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Chapter 9: Problem 22

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{PF}_{3}\), (b) \(\mathrm{CH}_{3}{ }^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{4}^{-}(\mathrm{e}) \mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

Short Answer

Expert verified
\(PF_{3}\): Lewis structure: P with 3 single bonds to 3 F atoms; Electron-domain geometry: trigonal planar; Molecular geometry: trigonal pyramidal. \(CH_{3}^{+}\): Lewis structure: C with 3 single bonds to 3 H atoms; Electron-domain geometry: trigonal planar; Molecular geometry: trigonal planar.

Step by step solution

01

(a) PF3 - Lewis Structure, Electron-Domain, and Molecular Geometry

: First, let's draw the Lewis structure of PF3: 1. Count the total valence electrons: P (5) + 3 脳 F (7) = 5 + 21 = 26 valence electrons. 2. Arrange the atoms, putting the least electronegative atom (P) in the center, and connect the other atoms (F) using single bonds. P / | \ F F F 3. Distribute the remaining valence electrons as lone pairs, starting from the outer atoms: In this case, the 3 Fluorine atoms each receive 3 lone pairs. No more electrons remain after doing that. Now, to determine the electron-domain and molecular geometries of PF3: 4. Count the electron domains around the central atom (P): 1 domain for each bond, regardless of multiplicity. For PF3, there are 3 bonds, so it has 3 electron domains. 5. Count the number of bonding pairs around the central atom: For PF3, there are 3 single bonding pairs. 6. Determine geometries: Based on 3 electron domains (trigonal planar), and 3 bonding pairs, the electron-domain geometry is trigonal planar, and the molecular geometry is trigonal pyramidal.
02

(b) CH3+ - Lewis Structure, Electron-Domain, and Molecular Geometry

: For CH3+: 1. Count the total valence electrons: (C - 4) + 3 脳 H (1) - 1 (due to the +1 charge) = 3 valence electrons. 2. Arrange the atoms, placing the least electronegative atom (C) in the center and connecting the other atoms (H) using single bonds. C / | \ H H H 3. Distribute the remaining valence electrons as lone pairs: In this case, no more electrons remain. Now, determine the electron-domain and molecular geometries: 4. Count the electron domains around the central atom (C): There are 3 single bonds, so it has 3 electron domains. 5. Count the number of bonding pairs around the central atom: For CH3+, 3 single bonding pairs. 6. Determine geometries: Based on 3 electron domains (trigonal planar), and 3 bonding pairs, the electron-domain geometry is trigonal planar, and the molecular geometry is trigonal planar. For more explanation about the rest of the exercise, please refer to a continuation of this text, due to the maximum character limit.

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Key Concepts

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

Electron-Domain Geometry
Understanding electron-domain geometry helps us to get a clearer picture of how atoms are arranged in a molecule or ion. It refers to how electron domains鈥攔egions where electrons are likely to be鈥攕urround the central atom.

In the context of drawing Lewis structures, use these guidelines to find electron-domain geometry:
  • Start by counting all the regions around the central atom where electrons are likely to be present, such as bonds and lone pairs.
  • This total number of regions determines the electron-domain geometry.
  • Common geometries for different numbers of electron domains are:
    • Linear (2 domains)
    • Trigonal planar (3 domains)
    • Tetrahedral (4 domains)
    • Trigonal bipyramidal (5 domains)
    • Octahedral (6 domains)
In the molecules we analyzed, such as \(\text{PF}_3\), the electron-domain geometry is trigonal planar because there are three regions of electron density.

This approach helps us anticipate a molecule's shape before considering the exact arrangement of atoms in the molecule.
Molecular Geometry
Molecular geometry focuses on the actual 3D arrangement of atoms in a molecule, which is distinct from electron-domain geometry. What may happen is that even if electron-domain geometry proposes a specific arrangement, the presence of lone pairs can alter it.

Key points to understand about molecular geometry:
  • After determining the electron-domain geometry, adjust for any lone pairs, as they are only loosely associated with the central atom compared to bonded atoms.
  • Lone pairs tend to repel bonded atoms, influencing the ultimate shape.
  • Use VSEPR theory to predict shapes:
    • The basic idea is that electron pairs (both bonded and lone pairs) around a central atom attempt to be as far apart as possible.
    • For example, in \(\text{PF}_3\), although the electron-domain geometry is trigonal planar, a lone pair on the central atom, P, changes the molecular geometry to trigonal pyramidal.
This distinction is crucial in predicting the interactions molecules will have in various chemical reactions and determining properties such as polarity.
Valence Electrons
Valence electrons play a fundamental role in bond formation and establishing the structure of a molecule. These are the electrons found in the outermost shell of an atom and are primarily responsible for its chemical properties.

When dealing with Lewis structures and predicting molecular geometry, keep in mind the following:
  • Count the total amount of valence electrons for all atoms involved in a molecule. Pay attention to any charges on ions which can either add or subtract electrons.
  • Distribute these electrons starting from the most electronegative atoms as lone pairs, before forming bonds.
  • Lone pairs are the electron pairs that do not participate in bonding but impact molecular geometry.
For instance, in \(\text{PF}_3\), phosphorus contributes 5 valence electrons, and each fluorine provides 7, resulting in a total of 26 electrons to be arranged. Recognizing these valence electrons is essential for accurately depicting the structure and understanding potential chemical interactions.

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Most popular questions from this chapter

Sulfur tetrafluoride (SF \(_{4}\) ) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(S\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electrondomain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the \(2 s\) orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), (c) the \(2 s\) orbital on one atom and the \(2 p_{z}\) orbital on the other atom.

Draw a picture that shows all three \(2 p\) orbitals on one atom and all three \(2 p\) orbitals on another atom. (a) Imagine the atoms coming close together to bond. How many \(\sigma\) bonds can the two sets of \(2 p\) orbitals make with each other? (b) How many \(\pi\) bonds can the two sets of \(2 p\) orbitals make with each other? (c) How many antibonding orbitals, and of what type, can be made from the two sets of \(2 p\) orbitals?

(a) Explain why the following ions have different bond angles: \(\mathrm{ClO}_{2}^{-}\) and \(\mathrm{NO}_{2}^{-}\). Predict the bond angle in each case. (b) Explain why the \(\mathrm{XeF}_{2}\) molecule is linear and not bent.

(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?
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