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Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the 3 d orbitals (Figure 6.23 ) is most likely to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and \(\sigma_{3 d}^{*}\) antibonding MOs. (c) With reference to the "Closer Look" box on the phases oforbitals, explain why a node is generated in the \(\sigma_{3 d}^{*}\) MO. (d) Sketch the energylevel diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2} ?\)

Short Answer

Expert verified
The \(d_{z^2}\) orbital is most likely to form a σ bond between metal atoms. In the Sc2 molecule, the bond order is 1, and a node is generated in the σ(3d)* MO due to the cancellation of electron densities from the combination of two \(d_{z^2}\) orbitals with opposite phases.

Step by step solution

01

Identify the 3d orbital for σ bonding

To identify the 3d orbital most likely to form a σ bond, remember that the z-axis is the metal-metal bond axis. Among the 3d orbitals (\(d_{xy}\), \(d_{xz}\), \(d_{yz}\), \(d_{x^2-y^2}\), and \(d_{z^2}\)), the \(d_{z^2}\) orbital has the most significant electron density along the z-axis. Therefore, the \(d_{z^2}\) orbital is most likely to make a σ bond between metal atoms.
02

Sketch the σ(3d) bonding and σ(3d)* antibonding MOs

To sketch the σ(3d) bonding and σ(3d)* antibonding MOs, refer to the shape of the \(d_{z^2}\) orbital. With the electron density along the z-axis, the σ(3d) bonding MO will look like the combination of the two \(d_{z^2}\) orbitals from each metal atom along the z-axis in a constructive manner, with their electron density overlapping. The σ(3d)* antibonding MO will look like the combination of the two \(d_{z^2}\) orbitals from each metal atom along the z-axis in a destructive manner, meaning the electron densities would have opposite phases and will create a node on the bond axis.
03

Explain the node in the σ(3d)* MO

A node is generated in the σ(3d)* MO because when the two \(d_{z^2}\) orbitals combine with opposite phases, their electron densities cancel each other out along the metal-metal bond axis (z-axis). This cancellation of electron density creates a node, which is a region of zero electron density.
04

Sketch the energy-level diagram for the Sc2 molecule

To sketch the energy-level diagram for the Sc2 molecule, assuming only the \(d_{z^2}\) orbital from part (a) is important, place the σ(3d) bonding MO at a lower energy level than the isolated Sc atom's \(d_{z^2}\) orbital. Then, place the σ(3d)* antibonding MO at a higher energy level than the isolated Sc atom's \(d_{z^2}\) orbital. Fill in the electrons based on Sc, which has one electron in the 3d orbital.
05

Calculate the bond order in Sc2

To calculate the bond order in the Sc2 molecule, recall that the bond order is given by (number of electrons in bonding MOs - number of electrons in antibonding MOs)/2. In this case, each Sc atom contributes one electron to the σ(3d) bonding MO, and there are no electrons in the σ(3d)* antibonding MO. Therefore, the bond order in Sc2 would be: Bond order = (2 - 0)/2 = 1. The bond order in Sc2 is 1.

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Key Concepts

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

Sigma Bonds
Sigma (\(\sigma\)) bonds are fundamental components of chemical bonding. Created through the overlap of atomic orbitals along the internuclear axis, they represent the strongest form of covalent interaction. In the context of transition metals, which are known for their rich chemistry, sigma bonds become particularly fascinating. Transition metals involve not only s and p orbitals typical in sigma bonding but also the involvement of \(d\) orbitals.

Specifically, in the realm of \(d\) orbitals, the \(d_{z^2}\) orbital is of prime interest. This is because it has lobes extending along the axial direction, or the z-axis, making it suitable for \(\sigma\) bond formation.
  • The \(d_{z^2}\) orbital, having its electron density concentrated along this axis, provides a path for effective orbital overlap.
  • Such overlaps create strong, stable molecular orbitals essential for bonding in metal complexes.
Sigma bonds between transition metals are often found in metal-metal bonded clusters, an important structural motif in materials science.
3d Orbitals
The 3d orbitals are a set of five degenerate orbitals often involved in the intricate bonding of transition metals.

