Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom: \((\mathbf{a}), \mathbf{( b )} 4,(\mathbf{c}) 5,(\mathbf{d}) 6\)

Short Answer

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The characteristic electron-domain geometries for the given numbers of electron domains are as follows: a) 2 electron domains - Linear b) 4 electron domains - Tetrahedral c) 5 electron domains - Trigonal bipyramidal d) 6 electron domains - Octahedral

Step by step solution

01

Identifying the electron-domain geometry for 2 regions.

For a central atom with two electron domains, the most stable geometry will be one in which the electron domains are as far apart as possible to minimize the repulsion between them. In this case, the electron domains will be 180 degrees apart, which results in a linear geometry.
02

Identifying the electron-domain geometry for 4 regions.

For a central atom with four electron domains, the most stable geometry is where the electron domains are positioned at the vertices of a tetrahedron, evenly spaced from each other. Two geometries can be observed from four electron domains around a central atom: tetrahedral and square planar. However, the tetrahedral geometry, with bond angles of approximately 109.5 degrees, is the most common one as it has the least electron repulsion.
03

Identifying the electron-domain geometry for 5 regions.

For a central atom with five electron domains, the most stable geometry is one where the electron domains are arranged at the vertices of a trigonal bipyramid. This arrangement minimizes electron-electron repulsion and features two distinct bond angles: 90 degrees for axial-equatorial electron domains and 120 degrees for equatorial-equatorial electron domains.
04

Identifying the electron-domain geometry for 6 regions.

For a central atom with six electron domains, the most stable geometry is one where the electron domains are arranged at the vertices of an octahedron. This arrangement minimizes electron-electron repulsion and has equal bond angles of 90 degrees for all adjacent electron domain pairs. In conclusion, the characteristic electron-domain geometries for the given numbers of electron domains are as follows: a) 2 electron domains - Linear b) 4 electron domains - Tetrahedral c) 5 electron domains - Trigonal bipyramidal d) 6 electron domains - Octahedral

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Key Concepts

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

Molecular Geometry
Understanding the shapes molecules take is key in chemistry, and molecular geometry plays a crucial role in determining the properties and behavior of compounds. Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The physical and chemical properties of a molecule, such as reactivity, polarity, phase of matter, color, magnetism, as well as biological activity, can be influenced by even minute changes in molecular geometry.

As illustrated in the textbook solution, the number of electron domains around the central atom will shape the molecule's geometry. For instance, a molecule with two electron domains around the central atom exhibits a linear shape. This concept is pivotal since it explains why water (H2O), with its four electron domains and angular shape, has different properties compared to carbon dioxide (CO2), which has two electron domains and is linear.
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules based on the number of regions of high electron density surrounding a central atom. Electron pairs, including bonding pairs and lone pairs, repel each other due to their negative charge. VSEPR theory posits that the shape of a molecule is mainly derived from the tendency of electron domains to repel each other and to maximize the distance between them.

Using the VSEPR model provides a simple method to predict and explain the shapes of molecules. For example, when there are six electron domains around a central atom, as covered in step 4 of the textbook solution, the octahedral shape allows for these domains to be equally spaced apart, reducing repulsion.
Chemical Bonding
Chemical bonding is the force of attraction that holds atoms together within molecules. There are three primary types of chemical bonds: ionic, covalent, and metallic. Each type of bonding contributes to the molecular geometry and stability observed within different substances.

Ionic bonds occur between metals and nonmetals when electrons are transferred from one atom to another. Covalent bonds, more relevant to the original exercise, involve the sharing of electron pairs between atoms. These shared pairs, or bonding pairs, along with unshared pairs, or lone pairs, determine the electron-domain geometry that VSEPR theory describes. Metallic bonding, on the other hand, involves a sea of delocalized electrons surrounding a lattice of metal cations.
Electron Repulsion
Electron repulsion is a natural consequence of the negative charge carried by electrons. When electron domains are in close proximity, the like charges repel each other. This repulsion is a fundamental force in chemistry, influencing molecular shapes, bond angles, and molecular polarities.

The textbook exercise highlighted that the characteristic electron-domain geometries aim to minimize this repulsion. In a linear geometry with two electron domains, repulsion is minimized by spacing the domains 180 degrees apart. As the number of electron domains increases, the spatial challenge to minimize repulsion becomes more complex, which in turn dictates more varied molecular shapes such as tetrahedral, trigonal bipyramidal, and octahedral configurations.

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Most popular questions from this chapter

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.

For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. (c) To create a hybrid orbital, you could use the sorbital on one atom with a porbital on another atom. (d) Nonbonding electron pairs cannot occupy a hybrid orbital.

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 \(\pi_{2 p}-\pi_{2 p}^{\star}\) ,\(\sigma_{25}-\sigma_{25}^{*},\) 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 \(N_{2} .\) (b) Is \(N_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(N_{2}\) molecule has the lowest energy peak at 170 nm. To what orbital transition does this corre- spond? (a) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of kJ/mol. (e) Is the N-N bondin the first excited state stronger or weaker compared to that in the ground state?

(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 H and 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 Finteracting with a 15 orbital 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 HF. 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.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

How many nonbonding electron pairs are there in each of the following molecules: \((\mathrm{a})\left(\mathrm{CH}_{3}\right)_{2} \mathrm{S},(\mathbf{b}) \mathrm{HCN},(\mathbf{c}) \mathrm{C}_{2} \mathrm{H}_{2}\) \((\mathbf{d}) \mathrm{CH}_{3} \mathrm{F} ?\)

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