A-II, B-I, C-III, D-IV
(4) A-IV, B-II, C-III, D-I

Accepted Answer

Correct Answer: B — A-II, B-I, C-IV, D-IIIStep-by-step Solution:Xe has 8 valence electrons. Count total electron pairs (BP + LP) to determine hybridisation and shape:A. XeO₃: 3 bond pairs + 1 lone pair = 4 total → sp³ → electron geometry: tetrahedral → molecular shape: pyramidal (II)B. XeF₂: 2 bond pairs + 3 lone pairs = 5 total → sp³d → electron geometry: trigonal bipyramidal → 3 LPs in equatorial positions → molecular shape: linear (I)C. XeOF₄: 5 bond pairs (1 Xe=O + 4 Xe-F) + 1 lone pair = 6 total → sp³d² → electron geometry: octahedral → 1 LP → molecular shape: square pyramidal (IV)D. XeF₆: 6 bond pairs + 1 lone pair = 7 total → sp³d³ → electron geometry: pentagonal bipyramidal → 1 LP → molecular shape: distorted octahedral (III)Key Concepts:VSEPR theory: Shape depends on total electron pairs (BP + LP). Lone pairs cause more repulsion than bond pairs → distort geometry.Hybridisation: 4 pairs → sp³; 5 pairs → sp³d; 6 pairs → sp³d²; 7 pairs → sp³d³XeF₂ has 3 LPs in equatorial plane of trigonal bipyramid → linear molecular shape.

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