JEE Advanced · Chemistry · 4. Chemical Bonding
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.
| List - I | List - II |
| (P) \(XeF _2\) | (1) Trigonal bipyramidal and two lone pair of electrons |
| (Q) \(XeF _4\) | (2) Tetrahedral and one lone pair of electrons |
| (R) \(XeO _3\) | (3) Octahedral and two lone pair of electrons |
| (S) \(XeO _3 F_2\) | (4) Trigonal bipyramidal and no lone pair of electrons |
| (5) Trigonal bipyramidal and three lone pair of electrons |
- A \(\mathrm{P}-5, \mathrm{Q}-2, \mathrm{R}-3, \mathrm{~S}-1\)
- B \(\mathrm{P}-5, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-4\)
- C \(\mathrm{P}-4, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-1\)
- D \(\mathrm{P}-4, \mathrm{Q}-2, \mathrm{R}-5, \mathrm{~S}-3\)
Answer & Solution
Correct Answer
(B) \(\mathrm{P}-5, \mathrm{Q}-3, \mathrm{R}-2, \mathrm{~S}-4\)
Step-by-step Solution
Detailed explanation
\(\mathrm{XeF}_2 \Rightarrow 2\) 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.
\(\mathbf{S}-\mathbf{4}\)
\(\mathbf{P}-\mathbf{5}\)
\(\mathrm{XeF}_4 \Rightarrow 4\) 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.
\(\mathbf{S}-\mathbf{4}\)
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