Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 119

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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) 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?

How would we describe a substance that contains only paired electrons and is weakly repelled by a magnetic field? Which of the following ions would you expect to possess similar characteristics: $\mathrm{H}_{2}^{-}, \mathrm{Ne}_{2}^{+}, \mathrm{F}_{2}, \mathrm{O}_{2}^{2+} ?$

The molecule shown here is difluoromethane (CH_2F2), which is used as a refrigerant called R-32. (a) Based on the structure, how many electron domains surround the \(\mathrm{C}\) atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, which of the following describes the direction of the overall dipole moment vector in the molecule: (i) from the carbon atom toward a fluorine atom, (ii) from the carbon atom to a point midway between the fluorine atoms, (iii) from the carbon atom to a point midway between the hydrogen atoms, or (iv) from the earbon atom toward a hydrogen atom? [Sections 9.2 and 9.3\(]\)

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the \(2 s\) orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), \((\mathbf{c})\) the 2 s orbital on one atom and the \(2 p_{2}\) orbital on the other atom.

(a) Is the molecule \(\mathrm{BF}_{3}\) polar or nonpolar? (b) If you react \(\mathrm{BF}_{3}\) to make the ion \(\mathrm{BF}_{3}^{2-}\), is this ion planar? (c) Does the molecule \(\mathrm{BF}_{2} \mathrm{Cl}\) have a dipole moment?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation
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