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Chapter 9: Problem 85

(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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The physical basis for the VSEPR model is the repulsion between valence electron pairs due to their negative charge, causing them to arrange in a way that minimizes repulsions, ultimately determining the molecule's shape. When applying the VSEPR model, we consider double or triple bonds as a single electron domain because additional electron density from 蟺 bonds in multiple bonds does not significantly contribute to overall repulsion. This is due to the different spatial distribution of electron density in 蟺 bonds compared to 蟽 bonds, and considering multiple bonds as single electron domains simplifies the model's application for predicting molecular shapes.

Step by step solution

01

Understanding the VSEPR Model

The VSEPR (Valence Shell Electron Pair Repulsion) model is a widely used model for predicting the shapes of molecules based on the assumption that electron pairs will repel each other and arrange themselves in a way that minimizes these repulsive forces. The model primarily focuses on the arrangement of electron pairs around the central atom of a molecule. These electron pairs can be either bonding pairs (shared between two atoms) or lone pairs (not shared with any other atoms). The VSEPR model is built on the observation that electron pairs in the valence shell of an atom are negatively charged and will repel each other. They will arrange themselves in a way that minimizes these repulsions, leading to the observed shape of the molecule.
02

Electron Domains and Their Representation in the VSEPR Model

When applying the VSEPR model, electron domains are used to describe the regions of electron density around a central atom. An electron domain can be a single bond, a double bond, a triple bond, or a lone pair of electrons. It is essential to determine the number of electron domains around an atom in order to predict the corresponding molecular geometry using the VSEPR model. Now, let's address the second part of the exercise, which asks why we can treat a double or triple bond as a single electron domain.
03

Justification for Treating Multiple Bonds as a Single Electron Domain

The reason behind treating a double or triple bond as a single electron domain in the VSEPR model is that the additional electron density from double and triple bonds do not significantly contribute to the overall repulsion between electron pairs. This is because multiple bonds consist of one sigma (蟽) bond and additional pi (蟺) bonds, with 蟺 bonds having electron density distributed above and below the plane of the bond axis, rather than directly between the atoms. Consequently, the spatial extent of repulsion caused by pi (蟺) bonds is less significant than that of sigma (蟽) bonds, which are between atoms. Since the VSEPR model primarily focuses on the minimization of repulsive forces between electron pairs, it is justified to consider single, double, and triple bonds as occupying single electron domains. This simplification makes the application of the VSEPR model more straightforward and practical for predicting molecular shapes.

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Most popular questions from this chapter

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 p}-\pi_{2 p}^{\star}\) ,\(\sigma_{25}-\sigma_{25}^{*},\) and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(N_{2} .\) (b) Is \(N_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(N_{2}\) molecule has the lowest energy peak at 170 nm. To what orbital transition does this corre- spond? (a) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of kJ/mol. (e) Is the N-N bondin the first excited state stronger or weaker compared to that in the ground state?

(a) The PH \(_{3}\) molecule is polar. Does this offer experimental proof that the molecule cannot be planar? Explain. (b) It turns out that ozone, \(\mathrm{O}_{3},\) has a small dipole moment. How is this possible, given that all the atoms are the same?

The molecule shown here is diffuoromethane \(\left(\mathrm{CH}_{2} \mathrm{F}_{2}\right),\) which is used as a refrigerant called \(\mathrm{R}-32\) . (a) Based on the structure, how many electron domains surround the C atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, which of the following describes the direction of the overall dipole moment vector in the molecule: (i) from the carbon atom toward a fluorine atom, (ii) from the carbon atom to a point midway between the fluorine atoms, (iii) from the carbon atom to a point midway between the hydrogen atoms, or (iv) from the carbon atom toward a hydrogen atom? [Sections 9.2 and 9.3\(]\)

Predict whether each of the following molecules is polar or nonpolar: \((\mathbf{a}){CCl}_{4},(\mathbf{b}) \mathrm{NH}_{3},(\mathbf{c}) \mathrm{SF}_{4},(\mathbf{d}) \mathrm{XeF}_{4},(\mathbf{e}) \mathrm{CH}_{3} \mathrm{Br}\)

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN},(\mathbf{b}) \mathrm{SO}_{3}^{2-},(\mathbf{c}) \mathrm{SF}_{4}\) \((\mathbf{d}) \mathrm{PF}_{6},(\mathbf{e}) \mathrm{NH}_{3} \mathrm{Cl}^{+},(\mathbf{f}) \mathrm{N}_{3}^{-}\)
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