91影视

Consider the \(\mathrm{H}_{2}^{+}\) ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\) ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\) ? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher- energy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\) ion to be stable or to fall apart? (f) Which of the following statements about part (e) 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?

Step by step solution

01

(a) Sketch the molecular orbitals of the H鈧傗伜 ion and its energy-level diagram.

To sketch the molecular orbitals of the H鈧傗伜 ion (a hydrogen molecule with one positive charge), we need to understand that the H鈧傗伜 ion has two atomic orbitals from the two hydrogen atoms. When they combine, they form two molecular orbitals: one bonding and one antibonding. The bonding molecular orbital is lower in energy (more stable), while the antibonding molecular orbital is higher in energy. The energy level diagram shows these MOs with their respective energy levels.

02

(b) Number of electrons in H鈧傗伜 ion

In a neutral hydrogen molecule, there are two electrons, one from each hydrogen atom. However, in a H鈧傗伜 ion, one electron has been removed, leaving only one electron.

03

(c) Electron configuration of the ion in terms of MOs

Since there is only one electron, it will occupy the lowest energy molecular orbital, which is the bonding orbital (denoted as 蟽). Thus, the electron configuration is \(\mathrm{1\sigma^1}\).

04

(d) Bond order of H鈧傗伜 ion

The bond order is given by the formula: \(Bond\,order = \frac{1}{2} (number\, of\, electrons\,in\,bonding\,MOs - number\,of\,electrons\,in\,antibonding\,MOs)\) In the case of H鈧傗伜, there is only one electron in the bonding MO with none in the antibonding MO. So, the bond order is: \( \frac{1}{2} (1-0) = \frac{1}{2}\).

05

(e) Stability of excited-state H鈧傗伜 ion

Suppose an electron moves from a lower-energy MO to a higher-energy MO when the ion is excited by light. In this case, the electron would move from the bonding orbital to the antibonding orbital. This would result in a weaker bond or no bond at all between the hydrogen atoms. Therefore, the excited-state H鈧傗伜 ion would be unstable and likely to fall apart.

06

(f) Choose the correct statement about part (e)

Out of the given options, (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, (iii) In the excited state, there are more bonding electrons than antibonding electrons, The correct statement is (i) The light excites an electron from a bonding orbital to an antibonding orbital, because when an electron in the H鈧傗伜 ion becomes excited, it moves from the lower-energy (bonding) orbital to the higher-energy (antibonding) orbital, weakening or breaking the bond between hydrogen atoms.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Bond Order

The concept of bond order is crucial in understanding the strength and stability of a molecule. Bond order refers to the number of chemical bonds between a pair of atoms. When considering molecular orbitals (MOs), the bond order can be calculated using the formula:

• Bond order = \( \frac{1}{2} \times (\text{Number of electrons in bonding MOs} - \text{Number of electrons in antibonding MOs}) \)

For the hydrogen cation \((\mathrm{H}_{2}^{+}) \), there is a single electron present, which is located in the bonding orbital. There are no electrons in the antibonding orbital. As a result, the bond order is calculated as:

• \( \frac{1}{2} \times (1 - 0) = \frac{1}{2} \)

This bond order indicates that \((\mathrm{H}_{2}^{+}) \) has a partial bond between the hydrogen atoms, which signifies that the bond is relatively weak as compared to a full bond (e.g., bond order = 1). A higher bond order typically corresponds to increased stability and a stronger bond.

Electron Configuration

Electron configuration helps us understand how electrons are distributed within molecular orbitals. For the \((\mathrm{H}_{2}^{+}) \) ion, which possesses only one electron, the configuration must reflect its occupancy of the molecular orbitals.

• The single electron in \((\mathrm{H}_{2}^{+}) \) occupies the lower energy bonding orbital, termed as \(\sigma\).

Therefore, the electron configuration for \((\mathrm{H}_{2}^{+}) \) in terms of molecular orbitals is expressed as \(\mathrm{1\sigma^1}\). This means that one electron is present in the bonding orbital, which influences the properties of the ion significantly.
Understanding this electron configuration helps predict how the molecule will behave chemically and physically. The lower energy or ground state configuration offers insight into the basic stability and bonding characteristics of the molecule before any excitations or reactions occur.

Excited State Stability

Excited state stability is an intriguing topic when discussing molecular orbitals. This refers to how stable a molecule remains when an electron is promoted to a higher energy level compared to its ground state. In the case of \((\mathrm{H}_{2}^{+}) \), if the lone electron absorbs light energy, it can move from the bonding orbital to a higher energy antibonding orbital.
This transition results in an excited state where the bond is destabilized. The cause of this instability is:

• The reduction of bonding interactions as the electron resides in an antibonding orbital.
• The antibonding orbital adds repulsion between the nuclei, which can weaken or destroy the existing bond entirely.

Consequently, the excited \((\mathrm{H}_{2}^{+}) \) ion is generally unstable and tends to dissociate or fall apart into individual hydrogen atoms. Understanding the dynamics of excited state stability is crucial for predicting molecular behavior upon energy absorption.

Antibonding and Bonding Orbitals

Antibonding and bonding orbitals form the foundation of molecular orbital theory. They are critical in explaining how molecules form and the stability of their bonds. In any diatomic molecule, two types of molecular orbitals are formed when atomic orbitals combine:

• Bonding Orbitals: These are the lower energy orbitals that result when atomic orbitals constructively combine. They enhance electron density between the nuclei, stabilizing the molecule.
• Antibonding Orbitals: These are formed when atomic orbitals destructively interfere, resulting in higher energy levels and diminished electron density between nuclei, leading to destabilization.

For \((\mathrm{H}_{2}^{+}) \), the bonding orbital is occupied in its ground state, which accounts for its bond, albeit weak. If an electron is excited into an antibonding orbital, the molecule's stability decreases, often resulting in bond breakage.
Understanding these types of orbitals is essential for comprehending the formation and dissociation of chemical bonds, particularly when exploring how light or other stimuli can alter electronic states in molecules.