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The following is part of a molecular orbital energy-level diagram for MOs constructed from \(1 s\) atomic orbitals. (a) What labels do we use for the two MOs shown? (b) For which of the following molecules or ions could this be the energy-level diagram: \(\mathrm{H}_{2}, \mathrm{He}_{2}, \mathrm{H}_{2}{ }^{+}, \mathrm{He}_{2}{ }^{+}\), or \(\mathrm{H}_{2}{ }^{-}\)? (c) What is the bond order of the molecule or ion? (d) If an electron is added to the system, into which of the MOs will it be added? [Section 9.7]

Short Answer

Expert verified
The given molecular orbital energy-level diagram is for the H鈧 molecule, with two MOs labeled 蟽 (bonding) and 蟽* (antibonding). The bond order is 1. If an electron is added, it will enter the antibonding MO (蟽*).

Step by step solution

01

Identify the MOs and their labels

The molecular orbital energy-level diagram is constructed from 1s atomic orbitals. Hence, there will be two molecular orbitals formed, one bonding and one antibonding. The bonding MO is called sigma (蟽) and the antibonding MO is called sigma star (蟽*).
02

Determine which molecule/ion the diagram represents

For this step, count the number of electrons for each given molecule or ion and compare it to the electron count in the given energy-level diagram to find the match. 1. H鈧: 2 electrons (1 from each H atom) 2. He鈧: 4 electrons (2 from each He atom) 3. H鈧傗伜: 1 electron (2 from H鈧 and removing 1) 4. He鈧傗伜: 3 electrons (4 from He鈧 and removing 1) 5. H鈧傗伝: 3 electrons (2 from H鈧 and adding 1) The energy level diagram has 2 electrons, so it's for the H鈧 molecule.
03

Calculate the bond order

To calculate the bond order, use the formula: Bond order = (Number of electrons in bonding MOs - Number of electrons in antibonding MOs) / 2 Since the energy level diagram is for H鈧, there are two electrons in the bonding MO (蟽) and none in the antibonding MO (蟽*). Thus: Bond order = (2 - 0) / 2 = 1 Therefore, the bond order of the H鈧 molecule is 1.
04

Identify into which MO an added electron will be placed

If an electron is added to the system, it will enter an orbital with the lowest possible energy. In this case, both MOs in the energy level diagram are completely filled with electrons. Therefore, if we add an electron, it will go into the higher energy antibonding MO (蟽*).

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Key Concepts

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

Molecular Orbitals
Understanding molecular orbitals is crucial for comprehending how atoms bond together to form molecules. In essence, molecular orbitals (MOs) are combinations of atomic orbitals, where electrons reside in a molecule. When two atomic orbitals merge, they create two MOs - one bonding and one antibonding. The bonding molecular orbital, symbolized as \( \sigma \), is lower in energy and facilitates the bond between atoms. Electrons in this orbital help stabilize a molecule.

On the other hand, the antibonding molecular orbital, designated as \( \sigma^* \), is higher in energy and can destabilize a molecule if it contains electrons. These electrons work against the attractive forces holding the nuclei together. Pictorially, such interactions are often represented using a molecular orbital energy-level diagram, which showcases the relative energies of the MOs and the distribution of electrons within them.

The construction of these diagrams follows specific rules, and the electrons fill the MOs similar to how they would in isolated atoms, adhering to the Pauli exclusion principle and Hund's rule. However, it's important to note that the MOs are not just a simple addition or subtraction of energy levels; they involve complex interactions between the atomic orbitals.
Bond Order Calculation
Bond order is a numerical representation of the stability and strength of a chemical bond in a molecule. It can offer insight into the bond length and the magnetic properties of a molecule. To calculate the bond order, you can apply a straightforward formula:

\[ \text{Bond order} = \frac{(\text{Number of electrons in bonding MOs} - \text{Number of electrons in antibonding MOs})}{2} \]

If the bond order is 1, like in the hydrogen molecule discussed in the exercise, it indicates a single bond between atoms. A bond order of 2 would suggest a double bond, and so on. A positive bond order means a stable molecule, whereas a bond order of zero implies that a molecule is not likely to form. Understanding how to calculate bond order is not only critical for predicting the strength and presence of bonds but also for evaluating the physical properties of molecules.
Sigma and Sigma Star Orbitals
Diving deeper into the types of molecular orbitals, \( \sigma \) and \( \sigma^* \) orbitals are intimately linked to single bonds formed between atoms. \( \sigma \) orbitals are the result when atomic orbitals overlap end-to-end. They are cylindrically symmetric around the bond axis and are the most stable type of molecular orbitals because they allow for effective overlapping of atomic orbitals.

Contrastingly, \( \sigma^* \) orbitals, or the sigma antibonding orbitals, result from the out-of-phase combination of atomic orbitals. Visually, they feature a nodal plane where the electron probability density is zero, passing between the two nuclei. This feature is responsible for their higher energy and destabilizing effect in a molecule. The stark differences in energy and properties between \( \sigma \) and \( \sigma^* \) orbitals are pivotal for bonding theories and explain the dynamic behaviors of molecules during chemical reactions.

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Most popular questions from this chapter

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{6}\), is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene 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?

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.

Indicate the hybridization of the central atom in (a) \(\mathrm{BCl}_{3}\), (b) \(\mathrm{AlCl}_{4}^{-}\), (c) \(\mathrm{CS}_{2}\), (d) \(\mathrm{GeH}_{4}\) -

(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 \(\mathrm{H}\) and \(\mathrm{F}\) 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 \(\mathrm{F}\) interacting with a 1 s 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?

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}{ }^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p} \mathrm{MO}\). (c) The \(\mathrm{O}_{2}{ }^{2+}\) ion has a stronger \(\mathrm{O}-\mathrm{O}\) bond than \(\mathrm{O}_{2}\) itself.

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