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(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

Short Answer

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(a) The VSEPR model is based on minimizing electron pair repulsion to determine molecule shape. (b) Double or triple bonds are counted as single electron domains because they act as one region of electron density.

Step by step solution

01

Understanding the VSEPR Model

The VSEPR (Valence Shell Electron Pair Repulsion) model is based on the principle that electron pairs around a central atom in a molecule will adopt a geometry that minimizes repulsion. This results in a spatial arrangement where electron pairs are as far apart as possible, thus determining the molecular shape.
02

Repulsion and Electron Pair Arrangement

Repulsion occurs because electron pairs are negatively charged and, like charges repel. By arranging themselves as far apart as possible, the electrons minimize potential energy in the molecule. This arrangement ultimately defines the molecule's shape.
03

Counting Electron Domains in VSEPR

In VSEPR, any bond (single, double, or triple) is treated as a single electron domain. This is justified because the entire bond acts as a singular region of electron density which exerts repulsion on other regions in a similar way, regardless of whether it contains multiple pairs or one pair of electrons.
04

Effect of Multiple Bonds

Multiple bonds, although having more electron density, still occupy one location in space and impose similar repulsive forces as single bonds. The VSEPR model simplifies calculations by considering them as a single domain, focusing on spatial positioning rather than bond order.

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Key Concepts

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

Electron Pair Repulsion
In chemistry, understanding how electrons interact is crucial for predicting molecular shapes. The principle of electron pair repulsion is at the heart of the VSEPR (Valence Shell Electron Pair Repulsion) model. Electrons are negatively charged, meaning they repel each other due to similar charges. This is a fundamental law of electrostatics: like charges repel. In a molecule, electron pairs tend to position themselves as far apart as possible around the central atom to minimize repulsion. This distribution of electron pairs dictates the geometry of a molecule. By reducing the repulsion, a molecule can achieve a more stable and low-energy configuration. Thus, the arrangements that molecules assume under the VSEPR model reflect the natural effort to minimize electrostatic repulsions.
Molecular Geometry
The term molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. Using the VSEPR model, we can predict molecular geometry by considering electron pair repulsion. When electron pairs arrange themselves to minimize repulsive forces, they define specific, predictable shapes:
  • Linear
  • Trigonal Planar
  • Tetrahedral
  • Trigonal Bipyramidal
  • Octahedral
These shapes depend on the number of electron pairs and their spatial arrangement. For example, a molecule with two electron pairs will likely form a linear shape. By understanding molecular geometry, we can infer properties like polarity, reactivity, and biological activity.
Electron Domains
The concept of electron domains is vital in understanding molecular shapes via the VSEPR model. An electron domain is any area where electrons are likely to be found, such as a single bond, double bond, triple bond, or lone pair. In the context of the VSEPR model, each of these is considered a single electron domain. Even though a double or triple bond contains more electron density than a single bond, it still counts as one domain. This is because it occupies one region of space around the central atom. This simplification allows chemists to focus on the geometry and three-dimensional arrangement of these domains, rather than the specific types of bonds involved.
Bonding and Repulsion
Bonding and repulsion are fundamental concepts intertwined with electron domains and molecular geometry. When atoms bond in a molecule, they share electrons. This process creates bonds composed of shared electron pairs. Meanwhile, lone pairs of electrons remain unbonded and can influence the shape of a molecule. Both bonding and lone pairs create repulsion forces. However, lone pairs are generally more repulsive than bonding pairs due to their proximity to the central atom. Thus, they may cause slight distortions in molecular geometry. Understanding how bonding and repulsion interact allows for better predictions of molecular shapes and behaviors. This insight is essential in fields such as drug design and material science, where the shape and stability of molecules can be of great importance.

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Most popular questions from this chapter

Ethyl propanoate, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3},\) gives a fruity pineapple-like smell. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\sigma\) and how many \(\pi\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

Ammonia, \(\mathrm{NH}_{3}\), reacts with incredibly strong bases to produce the amide ion, \(\mathrm{NH}_{2}^{-}\). Ammonia can also react with acids to produce the ammonium ion, \(\mathrm{NH}_{4}^{+}\). (a) Which species (amide ion, ammonia, or ammonium ion) has the largest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle? \((\mathbf{b})\) Which species has the smallest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle?

Give the electron-domain and molecular geometries for the following molecules and ions: \((\mathbf{a}) \mathrm{BeF}_{2},(\mathbf{b}) \mathrm{AsCl}_{5},(\mathbf{c}) \mathrm{NO}_{2}^{-},\) (d) \(\mathrm{CS}_{2}\), (e) \(\mathrm{SF}_{4}\) (f) \(\mathrm{BrF}_{5}\).

Indicate whether each statement is true or false. (a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital. (c) Molecules containing electrons that occupy antibonding orbitals must be unstable. (d) Electrons cannot occupy a nonbonding orbital.

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower- energy to a higher-energy molecular orbital. Would you expect the excited- state \(\mathrm{H}_{2}^{-}\) ion to be stable? (e) Which of the following statements about part (d) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

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