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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 H₂ molecule, with two MOs labeled σ (bonding) and σ* (antibonding). The bond order is 1. If an electron is added, it will enter the antibonding MO (σ*).

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 (σ) and the antibonding MO is called sigma star (σ*).
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. H₂: 2 electrons (1 from each H atom) 2. He₂: 4 electrons (2 from each He atom) 3. H₂⁺: 1 electron (2 from H₂ and removing 1) 4. He₂⁺: 3 electrons (4 from He₂ and removing 1) 5. H₂⁻: 3 electrons (2 from H₂ and adding 1) The energy level diagram has 2 electrons, so it's for the H₂ 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 H₂, there are two electrons in the bonding MO (σ) and none in the antibonding MO (σ*). Thus: Bond order = (2 - 0) / 2 = 1 Therefore, the bond order of the H₂ 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 (σ*).

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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

What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\), (b) \(\mathrm{HCN}^{\text {, }}\) (c) \(\mathrm{SO}_{3}\), (d) \(\mathrm{TeCl}_{2}\).

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{*}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\)-axis is vertical in the plane of the paper and the \(x\)-axis horizontal. Write " \(\mathrm{M}^{"}\) at the origin to denote a metal atom. (b) Now, on the \(x\) axis to the right of \(M\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(M\). The CO bond axis should be on the \(x\)-axis. (c) Draw the \(\mathrm{CO} \pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2}^{*}\) orbital of \(\mathrm{CO}\) ? (e) What kind of bond is being made with the orbitals between \(M\) and \(\mathrm{C}_{,} \sigma\) or \(\pi\) ? (f) Predict what will happen to the strength of the CO bond in a metal\(\mathrm{CO}\) complex compared to \(\mathrm{CO}\) alone.

For each of these contour representations of molecular orbitals, identify (a) the atomic orbitals (s or \(p\) ) used to construct the MO (b) the type of MO ( \(\sigma\) or \(\pi\) ), (c) whether the MO is bonding or antibonding, and (d) the locations of nodal planes. [Sections \(9.7\) and 9.8]

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3}\), (b) \(\mathrm{CH}_{3}^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{3}^{-}\), (e) \(\mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom: (a) 3 , (b) 4 , (c) 5, (d) 6 .

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