Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

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 Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that theion is excited by light, sothat an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? (e) Which of the following statements about part (d) 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?

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}^{-}\) localized or delocalized?

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same?

(a) Predict the electron-domain geometry around the central \(\mathrm{S}\) atom in \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). ( \(\mathbf{b}\) ) The anion \(\mathrm{IO}_{4}^{-}\) has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of \(\mathrm{IO}_{6}{ }^{5-}\).

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free