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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

Predict whether each of the following molecules is polar or nonpolar: (a) IF, (b) \(\mathrm{CS}_{2}\), (c) \(\mathrm{SO}_{3}\), (d) \(\mathrm{PCl}_{3}\), (e) \(\mathrm{SF}_{6}\), (f) \(\mathbb{F}_{5}\).

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.

In ozone, \(\mathrm{O}_{3}\), the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

\((\mathrm{a})\) If the valence atomic orbitals of an atom are \(s p\) hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double ( \(\sigma\) plus \(\pi\) ) bond, or would they be the same? Explain.

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about Si in terms of hydrid orbitals? Does it make a difference that Si lies in the row below \(\mathrm{C}\) in the periodic table? Explain.
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