(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 of the Valence Shell Electron Pair Repulsion (VSEPR) model is the idea that electron pairs in the valence shell of an atom repel one another due to their negative charge, arranging themselves in space to minimize repulsive forces. This model predicts molecular geometry by considering the arrangement of valence electron domains (bonding pairs and lone pairs) around the central atom. (b) In the VSEPR model, double and triple bonds are treated as single electron domains because they occupy roughly the same space as single bonds due to the nature of the overlapping orbitals involved (one σ bond and additional π bonds in parallel orbitals). This simplifies geometry predictions without sacrificing accuracy.

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(a) Physical Basis of VSEPR Model

The Valence Shell Electron Pair Repulsion (VSEPR) model is based on the idea that electron pairs in the valence shell of an atom repel one another, thereby arranging themselves in space so that their repulsion is minimized. This is because electron pairs are negatively charged, and thus they experience repulsive forces when in close proximity to each other. To predict the geometry of molecules, the VSEPR model considers the arrangement of valence electrons around the central atom in a molecule. The regions where electrons are most likely found are referred to as electron domains, which can be bonding electron pairs (shared between atoms) or lone pairs (unshared electrons). The electron domains are arranged in space to minimize the repulsive forces between them, determining the molecular geometry.
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(b) Counting Double and Triple Bonds as Single Electron Domains

In the VSEPR model, we treat double and triple bonds as single electron domains because they occupy roughly the same space as single bonds. This is due to the nature of the overlapping orbitals in double and triple bonds. In a single bond, one sigma (σ) bond is formed by the end-to-end overlap of orbitals. In a double bond, there is one σ bond and one pi (π) bond, which is formed by the side-by-side overlap of parallel orbitals. Similarly, in a triple bond, there is one σ bond and two π bonds. The important aspect to consider for the VSEPR model is the spatial arrangement of electron domains. In double and triple bonds, the additional π bonds are formed by parallel orbitals, which don't significantly affect the shape or size of the electron domain compared to a single bond. Therefore, treating double and triple bonds as single electron domains in the VSEPR model is justified because it simplifies geometry predictions without sacrificing accuracy.

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Most popular questions from this chapter

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (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 \(\mathrm{C}-\mathrm{C}\) bond length in butadiene $(148 \mathrm{pm})\( is a little shorter than the average \)\mathrm{C}-\mathrm{C}$ single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

(a) The nitric oxide molecule, NO, readily loses one electron to form the NO \(^{+}\) ion. Which of the following is the best explanation of why this happens: (i) Oxygen is more electronegative than nitrogen, (ii) The highest energy electron in NO lies in a \(\pi_{2 p}^{*}\) molecular orbital, or (iii) The \(\pi_{2 p}^{*}\) MO in NO is completely filled. (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in NO, NO^, and NO', and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the NO \(^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

In which of the following molecules can you confidently predict the bond angles about the central atom, and for which would you be a bit uncertain? Explain in each case. (a) \(\mathrm{H}_{2} \mathrm{~S},\) (b) \(\mathrm{BCl}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{I}\) (d) \(\mathrm{CBr}_{4}\) (e) TeBr \(_{4}\)

Dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has three forms (isomers), each of which is a different substance. (a) Draw Lewis structures of the three isomers, all of which have a carbon-carbon double bond. \((\mathbf{b})\) Which of these isomers has a zero dipole moment? (c) How many isomeric forms can chloroethylene, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}$, have? Would thev be expected to have dipole moments?

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of \(p\) orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?

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