/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 (a) Sketch the molecular orbital... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}{ }^{-}\)ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\)ion to be stable? (e) Which of the following statements about part (d) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

Short Answer

Expert verified
The molecular orbitals and energy-level diagram for H2- consist of one bonding (sigma) and one antibonding (sigma*) orbital. The electron configuration is \(\sigma^2\sigma^{*1}\), and the bond order is 1/2. The excited-state H2- ion is not stable due to a bond order of zero, and the correct statement about part (d) is that the light excites an electron from a bonding orbital to an antibonding orbital.

Step by step solution

01

Sketch molecular orbitals and energy-level diagram for H2-

Draw the molecular orbitals associated with the H2- ion, consisting of one bonding (sigma) and one antibonding (sigma*) orbital. Place these orbitals on an energy-level diagram, with the bonding orbital below the antibonding orbital.
02

Write the electron configuration

The H2- ion has two hydrogen atoms, each with one electron, and an additional electron from the negative charge. In terms of molecular orbitals, the electron configuration of H2- is \(\sigma^2\sigma^{*1}\).
03

Calculate the bond order

The bond order, which represents the number of chemical bonds between atoms, can be calculated using the formula: \[Bond \, Order = \frac{1}{2}(number \, of \, electrons \, in \, bonding \, orbitals - number \, of \, electrons \, in \, antibonding \, orbitals)\] For H2-, this calculation becomes: \[Bond \, Order = \frac{1}{2}(2 - 1) = \frac{1}{2}\]
04

Determine the stability of the excited-state H2- ion

In the excited state, an electron is promoted from the lower-energy bonding orbital to the higher-energy antibonding orbital. This results in an electron configuration of \(\sigma^1\sigma^{*2}\). This situation indicates that there is now an equal number of electrons in bonding and antibonding orbitals, leading to a bond order of zero. Therefore, the excited-state H2- ion would not be stable.
05

Choose the correct statement about part (d)

Based on the analysis in Step 4, the excited state is achieved by promoting an electron from a bonding orbital to an antibonding orbital. Thus, the correct statement about part (d) is: (i) The light excites an electron from a bonding orbital to an antibonding orbital.

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Key Concepts

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

Molecular Orbital Theory
Molecular orbital theory informs us that electrons in a molecule are not confined to individual atoms but are distributed over the entire molecule in regions known as molecular orbitals. This theory helps us understand molecular structure and bonding by considering the wave-like properties of electrons. In the case of the hydrogen molecule ion, H2-, molecular orbitals are formed by the mathematical combination of the atomic orbitals of the two hydrogen atoms.

When two hydrogen atoms approach each other, their atomic orbitals overlap to form a lower-energy bonding molecular orbital (sigma) and a higher-energy antibonding molecular orbital (sigma*). The bonding orbital, which is symmetrical along the axis connecting the two nuclei, can host two electrons with opposite spins. The antibonding orbital, on the other hand, has a node between the nuclei and is less stable. Electrons in the antibonding orbitals work against bond formation, hence the name 'antibonding'.
Electron Configuration
The electron configuration of a molecule reveals how electrons are distributed among the available molecular orbitals. In our H2- ion, the electron configuration is represented as \(\sigma^2\sigma^{*1}\), where the sigma orbital is filled with two electrons and the sigma* orbital contains one electron. The superscripts designate the number of electrons in each orbital.

Understanding electron configuration is crucial because it determines the properties and behavior of the molecule. Counting the electrons in bonding and antibonding orbitals will not only allow us to predict molecular stability but also its magnetic properties and how it might interact with light or other molecules.
Bond Order Calculation
The bond order of a molecule gives us a simple measurement of its bond strength and stability. It is calculated by subtracting the number of electrons in antibonding orbitals from the number of electrons in bonding orbitals and then dividing by two. In mathematical terms for H2-, the bond order is given by the formula: \[Bond \, Order = \frac{1}{2}(2 - 1) = \frac{1}{2}\]. A larger bond order typically indicates a stronger and more stable bond. A bond order of zero signifies that a molecule will have no stability and therefore not likely exist under normal conditions. The fractional bond order of 0.5 for H2- suggests that this ion possesses a relatively weak and less stable bond compared to H2 with a bond order of 1.
Excited-State Stability
The concept of excited-state stability is crucial when discussing how molecular systems interact with energy, for instance, light absorption. An excited-state occurs when electrons absorb energy and transition to a higher-energy orbital, typically to an antibonding orbital. For H2-, when light is absorbed, one electron from the bonding orbital (sigma) is excited to the antibonding orbital (sigma*), changing the electron configuration to \(\sigma^1\sigma^{*2}\).

This electron rearrangement leads to an excited-state with a bond order of zero \[Bond \, Order = \frac{1}{2}(1 - 2) = 0\], indicating that the excited-state H2- ion has no net bonding interactions between the two hydrogen nuclei, and therefore, is not stable. It highlights that provided enough energy, even a weakly bonding system can be broken apart. This instability has crucial implications in fields such as photochemistry and in understanding chemical reactions under the influence of light.

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Most popular questions from this chapter

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 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 \(\mathrm{XeF}_{6}\).

The orbital diagram that follows presents the final step in the formation of hybrid orbitals by a silicon atom. (a) Which of the following best describes what took place before the step pictured in the diagram: (i) Two 3 p electrons became unpaired, (ii) An electron was promoted from the \(2 p\) orbital to the \(3 s\) orbital, or (iii) An electron was promoted from the \(3 s\) orbital to the \(3 p\) orbital? (b) What type of hybrid orbital is produced in this hybridization? [Section 9.5]

The three species \(\mathrm{NH}_{2}^{-}, \mathrm{NH}_{3}\), and \(\mathrm{NH}_{4}{ }^{+}\)have \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angles of \(105^{\circ}, 107^{\circ}\), and \(109^{\circ}\), respectively. Explain this variation in bond angles.

Suppose that silicon could form molecules that are precisely the analogs of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). How would you describe the bonding about \(\mathrm{Si}\) in terms of hydrid orbitals? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

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