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An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

Short Answer

Expert verified
(a) There are 2 nonbonding domains on atom A. (b) The molecular geometry of the \(\mathrm{AB}_{2}\) molecule is (iv) tetrahedral.

Step by step solution

01

Identifying the number of nonbonding domains on atom A

In an \(\mathrm{AB}_{2}\) molecule, there are two bonding domains surrounding atom A, as it is connected to two B atoms. Each bonding domain consists of a pair of electrons shared between atom A and one of the B atoms. To determine how many nonbonding domains are present on atom A, we need to find out how many electron pairs are not involved in forming bonds. The molecule is described as having a tetrahedral geometry, which implies that there must be four total domains surrounding atom A (including both bonding and nonbonding domains). Thus, the number of nonbonding domains can be calculated as follows: \[Number\,of\,nonbonding\,domains = Total\,domains - Bonding\,domains\]
02

Calculate the number of nonbonding domains on atom A

Using the formula from Step 1, we can calculate the number of nonbonding domains on atom A: \[Number\,of\,nonbonding\,domains = 4 - 2\] \[Number\,of\,nonbonding\,domains = 2\] So there are 2 nonbonding domains on atom A. Answer (a): 2
03

Identifying the molecular geometry of the molecule

Since we know the total domains surrounding atom A and the number of nonbonding domains, we can now identify the molecular geometry of the \(\mathrm{AB}_{2}\) molecule. The molecular geometry can be one of the following based on the information given: (i) linear (ii) bent (iii) trigonal planar (iv) tetrahedral However, recall that we are given that the molecule has a tetrahedral geometry, implying that the molecular geometry must be tetrahedral. In the case of an \(\mathrm{AB}_{2}\) molecule with a tetrahedral geometry, the structure consists of 2 bonding domains and 2 nonbonding domains, and the B-A-B angle would be less than that of a regular tetrahedron, which is 109.5°. Answer (b): (iv) tetrahedral

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Key Concepts

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

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is determined by considering all electron pairs surrounding a central atom, including both bonding and nonbonding pairs. The concept is crucial because it influences properties like reactivity, polarity, and biological activity.
Different types of molecular geometry include linear, bent, trigonal planar, tetrahedral, and others. The arrangement is predicted using the VSEPR (Valence Shell Electron Pair Repulsion) theory, which states that electron pairs around a central atom tend to orient themselves as far apart as possible to minimize repulsion.
By counting and categorizing the electron pairs (bonding and nonbonding), one can predict the shape or geometry, which dictates molecular function and behavior.
Tetrahedral Geometry
Tetrahedral geometry is a specific type of molecular shape characterized by four electron domains surrounding a central atom. This shape occurs when the central atom forms four bonds with other atoms, or when there are a combination of bonds and lone pairs, totaling four domains.
  • The ideal bond angle in a tetrahedral molecule is 109.5°.
  • Tetrahedral geometry contributes to the stability and symmetry of many molecular structures.
  • Common compounds like methane ( ext{CH}_4) exemplify tetrahedral geometry.
When nonbonding pairs replace some of the bonding pairs in molecules like water ( ext{H}_2 ext{O}), the angle gets distorted from the ideal value. Such variations still arise under the umbrella of tetrahedral electron geometry.
Nonbonding Domains
Nonbonding domains, also known as lone pairs, consist of electron pairs that do not partake in bonding. These pairs are significant as they afford molecules flexibility and influence molecular shape and polarity.
Lone pairs occupy more space than bonding pairs because they are solely influenced by one nucleus. They are key in determining a molecule's geometry alongside bonding domains. For instance, in a water molecule, the two nonbonding domains significantly alter the geometry to a bent shape, although the electron geometry remains tetrahedral.
  • Lone pairs can strongly affect molecular angles by repelling bonding pairs.
  • They play a crucial role in chemical reactivity and intermolecular interactions.
Bonding Domains
Bonding domains result from electrons shared between atoms, forming a chemical bond. These bonds are essential in dictating the framework of a molecule. In the context of VSEPR theory, bonding domains contribute to the total electron domains around a central atom.
Understanding bonding domains is fundamental in predicting how atoms will organize around a central atom. For example, in an ext{AB}_2 molecule with two bonding domains, like in the linear carbon dioxide molecule, the arrangement creates distinct molecular geometries that impact the molecule's properties.
  • Bonding domains typically form bonds such as sigma or pi bonds.
  • They are instrumental in determining the basic skeleton and function of molecules in a variety of chemical contexts.
Recognizing the interplay between bonding and nonbonding domains allows for a deeper comprehension of molecular architecture and interactions.

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Most popular questions from this chapter

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

(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.

What hybridization do you expect for the atom that is underlined in each of the following species? (a) \(\underline{\mathrm{O}}_{2}^{-} ;(\mathbf{b}) \underline{\mathrm{N}} \mathrm{H}_{4}^{+} ;\) (c) \(\mathrm{SCN}^{-}\) (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\)

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 \mathrm{p}}-\pi_{2 \mathrm{p}}{ }^{*}\), \(\sigma_{2 s}-\sigma_{2 s}{ }^{*},\) and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(\mathrm{N}_{2}\). (b) Is \(\mathrm{N}_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(\mathrm{N}_{2}\) molecule has the lowest energy peak at \(170 \mathrm{nm}\). To what orbital transition does this correspond? (d) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of \(\mathrm{kJ} / \mathrm{mol}\). (e) Is the \(\mathrm{N}-\mathrm{N}\) bond in the first excited state stronger or weaker compared to that in the ground state?

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\) A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2}\) (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(\mathrm{C}\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

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