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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Chemistry
Chapter
Chemical Bonding and Molecular Structure

JEE Advanced 2024 · Paper 1 · Chemistry · Chemical Bonding and Molecular Structure

Based on VSEPR model, match the xenon compounds given in List-I...

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. Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of

Options

  1. AP−5,Q−2,R−3, S−1\mathrm{P}-5, \mathrm{Q}-2, \mathrm{R}-3, \mathrm{~S}-1
  2. BP−5,Q−3,R−2, S−4\mathrm{P}-5, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-4
  3. CP−4,Q−3,R−2, S−1\mathrm{P}-4, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-1
  4. DP−4,Q−2,R−5, S−3\mathrm{P}-4, \mathrm{Q}-2, \mathrm{R}-5, \mathrm{~S}-3

Correct answer

B. P−5,Q−3,R−2, S−4\mathrm{P}-5, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-4

Step-by-step solution

XeF2⇒2\mathrm{XeF}_2 \Rightarrow 2 sigma bonds and 3 lone pairs on Xe\mathrm{Xe}, number of hybrid orbitals =5,sp3 d=5, \mathrm{sp}^3 \mathrm{~d} hybridisation geometry will be trigonal bipyramidal. P−5\mathbf{P}-\mathbf{5} XeF4⇒4\mathrm{XeF}_4 \Rightarrow 4 sigma bonds and 2 lone pairs on Xe\mathrm{Xe}, number of hybrid orbitals =6,sp3 d2=6, \mathrm{sp}^3 \mathrm{~d}^2 hybridisation , geometry will be octahedral. Q−3\mathbf{Q}-\mathbf{3} XeO3⇒3\mathrm{XeO}_3 \Rightarrow 3 sigma bonds and 1 lone pairs on Xe\mathrm{Xe}, number of hybrid orbitals =4,sp3=4, \mathrm{sp}^3 hybridisation , geometry will be tetrahedral. R−2\mathbf{R}-\mathbf{2} XeO3 F2⇒5\mathrm{XeO}_3 \mathrm{~F}_2 \Rightarrow 5 sigma bonds and 0 lone pairs on Xe\mathrm{Xe}, number of hybrid orbitals =5,sp3 d=5, \mathrm{sp}^3 \mathrm{~d} hybridisation , geometry will be trigonal bipyramidal. S−4\mathbf{S}-\mathbf{4}
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