These orbitals, namely \(d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}, \text{and } d_{z^2}\), play pivotal roles in forming bonds depending on their orientation and symmetry relative to nearby atoms.

The \(d_{z^2}\) orbital is a standout example owing to its alignment along the z-axis:
  • This alignment makes it particularly suited for \(\sigma\) bond formation when the bond axis aligns with the z-axis.
  • The lobes directed along the z-axis lead to constructive and destructive interferences needed for bonding and antibonding molecular orbitals.
  • Due to its symmetry, it can form a head-on overlap with another \(d_{z^2}\) orbital.
Transition metal compounds that exploit 3d orbitals often exhibit unique electronic, magnetic, and catalytic properties.
Metal-Metal Bonds
Metal-metal bonds are intriguing features of transition metal chemistry, formed when two metal centers interact through their d orbitals. These bonds are crucial for the stability and properties of metal clusters and chains.

Often, metal-metal bonds involve direct overlapping of 3d orbitals:
  • This creates a pathway for electron sharing between metal atoms, akin to covalent bonds but between metals.
  • In particular, the \(d_{z^2}\) orbitals create strong \(\sigma\) bonds, while other \(d\) orbitals like \(d_{xy}\) or \(d_{xz}\) may form \(\pi\) or \(\delta\) bonds.
The bond order, which quantifies the strength and number of bonds, can be determined by analyzing the number of electrons in bonding versus antibonding molecular orbitals. Metal-metal bonds contribute significantly to the electronic, magnetic, and structural properties of metallic complexes.
Molecular Orbital Theory
Molecular Orbital Theory (MOT) provides a comprehensive model for understanding the electronic structure and bonding in molecules, including those involving transition metals.

For complex compounds like \(Sc_2\) with metal-metal interactions, MOT is particularly useful:
  • MOT describes how atomic orbitals combine to form molecular orbitals, which are spread over the entire molecule.
  • This theory allows for the depiction of bonding (\(\sigma\)), antibonding (\(\sigma^*\)), and non-bonding orbitals.
  • A node, or a region with zero electron density, appears in antibonding orbitals due to the destructive interference of atomic orbitals.
Through MOT, scientists can derive conclusions about molecule stability, reactivity, and magnetic properties, assisting in the prediction of compound formation and behavior. For transition metals, this often involves careful consideration of d-orbital interactions in these calculations.

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Most popular questions from this chapter

Consider the molecule \(\mathrm{PF}_{4}\) Cl. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl}\). How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

Vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), is a gas that is used to form the important polymer called polyvinyl chloride (PVC). Its Lewis structure is (a) What is the total number of valence electrons in the vinyl chloride 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?

Ammonia, \(\mathrm{NH}_{3}\), reacts with incredibly strong bases to produce the amide ion, \(\mathrm{NH}_{2}\) -. Ammonia can also react with acids to produce the ammonium ion, \(\mathrm{NH}_{4}^{+}\). (a) Which species (amide ion, ammonia, or ammonium ion) has the largest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle? (b) Which species has the smallest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle?

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 \mathrm{p}}-\pi_{2 \mathrm{p}}{ }^{*}\), \(\sigma_{2 s}-\sigma_{2 s}{ }^{*},\) and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(\mathrm{N}_{2}\). (b) Is \(\mathrm{N}_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(\mathrm{N}_{2}\) molecule has the lowest energy peak at \(170 \mathrm{nm}\). To what orbital transition does this correspond? (d) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of \(\mathrm{kJ} / \mathrm{mol}\). (e) Is the \(\mathrm{N}-\mathrm{N}\) bond in the first excited state stronger or weaker compared to that in the ground state?

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