Chapter 4: Problem 53
Which one of the following is a correct set with respect to molecule, hybridization and shape? (a) \(\mathrm{BeCl}_{2}, \mathrm{sp}^{2}\), linear (b) \(\mathrm{BeCl}_{2}, \mathrm{sp}^{2}\), triangular planar (c) \(\mathrm{BCl}_{3}, \mathrm{sp}^{2}\), triangular planar (d) \(\mathrm{BCl}_{3}, \mathrm{sp}^{3}\), tetrahedral
Short Answer
Step by step solution
Analyze the First Option
Analyze the Second Option
Analyze the Third Option
Analyze the Fourth Option
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molecular Geometry
- Molecular geometry is determined by the number and arrangement of electron pairs around the central atom.
- The shape of a molecule considers only the positions of the atoms, not the electron pairs.
Electron Pairs
- Bonding pairs: Electrons shared between two atoms to form a covalent bond.
- Lone pairs: Electrons found on an atom but not involved in bonding.
Hybridization Types
- \( \mathrm{sp}^3 \) hybridization: Involves one s orbital and three p orbitals, resulting in four hybrid orbitals. This type is often seen in tetrahedral geometries.
- \( \mathrm{sp}^2 \) hybridization: Involves one s orbital and two p orbitals, creating three hybrid orbitals. It is commonly associated with triangular planar shapes, as seen in \( \mathrm{BCl}_3 \).
- \( \mathrm{sp} \) hybridization: Involves one s and one p orbital, forming two hybrid orbitals, typically resulting in a linear configuration.
Triangular Planar Shape
- In this configuration, the bond angles between adjacent atoms are ideally 120 degrees, as the atoms are symmetrically arranged around the central atom.
- The absence of lone pairs enhances the molecule's stability and keeps the three bonded atoms in a planar arrangement.