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Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

Short Answer

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(a) Electron-domain geometry: tetrahedral, Molecular geometry: tetrahedral (e.g. CH4) (b) Electron-domain geometry: trigonal bipyramidal, Molecular geometry: T-shaped (e.g. ClF3) (c) Electron-domain geometry: octahedral, Molecular geometry: square pyramidal (e.g. SF6) (d) Electron-domain geometry: octahedral, Molecular geometry: square planar (e.g. XeF4)

Step by step solution

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(a) Four bonding domains and no nonbonding domains

With four bonding domains and no nonbonding domains, the electron-domain geometry is tetrahedral, as all electron pairs are equally distributed around the central atom. Since all the electrons are involved in bonding, the molecular geometry is also tetrahedral. An example of a molecule with this configuration would be methane (CH4).
02

(b) Three bonding domains and two nonbonding domains

With three bonding domains and two nonbonding domains, the electron-domain geometry is trigonal bipyramidal. In this case, due to the presence of two nonbonding domains, the molecular geometry will be T-shaped. An example of a molecule with this configuration would be chlorine trifluoride (ClF3).
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(c) Five bonding domains and one nonbonding domain

With five bonding domains and one nonbonding domain, the electron-domain geometry is octahedral. The molecular geometry will be square pyramidal, as only five electron domains are involved in bonding. An example of a molecule with this configuration would be sulfur hexafluoride (SF6).
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(d) Four bonding domains and two nonbonding domains

With four bonding domains and two nonbonding domains, the electron-domain geometry is octahedral. However, due to the presence of two nonbonding domains, the molecular geometry will be square planar. An example of a molecule with this configuration would be xenon hexafluoride (XeF4).

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Key Concepts

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

Electron-Domain Geometry
Electron-domain geometry is a way to describe the arrangement of electron pairs around a central atom in a molecule. This includes both bonding and nonbonding electrons, also known as electron domains. It essentially provides a view of the electron landscape around a central atom and is crucial in determining the overall shape of the molecule.

Different electron-domain geometries arise depending on the number of these electron pairs. For example, two electron domains will arrange themselves linearly, three will form a trigonal planar shape, while four will create a tetrahedral geometry. As you increase the number of electron domains, the geometrical arrangement changes to accommodate these additional electron pairs. Knowing the electron-domain geometry helps predict the molecular geometry, as it gives insights into how the electrons will influence the positioning of the atoms.
Bonding Domains
Bonding domains refer to regions in a molecule where electron pairs are shared between atoms, leading to the formation of chemical bonds. These domains are critical as they define how atoms are connected within a molecule, contributing to its stability and reactivity.

The number of bonding domains around a central atom influences the electron-domain geometry. For instance, methane ( ext{CH}_4) has four bonding domains with no nonbonding domains, resulting in a tetrahedral shape. In such cases, all electron domains are used for forming bonds, affecting both electron-domain and molecular geometry. Understanding bonding domains is essential for predicting molecular geometry and thereby determining the molecule's physical properties and chemical behavior.
Nonbonding Domains
Nonbonding domains, often called lone pairs, are pairs of valence electrons that are not involved in bonding but still occupy space around the central atom. These domains influence the shape of a molecule, even though they do not contribute directly to bonding.

Nonbonding domains repel bonding domains, causing changes in molecular geometry. For example, in water ( ext{H}_2 ext{O}), two of the four electron domains are nonbonding, leading to a bent molecular shape despite its tetrahedral electron-domain geometry. These nonbonding electrons can also lead to asymmetrical charge distribution within the molecule, which can affect polarity and intermolecular interactions.
Tetrahedral
The tetrahedral electron-domain geometry occurs when a central atom is surrounded by four electron domains. This shape is named for the geometric figure, which resembles a pyramid with a triangular base. In a perfect tetrahedron, the angles between the bonds are approximately 109.5 degrees.

A molecule such as methane ( ext{CH}_4) is an excellent example of tetrahedral geometry, where four hydrogen atoms are equally spaced around a central carbon atom. In this case, as there are no nonbonding domains, the electron-domain and molecular geometries are the same. Tetrahedral geometry is common in organic compounds and is associated with sp^3 hybridization of orbitals.
Trigonal Bipyramidal
A trigonal bipyramidal electron-domain geometry features a central atom surrounded by five electron domains. This configuration results from three domains forming an equatorial plane with two axial atoms perpendicular to it. The bond angles can be 90 degrees (axial to equatorial) and 120 degrees (equatorial to equatorial).

In phosphorus pentachloride ( ext{PCl}_5), the central phosphorus is surrounded by five chlorine atoms, creating this shape. However, the presence of nonbonding domains can alter the geometry. For example, if some domains are lone pairs, the structure can change to a T-shaped or seesaw geometry, reflecting changes in molecular geometry, while maintaining trigonal bipyramidal electron-domain geometry.
Octahedral
Octahedral electron-domain geometry is seen when a molecule has six electron domains around a central atom. The atoms form a shape resembling an octahedron, where all positions are equivalent and opposite each other. This geometry typically forms 90-degree bond angles between adjacent domains.

An example of octahedral geometry can be seen in sulfur hexafluoride ( ext{SF}_6), where six fluorine atoms surrond the central sulfur atom. If there are nonbonding domains, the geometry can alter. For instance, in xenon hexafluoride ( ext{XeF}_4), two nonbonding domains result in a square planar molecular geometry while the electron-domain geometry remains octahedral. This distinction between domain and molecular geometry is crucial for understanding how nonbonding pairs influence molecular shape.

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Most popular questions from this chapter

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?

(a) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}{ }^{+} ?(\mathbf{b})\) What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

Vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), is a gas that is used to form the important polymer called polyvinyl chloride (PVC). Its Lewis structure is (a) What is the total number of valence electrons in the vinyl chloride molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(96 \mathrm{pm}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? \(\mathrm{In}\) what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxvgen?

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of \(p\) orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?

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