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Chapter 9: Problem 19

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (e) four bonding domains and two nonbonding domains.

Short Answer

Expert verified
(a) Electron-domain geometry: Tetrahedral, Molecular geometry: Tetrahedral. (b) Electron-domain geometry: Trigonal Bipyramidal, Molecular geometry: T-shaped. (c) Electron-domain geometry: Octahedral, Molecular geometry: Square Pyramidal. (e) Electron-domain geometry: Octahedral, Molecular geometry: Square Planar.

Step by step solution

01

(a) Four bonding domains and no nonbonding domains

In this case, the central atom has four bonding domains and no nonbonding domains. According to the VSEPR theory, the electrons will arrange themselves in a tetrahedral arrangement to minimize repulsion. Electron-domain geometry: Tetrahedral Molecular geometry: Tetrahedral
02

(b) Three bonding domains and two nonbonding domains

Here, the central atom has three bonding domains and two nonbonding domains. In this situation, the electron pairs will arrange themselves in a trigonal bipyramidal arrangement. The nonbonding domains will occupy the two equatorial positions since they experience less repulsion here as compared to the axial positions. Electron-domain geometry: Trigonal Bipyramidal Molecular geometry: T-shaped
03

(c) Five bonding domains and one nonbonding domain

In this scenario, the central atom has five bonding domains and one nonbonding domain. The electron pairs will adopt an octahedral arrangement to minimize repulsion. The nonbonding domain will take up one of the equatorial positions. Electron-domain geometry: Octahedral Molecular geometry: Square Pyramidal
04

(e) Four bonding domains and two nonbonding domains

In this case, the central atom has four bonding domains and two nonbonding domains. The electron pairs will arrange themselves in an octahedral arrangement to minimize repulsion. The two nonbonding domains will occupy the opposite equatorial positions. Electron-domain geometry: Octahedral Molecular geometry: Square Planar

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Most popular questions from this chapter

Figure \(9.47\) shows how the magnetic properties of a compound can be measured experimentally. When such measurements are made, the sample is generally covered by an atmosphere of pure nitrogen gas rather than air. Why do you suppose this is done?

For both atoms and molecules, ionization energies (Section 7.4) are related to the energies of orbitals: The lower the energy of the orbital, the greater the ionization energy. The first ionization energy of a molecule is therefore a measure of the energy of the highest occupied molecular orbital (HOMO). See the "Chemistry Put to Work" box on Orbitals and Energy. The first ionization energies of several diatomic molecules are given in electron-volts in the following table: \begin{tabular}{ll} \hline Molecule & \(I_{1}(\mathrm{eV})\) \\ \hline \(\mathrm{H}_{2}\) & \(15.4\) \\ \(\mathrm{~N}_{2}\) & \(15.6\) \\ \(\mathrm{O}_{2}\) & \(12.1\) \\ \(\mathrm{~F}_{2}\) & \(15.7\) \\ \hline \end{tabular} (a) Convert these ionization energies to \(\mathrm{kJ} / \mathrm{mol} .(\mathrm{b})\) On the same plot, graph \(I_{1}\) for the \(\mathrm{H}, \mathrm{N}, \mathrm{O}\), and \(\mathrm{F}\) atoms (Figure 7.11) and \(I_{1}\) for the molecules listed. (c) Do the ionization energies of the molecules follow the same periodic trends as the ionization energies of the atoms? (d) Use molecular orbital energy-level diagrams to explain the trends in the ionization energies of the molecules.

The molecule shown here is difluaromethane \(\left(\mathrm{CH}_{2} \mathrm{~F}_{2}\right)\) which is used as a refrigerant called 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, in what direction will the overall dipole moment vector point in the molecule? [Sections \(9.2\) and 9.3]

Explain the following (a) The peroxide ion, \(\mathrm{O}_{2}{ }^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-} \cdot\) (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger \(\mathrm{O}\) - \(\mathrm{O}\) bond than \(\mathrm{O}_{2}\) itself.

You can think of the bonding in the \(\mathrm{Cl}_{2}\) molecule in several ways. For example, you can picture the Cl- -Cl bond containing two electrons that each come from the \(3 p\) orbitals of a \(\mathrm{Cl}\) atom that are pointing in the appropriate direction. However, you can also think about hybrid orbitals. (a) Draw the Lewis structure of the \(\mathrm{Cl}_{2}\) molecule. (b) What is the hybridization of each \(\mathrm{Cl}\) atom? (c) What kind of orbital overlap, in this view, makes the Cl- -Cl bond? (d) Imagine if you could measure the positions of the lone pairs of electrons in \(\mathrm{Cl}_{2}\). How would you distinguish between the atomic orbital and hybrid orbital models of bonding using that knowledge? (e) You can also treat \(\mathrm{Cl}_{2}\) using molecular orbital theory to obtain an energy level diagram similar to that for \(\mathrm{F}_{2}\). Design an experiment that could tell you if the MO picture of \(\mathrm{Cl}_{2}\) is the best one, assuming you could easily measure bond lengths, bond energies, and the light absorption properties for any ionized species.
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