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What are the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom? (a) Three bonding domains and no nonbonding domains, (b) three bonding domains and one nonbonding domain, (c) two bonding domains and two nonbonding domains.

Short Answer

Expert verified
The electron-domain and molecular geometries for each case are as follows: (a) For three bonding domains and no nonbonding domains, both the electron-domain and molecular geometries are trigonal planar. (b) For three bonding domains and one nonbonding domain, the electron-domain geometry is tetrahedral, while the molecular geometry is trigonal pyramidal. (c) For two bonding domains and two nonbonding domains, the electron-domain geometry is tetrahedral, and the molecular geometry is bent or V-shaped.

Step by step solution

01

Determine electron-domain geometry

Since there are three bonding domains and no nonbonding domains, the electron-domain geometry is based on a total of three electron domains (three electron pairs). The three electron pairs will repel each other and try to be as far apart as possible, resulting in a trigonal planar electron-domain geometry.
02

Determine molecular geometry

The molecular geometry refers to the actual shape of the molecule, considering only the positions of the atoms. In this case, the molecular geometry is the same as the electron-domain geometry: trigonal planar. #Case B: Three bonding domains and one nonbonding domain#
03

Determine electron-domain geometry

With three bonding domains and one nonbonding domain, we have a total of four electron domains (four electron pairs). The four electron pairs will arrange themselves to be as far apart as possible, which results in a tetrahedral electron-domain geometry.
04

Determine molecular geometry

Since there are three bonding domains and one nonbonding domain, the molecular geometry will be different from the electron-domain geometry. The nonbonding domain will occupy one of the tetrahedral positions, leaving the three bonding domains distributed around the central atom, forming a trigonal pyramidal molecular geometry. #Case C: Two bonding domains and two nonbonding domains#
05

Determine electron-domain geometry

For two bonding domains and two nonbonding domains, we again have a total of four electron domains (four electron pairs). As in case B, the electron pairs will be arranged to minimize repulsion, which leads to a tetrahedral electron-domain geometry.
06

Determine molecular geometry

Since there are two bonding domains and two nonbonding domains, the molecular geometry will again be different from the electron-domain geometry. The two nonbonding domains will occupy two of the tetrahedral positions, with the remaining two positions being occupied by the bonding domains. This arrangement results in a bent or V-shaped molecular geometry.

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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 a crucial concept in chemistry because the shape of a molecule can determine its physical and chemical properties, including reactivity, polarity, and color. Understanding molecular geometry is essential for predicting how molecules will interact with one another.

For example, water (H2O) has a bent molecular shape, which is a key factor in its ability to form hydrogen bonds and its high boiling point compared to other similar-sized molecules.
Bonding Domains
Bonding domains are regions where electron pairs are shared between atoms, forming covalent bonds. In determining the shape of a molecule, we only consider the bonding domains and nonbonding domains of the central atom.

The number of bonding domains directly influences the molecular geometry, since electrons repel each other and the molecule will adopt a shape that minimizes this repulsion, leading to specific geometric configurations.
Nonbonding Domains
Nonbonding domains, also known as lone pairs, are regions where electrons are not shared with another atom and belong solely to one atom. These domains also contribute to the shape of a molecule because, like bonding domains, they occupy space and exert repulsive force.

However, since they are not shared, nonbonding domains can cause the molecule to have a different shape from the electron-domain geometry—where both bonding and nonbonding domains are considered.
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the geometry of individual molecules based on the extent of electron-pair electrostatic repulsion. It states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.

The VSEPR theory is fundamental in determining the electron-domain geometry and subsequently the molecular geometry. By applying VSEPR theory, we can predict whether a molecule will be linear, trigonal planar, tetrahedral, trigonal pyramidal, bent, and other shapes.
Trigonal Planar
Trigonal planar is an electron-domain geometry that occurs when there are three electron domains—either bonding domains or combinations of bonding and nonbonding domains—around a central atom.

With three electron pairs, they repel each other to lie at the corners of an equilateral triangle centered on the central atom, resulting in a flat, triangular shape. This arrangement is seen in molecules like boron trifluoride (BF3), where all electron domains are bonding.
Tetrahedral
A tetrahedral electron-domain geometry arises when a central atom is surrounded by four electron domains. These could be all bonding domains, as seen in methane (CH4), or a combination of bonding and nonbonding domains.

The tetrahedral geometry allows the electron domains to be equally spaced in a three-dimensional structure, minimizing repulsion and providing the base electron-domain geometry for other shapes like trigonal pyramidal and bent molecular geometries.
Trigonal Pyramidal
The trigonal pyramidal shape is a molecule's molecular geometry that has a tetrahedral electron-domain geometry with one nonbonding domain. With three bonding domains and one lone pair, the molecule takes on a pyramid shape with a triangular base.

This geometry is prevalent in molecules such as ammonia (NH3), where the nitrogen atom is at the apex of the pyramid, and the hydrogen atoms form the base.
Bent Molecular Shape
The bent molecular shape, also known as the V-shaped or angular shape, is typical for molecules with a tetrahedral electron-domain geometry and two nonbonding domains. The two bonding domains are repelled by the lone pairs, which results in a bending of the shape away from linear.

Water is the most widely known example of a molecule with a bent shape, crucial for its unique properties.

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Most popular questions from this chapter

Consider a molecule with formula \(\mathrm{AX}_{3}\). Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{3}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle increases from \(100^{\circ}\) to \(120^{\circ}\) ?

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$ \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \rightarrow \mathrm{XeF}_{6}(s) $$ (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) The molecule IF h 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 \(\mathrm{XeF}_{6}\).

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\mathrm{SiH}_{4}\), (b) \(\mathrm{PF}_{3}\), (c) \(\mathrm{HBr}\), (d) \(\mathrm{HCN}\), (e) \(\mathrm{SO}_{2}\).

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, $\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{N}_{2}$ . The Lewis structures of these two substances are (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the N atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the N and the C atoms in each of the substances? (c) Predict the \(N-N-C\) 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.)

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.

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