Q15 / 51 · JEE Advanced 2024 (Paper 1) · Chemistry
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.
- A$\mathrm{P}-5, \mathrm{Q}-2, \mathrm{R}-3, \mathrm{~S}-1$
- Correct answerB$\mathrm{P}-5, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-4$
- C$\mathrm{P}-4, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-1$
- D$\mathrm{P}-4, \mathrm{Q}-2, \mathrm{R}-5, \mathrm{~S}-3$
Solution
$\mathrm{XeF}_2 \Rightarrow 2$ sigma bonds and 3 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=5, \mathrm{sp}^3 \mathrm{~d}$ hybridisation geometry will be trigonal bipyramidal.
$\mathbf{P}-\mathbf{5}$
$\mathrm{XeF}_4 \Rightarrow 4$ sigma bonds and 2 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=6, \mathrm{sp}^3 \mathrm{~d}^2$ hybridisation , geometry will be octahedral.
$\mathbf{Q}-\mathbf{3}$
$\mathrm{XeO}_3 \Rightarrow 3$ sigma bonds and 1 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=4, \mathrm{sp}^3$ hybridisation , geometry will be tetrahedral.
$\mathbf{R}-\mathbf{2}$
$\mathrm{XeO}_3 \mathrm{~F}_2 \Rightarrow 5$ sigma bonds and 0 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=5, \mathrm{sp}^3 \mathrm{~d}$ hybridisation , geometry will be trigonal bipyramidal.
$\mathbf{S}-\mathbf{4}$
Hence, the correct answer is option: 'B'.