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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 physical basis for the VSEPR model is the electrostatic repulsion between electron pairs in the valence shell of an atom, which forces bonding and non-bonding electron pairs to arrange themselves in a way that minimizes the repulsion, ultimately defining the molecular geometry. (b) In the VSEPR model, double or triple bonds are counted as single electron domains because the electron pairs involved in multiple bonds are concentrated in a particular region of space due to the overlap of atomic orbitals. This assumption simplifies the application of VSEPR theory and still provides accurate predictions of molecular geometry.

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(a) Physical basis of VSEPR model

VSEPR (Valence Shell Electron Pair Repulsion) model is a widely used method to predict the molecular geometry of a given molecule by considering the repulsion between electron pairs. The physical basis for the VSEPR model lies in the electrostatic repulsion between electron pairs in the valence shell of an atom. Since electrons are negatively charged, they repel one another. This repulsion forces the bonding and non-bonding electron pairs in the valence shell of the central atom to arrange themselves in a way that minimizes the repulsion, ultimately defining the molecular geometry. In VSEPR theory, it is assumed that electron pairs will tend to minimize the repulsive forces by staying as far apart from one another as possible within the confines of the valence shell, thus determining the shape of the molecule.
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(b) Considering double and triple bonds as single electron domains

In the VSEPR model, a double or triple bond is counted as a single electron domain, despite having multiple electron pairs involved in the bond. This is justified because the electron pairs involved in multiple bonds (double or triple) are concentrated in a particular region of space due to the overlap of atomic orbitals. The multiple bonds are localized in the same region between the two bonding atoms. Since the VSEPR model is focused on the repulsive forces between electron domains, the presence of multiple bonds in the same region won't increase the repulsion to a significant extent. Moreover, considering multiple bonds as single electron domains simplifies the application of VSEPR theory and still provides accurate predictions of molecular geometry. The VSEPR model is an empirical model, and it has been observed that counting a double or triple bond as a single electron domain provides better results for predicting molecular shapes than treating them as separate electron domains.

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Key Concepts

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

Molecular Geometry
Understanding the shape of molecules is crucial in chemistry because it influences how molecules interact with each other and with other substances. Molecular geometry is the three-dimensional arrangement of atoms within a molecule. The Valence Shell Electron Pair Repulsion (VSEPR) model is the key tool used to predict this geometry. According to VSEPR theory, the shape of a molecule is primarily determined by the need to minimize repulsion between all pairs of valence electrons, both bonding and non-bonding, around the central atom.

Think of it as a social gathering where guests try to keep a comfortable distance from each other; molecules adjust their shape so that electron pairs are as far apart as possible, leading to specific molecular geometries such as linear, bent, trigonal planar, tetrahedral, and others. For example, carbon dioxide (CO2) is linear, while water (H2O) has a bent shape. The simplicity and effectiveness of the VSEPR model make it an essential concept in chemistry for predicting and understanding the structures of molecules.
Electron Pair Repulsion
The principle of electron pair repulsion is what drives the VSEPR model. Since electrons have negative charges, they naturally repel each other due to the electrostatic force between like-charged particles. This force manifests within a molecule's valence shell, compelling electron pairs to adopt an orientation that maximizes the distance between them, consequently lowering the system's overall energy.

The repulsion isn't just among lone electron pairs, but also between lone pairs and bonding pairs, and between bonding pairs themselves, although the repulsion strength varies in intensity. Lone pair-lone pair repulsion is the strongest, followed by lone pair-bonding pair repulsion, and bonding pair-bonding pair repulsion is the weakest. This hierarchy influences molecular shapes since atoms will adjust their positions to ensure that this repulsion is minimized, leading to observable, predictable patterns in molecular geometry.
Electron Domains
Electron domains are regions in a molecule where electrons are likely to be found. These include both bonding electron pairs, such as those found in single, double, and triple bonds, as well as non-bonding electron pairs or lone pairs. In the VSEPR model, all bonds, whether single, double, or triple, are treated as a single electron domain. This simplification is justified because the multiple bonds between two atoms effectively occupy the same region of space and thus exert similar repulsive forces as a single bond would.

When counting electron domains, we don't differentiate between types of bonds; rather, we tally electron groups to predict molecular geometry. For instance, methane (CH4) has four electron domains, each a single bond, leading to a tetrahedral geometry. Treating multiple bonds as one domain helps streamline the VSEPR model and allows it to robustly predict the three-dimensional structures that molecules adopt.

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Most popular questions from this chapter

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on \(\mathrm{F}\) interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}\) is a planar molecule that has the following carbocarbon bond lengths: (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(C-\) bond length in butadiene \((1.48\) A) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length \((1.54 \hat{\mathrm{A}}) .\) Does this imply that the middle \(\mathrm{C}-\mathrm{Cbond}\) in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\)? (\mathbf{d} ) Based on your answer for part ( c ),discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\) C bond.

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) ? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)

Shown here are three pairs of hybrid orbitals, with each set at a characteristic angle. For each pair, determine the type of hybridization, if any, that could lead to hybrid orbitals at the specified angle.

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.

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