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Compare and contrast bonding molecular orbitals with antibonding molecular orbitals.

Short Answer

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Bonding molecular orbitals are formed by the in-phase, constructive combination of atomic orbitals, resulting in increased electron density between atoms and creating a bond. Antibonding molecular orbitals arise from the out-of-phase, destructive combination, causing regions of electron density to cancel each other out and leading to repulsion. Bonding MOs are lower in energy and more stable, while antibonding MOs are higher in energy and less stable. Electrons fill available bonding MOs before occupying antibonding MOs. Bond order, a measure of bond strength, is calculated as half the difference between the number of electrons in bonding and antibonding MOs. A positive bond order indicates bond formation, while negative or zero bond orders suggest no bond exists.

Step by step solution

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1. Definitions

Bonding molecular orbitals (MO) are formed when atomic orbitals combine in-phase, resulting in an increased electron density between the two atoms. This leads to the attractive force that holds them together, forming a bond. Antibonding molecular orbitals, on the other hand, are formed when atomic orbitals combine out-of-phase, resulting in regions of electron density that cancel each other out. This causes a decrease in electron density between the atoms, leading to repulsion and destabilizing the molecule.
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2. Formation

When atomic orbitals overlap, they can interact either constructively or destructively. Bonding MOs are formed by the in-phase, constructive combination of atomic orbitals, while antibonding MOs arise from the out-of-phase, destructive combination. For example, in the hydrogen molecule (H2), the 1s orbitals of both hydrogen atoms overlap constructively, forming a bonding MO (denoted as σ1s). If the 1s orbitals were to overlap destructively, an antibonding MO (σ1s*) would be formed instead.
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3. Electron Configuration and Stability

Bonding MOs are lower in energy and more stable than the original atomic orbitals, while antibonding MOs are higher in energy and less stable. Electrons will first fill the available bonding MOs before occupying antibonding MOs, as electrons seek the lowest possible energy states. An important principle, known as the Pauli Exclusion Principle, states that each MO can hold a maximum of two electrons with opposite spins. This principle helps determine electron configurations in molecules.
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4. Bonding Order

Bond order is a quantitative measure of the strength and stability of a bond. It is calculated as half the difference between the number of electrons in bonding MOs (NB) and antibonding MOs (NA): Bond order = \( \frac{_(NB} - {NA)}{2} \) A positive bond order indicates that a bond is being formed, while a negative or zero bond order suggests that no bond exists.
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5. Examples

To provide specific examples, let us examine the molecular orbitals of H2 and He2. In H2, both hydrogen atoms have one electron each in their 1s orbitals. When these orbitals overlap to form the H2 molecule, there are two bonding electrons (1σ1s) and no antibonding electrons, resulting in a bond order of 1 and a stable single bond. In He2, there are two electrons per helium atom in the 1s orbitals. When they overlap, two bonding electrons are present in the σ1s orbital and two antibonding electrons are found in the σ1s* orbital. This results in a bond order of 0, indicating that no bond is formed between the two helium atoms.

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Most popular questions from this chapter

Using molecular orbital theory, explain why the removal of one electron in \(\mathrm{O}_{2}\) strengthens bonding, while the removal of one electron in \(\mathrm{N}_{2}\) weakens bonding.

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lower-energy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of \(\mathrm{OH}\). Explain.

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and NO than the hybrid orbital model?

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

What type of molecular orbital would result from the in-phase combination of two \(d_{x z}\) atomic orbitals shown below? Assume the \(x\) -axis is the internuclear axis.

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