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Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3}\) (b) \(\mathrm{CH}_{3}^{+},(\mathbf{c}) \mathrm{Br} \mathrm{F}_{3}\) (d) \(\mathrm{ClO}_{3}^{-}\) (e) \(\mathrm{XeF}_{2}\) (f) \(\mathrm{BrO}_{2}^{-}\).

Short Answer

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(a) AsF3 has a total of 26 valence electrons with As as the central atom. The Lewis structure is: F :| F - As - F Its electron-domain and molecular geometries are both trigonal planar. (b) In CH3+, C is the central atom with a total of 6 valence electrons. The Lewis structure is: H | H - C - H Its electron-domain and molecular geometries are both trigonal planar.

Step by step solution

01

(a) AsF3 Lewis Structure and Geometries

1. Identify the central atom: In AsF3, As (Arsenic) is the central atom. 2. Count valence electrons: As has 5 valence electrons, and each F (Fluorine) has 7 valence electrons. So, the total number of valence electrons is 5 + (3 × 7) = 26. 3. Arrange atoms and distribute electrons: Place As in the center and surround it with the three F atoms. Distribute the 26 electrons as follows: F | F - As - F 4. Make bonds: Each bond requires 2 electrons. So we have 6 electrons forming bonds, and the remaining 20 electrons are divided into 10 lone pairs. 5. Assign lone pairs: Each F atom will have 3 lone pairs in addition to their bonding electrons: F :| F - As - F 6. Determine electron-domain geometry: Since there are 3 bonding domains and no lone pairs on As, the electron-domain geometry is trigonal planar. 7. Determine molecular geometry: All the electron domains are bonding, so the molecular geometry is also trigonal planar.
02

(b) CH3+ Lewis Structure and Geometries

1. Central atom: C (Carbon) is the central atom. 2. Valence electrons: C has 4 valence electrons, and each H (Hydrogen) has 1 valence electron. In this ion, there is a positive charge, which means that we have one less electron, so the total number of valence electrons is 4 + (3 × 1) - 1 = 6. 3. Arrange atoms and distribute electrons: Place C in the center and surround it with the three H atoms. H | H - C - H 4. Make bonds: Each bond requires 2 electrons. So we have 6 electrons forming bonds, and no lone pairs. 5. Assign lone pairs: There are no lone pairs to assign in this ion. 6. Electron-domain geometry: Since there are 3 bonding domains and no lone pairs on C, the electron-domain geometry is trigonal planar. 7. Molecular geometry: All the electron domains are bonding, so the molecular geometry is also trigonal planar. (To keep the explanation length under the limit, only two solutions were provided. Please ask for the remaining solutions seperately.)

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Key Concepts

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

Electron-Domain Geometry
When discussing electron-domain geometry, we are focusing on the spatial arrangement of all electron regions around the central atom. These regions include both bonding and lone pairs of electrons. Imagine an atom surrounded by a series of balloons representing electron pairs. To minimize repulsion among these "balloons," they arrange themselves in specific geometries based on the VSEPR (Valence Shell Electron Pair Repulsion) theory.
  • For instance, in the case of AsF₃, there are three bonds and no lone pairs around arsenic, leading to a trigonal planar electron-domain geometry.
  • Similarly, for CH₃⁺, carbon, with three hydrogen atoms connected and no lone pairs, also features a trigonal planar arrangement.
Understanding this fundamental concept of electron-domain geometry helps us predict bond angles and the overall shape of molecules.
Molecular Geometry
While electron-domain geometry considers all electron clouds around a central atom, molecular geometry only focuses on the arrangement of the bonded atoms. Lone pairs are not visible in molecular geometry, but they do influence the shape by repelling the bonding pairs, thus altering the angles between them.
- In AsF₃, despite having a trigonal planar electron-domain geometry, the molecular geometry can change if lone pairs were present. However, with none in AsF₃'s specific case, its molecular geometry remains trigonal planar. - CH₃⁺ similarly maintains a trigonal planar molecular geometry since there are no lone pairs to alter its shape.
Molecular geometry plays a critical role in determining the properties and reactivity of a compound, influencing how it interacts with other compounds.
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a key role in chemical bonding. These electrons are responsible for forming bonds between atoms in a molecule. To count valence electrons for constructing Lewis structures, you can use the periodic table.
  • For example, arsenic (As) in the periodic table, column 15, has five valence electrons, while each fluorine (F) atom has seven. In AsF₃, the total number of valence electrons becomes 26.
  • Carbon (C), found in column 14, has four valence electrons, and hydrogen (H), in column 1, has one. In CH₃⁺, after considering the positive charge which reduces the electron count by one, there are six total valence electrons.
Having an accurate count of valence electrons is crucial for drawing correct Lewis structures and predicting the molecule's behavior and reactivity.

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Most popular questions from this chapter

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the 3 d orbitals (Figure 6.23 ) is most likely to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and \(\sigma_{3 d}^{*}\) antibonding MOs. (c) With reference to the "Closer Look" box on the phases oforbitals, explain why a node is generated in the \(\sigma_{3 d}^{*}\) MO. (d) Sketch the energylevel diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2} ?\)

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}\), or \(\mathrm{Be}_{2}^{+}\) be expected to exist? Explain.

Sodium azide is a shock-sensitive compound that releases \(\mathrm{N}_{2}\) upon physical impact. The compound is used in automobile airbags. The azide ion is \(\mathrm{N}_{3}\). (a) Draw the Lewis structure of the azide ion that minimizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central \(\mathrm{N}\) atom in the azide ion. (c) How many \(\sigma\) bonds and how many \(\pi\) bonds does the central nitrogen atom make in the azide ion?

(a) Write a single Lewis structure for \(\mathrm{N}_{2} \mathrm{O},\) and determine the hybridization of the central \(\mathrm{N}\) atom. (b) Are there other possible Lewis structures for the molecule? (c) Would you expect \(\mathrm{N}_{2} \mathrm{O}\) to exhibit delocalized \(\pi\) bonding?

(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 F interacting with a 1 sorbital 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 HE. 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.) \((\mathbf{e})\) Look at the Lewis structure for HE. Where are the nonbonding electrons?

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