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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

Expert verified
The correct set is (c) \( \mathrm{BCl}_{3}, \mathrm{sp}^{2} \), triangular planar.

Step by step solution

01

Analyze the First Option

The first option is \( \mathrm{BeCl}_2, \mathrm{sp}^2, \text{linear} \). Generally, \( \mathrm{BeCl}_2 \) has two electron pairs distributed linearly around \( \mathrm{Be} \). Thus, its hybridization is \( \mathrm{sp} \), not \( \mathrm{sp}^2 \). Its shape is indeed linear, but the hybridization here is incorrect.
02

Analyze the Second Option

In the second option, \( \mathrm{BeCl}_2, \mathrm{sp}^2, \text{triangular planar} \), we still face the same hybridization issue as in the first option. \( \mathrm{BeCl}_2 \) actually has \( \mathrm{sp} \) hybridization, and it is linear, not triangular planar, due to its two electron pairs.
03

Analyze the Third Option

The third option is \( \mathrm{BCl}_3, \mathrm{sp}^2, \text{triangular planar} \). \( \mathrm{BCl}_3 \) indeed has three electron pairs around the \( \mathrm{B} \), which results in \( \mathrm{sp}^2 \) hybridization. The geometry formed with \( \mathrm{sp}^2 \) hybridization is triangular planar. Therefore, this option is correct.
04

Analyze the Fourth Option

In the fourth option, \( \mathrm{BCl}_3, \mathrm{sp}^3, \text{tetrahedral} \), \( \mathrm{BCl}_3 \) was given an incorrect hybridization. Since it has three electron pairs, the correct hybridization is \( \mathrm{sp}^2 \) and the shape should be triangular planar, not tetrahedral.
05

Conclusion

After analyzing all the options, the only correct choice is the third option: \( \mathrm{BCl}_3 \), with \( \mathrm{sp}^2 \) hybridization and triangular planar shape. The other options have either incorrect hybridizations or incorrect shapes based on the molecular geometries.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. Understanding molecular geometry is crucial because it influences the physical and chemical properties of the molecule, such as polarity, reactivity, and color.
  • 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.
For example, the molecule \( \mathrm{BCl}_3 \) possesses a triangular planar shape due to the arrangement of its atoms, even though electron pairs dictate this geometry through the hybridization of orbitals.
Electron Pairs
Electron pairs play a pivotal role in dictating both the shape and hybridization of molecules. They can be categorized into two main types: bonding pairs and lone 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.
The Valence Shell Electron Pair Repulsion (VSEPR) theory explains that electron pairs arrange themselves to minimize repulsion forces, affecting the molecule's shape. For instance, in \( \mathrm{BCl}_3 \), there are three bonding pairs and no lone pairs, leading to a symmetrical triangular planar arrangement.
Hybridization Types
Hybridization refers to the mixing of atomic orbitals to form new hybrid orbitals that are better suited for bonding with other atoms. Various types of hybridization exist, depending on the number of orbitals involved.
  • \( \mathrm{sp}^3 \) 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.
Understanding hybridization types is essential for predicting the geometry and bonding capabilities of molecules.
Triangular Planar Shape
The triangular planar shape is a specific molecular geometry where the central atom is surrounded by three atoms arranged in a flat, triangular shape. This geometry is prevalent among molecules with \( \mathrm{sp}^2 \) hybridization, such as \( \mathrm{BCl}_3 \).
  • 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.
The triangular planar shape is significant as it facilitates understanding molecular interactions and bonding characteristics, impacting how these molecules interact in chemical reactions.

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Most popular questions from this chapter

The common features among the species \(\mathrm{CN}^{-}, \mathrm{CO}\) and \(\mathrm{NO}^{+}\) are: (a) Bond order three and isoelectronic (b) Bond order three and weak field ligands (c) Bond order two and \(\pi\) -acceptors (d) Isoelectronic and weak field ligands.

Iodine pentafluoride has which of the following hybridization? (a) \(\mathrm{d}^{2} \mathrm{sp}^{3}\) (b) \(\mathrm{dsp}^{3}\) (c) \(\mathrm{sp}^{3} \mathrm{~d}\) (d) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\)

The paramagnetism of \(\mathrm{O}_{2}{ }^{+}\) is due to the presence of an odd electron in the MO (a) \(\sigma^{*} 2 \mathrm{~s}\) (b) \(\pi 2 \mathrm{py}\) (c) \(\sigma^{*} 2 \mathrm{px}\) (d) \(\pi^{*} 2 \mathrm{py}\)

Dipole moment is shown by: (a) 2 -methyl propane (b) Trans-2-butene (c) Cis-2-butene (d) Carbon tetrachloride

Which contains both polar and non-polar bonds? (a) \(\mathrm{NH}_{4} \mathrm{Cl}\) (b) HCN (c) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{CH}_{4}\)
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