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Chapter 9: Problem 14

(a) How does one determine the number of electron domains in a molecule or ion? (b) What is the difference between a bonding electron domain and a nonbonding electron domain?

Short Answer

Expert verified
To determine the number of electron domains in a molecule or ion, draw its Lewis structure and count the bonding electron domains (single, double, or triple bonds) and nonbonding electron domains (lone pairs) around the central atom. The total number of electron domains helps predict the molecular geometry using the valence shell electron pair repulsion (VSEPR) theory. Bonding electron domains are associated with chemical bonds (single, double, or triple) between atoms and involve electron pairs holding the atoms together. Nonbonding electron domains correspond to lone pairs of electrons that are not involved in bonding. Both types of electron domains influence the molecular geometry in the VSEPR model.

Step by step solution

01

Answer to the question (a) - Determining the number of electron domains in a molecule or ion

To determine the number of electron domains in a molecule or ion, follow these steps: 1. Draw the Lewis structure of the molecule or ion. 2. Count the number of bonding electron domains (single, double, or triple bonds) around the central atom. 3. Count the number of nonbonding electron domains, which are lone pairs, around the central atom. 4. Add the number of bonding electron domains and nonbonding electron domains together to find the total number of electron domains in the molecule or ion. The total number of electron domains helps in predicting the molecular geometry using the valence shell electron pair repulsion (VSEPR) theory.
02

Answer to the question (b) - Difference between a bonding electron domain and a nonbonding electron domain

Bonding electron domain and nonbonding electron domain are two types of electron domains present around the central atom in a molecule or ion: 1. Bonding electron domain: These are electron domains associated with chemical bonds (single, double, or triple) formed between two atoms. They are occupied by bonding electron pairs, which are involved in holding the atoms together in a molecule or ion. In the context of VSEPR theory, each type of bond (single, double, or triple) counts as one electron domain regardless of the number of shared electrons. 2. Nonbonding electron domain: These are electron domains associated with lone pairs of electrons that are not involved in forming chemical bonds. These electron pairs exist around the central atom and are not shared with other atoms. In the context of VSEPR theory, each lone pair of electrons represents one nonbonding electron domain. The main difference between the two is that bonding electron domains involve electrons shared between atoms to form chemical bonds, whereas nonbonding electron domains correspond to lone pairs of electrons that are not involved in bonding. Both types of electron domains contribute to the overall electron domain geometry and influence the molecular geometry in the VSEPR model.

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Most popular questions from this chapter

You can think of the bonding in the \(\mathrm{Cl}_{2}\) molecule in several ways. For example, you can picturethe 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 Cl atom? (c) What kind of orbital overlap, in this view, makes the \(\mathrm{Cl}-\mathrm{Cl}\) 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.

The lactic acid molecule, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\), gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\pi\) and how many \(\sigma\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around each carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$ \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6,}\) you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) Suggest a hybridization scheme for the Xe atom in XeF \(_{6}\). (e) The molecule \(\mathrm{IF}_{7}\) has a pentagonal-bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\)

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: $$ \begin{array}{ll} \hline \text { Molecule } & I_{1}(\mathrm{eV}) \\ \hline \mathrm{H}_{2} & 15.4 \\ \mathrm{~N}_{2} & 15.6 \\ \mathrm{O}_{2} & 12.1 \\ \mathrm{~F}_{2} & 15.7 \\ \hline \end{array} $$ (a) Convert these ionization energies to \(\mathrm{kJ} / \mathrm{mol}\). (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.

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN}\), and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?
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