/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 On average, the \(d\) orbitals o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 36

On average, the \(d\) orbitals of a transition metal ion in an octahedral field are higher in energy than they are when the ion is in the gas phase. Why?

Short Answer

Expert verified
Answer: The energy levels of \(d\) orbitals in a transition metal ion are higher in an octahedral field compared to the gas phase due to the interaction between ligands and the ion's \(d\) orbitals, which causes electron repulsion and increases energy levels. Additionally, the \(d\) orbitals split into two groups with different energy levels, further increasing the overall energy.

Step by step solution

01

Understand the energy levels of \(d\) orbitals in the gas phase

In the gas phase, the five \(d\) orbitals of a transition metal ion are degenerate, meaning they have the same energy. When the ion is not interacting with ligands, these energy levels are relatively low.
02

Identify the change in energy levels when the ion is in an octahedral field

When a transition metal ion is in an octahedral field, the energy levels of its \(d\) orbitals change. The ligands in the octahedral field approach the ion along its \(x, y, z\) axis, which results in the \(d\) orbitals splitting into two energy levels: three lower-energy \(t_{2g}\) orbitals (\(d_{xy}, d_{yz}, d_{zx}\)) and two higher-energy \(e_g\) orbitals (\(d_{x^2-y^2}, d_{z^2}\)).
03

Understand the effect of ligand interaction on the \(d\) orbitals

As ligands approach the ion along the \(x, y, z\) axis, electron repulsion between the ligands and the ion's \(d\) orbitals increases. The ligands push the \(d\) electrons away, causing an increase in energy levels.
04

Explain the overall energy increase

Although the \(d\) orbitals split into two groups, with the \(t_{2g}\) orbitals having a lower energy than the \(e_g\) orbitals, the overall energy is increased for all five \(d\) orbitals in an octahedral field compared to the gas phase. This energy increase is mainly due to the electron repulsion between ligands and the transition metal ion's \(d\) orbitals. In conclusion, the \(d\) orbitals of a transition metal ion in an octahedral field are higher in energy compared to the gas phase due to the interaction between ligands and the ion's \(d\) orbitals, as well as the splitting of the orbitals into two different energy levels.

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 91影视!

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

What are the names of the following complex ions? a. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) b. \(\operatorname{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) c. \(\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}\)

What minimum number of different types of donor groups are required in order to have stereoisomers of a square planar complex?

Chromium(III) chloride forms six-coordinate complexes with bipyridine, including cis-[Cr(bipy)_ \(\left._{2} \mathrm{Cl}_{2}\right]^{+}\), which reacts slowly with water to produce two products, cis\(\left[\mathrm{Cr}(\text { bipy })_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) and \(c i s-\left[\mathrm{Cr}(\text { bipy })_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{3+} \cdot\) In which of these complexes should \(\Delta_{0}\) be the largest?

What are the names of the following coordination compounds? a. \(\left(\mathrm{NH}_{4}\right)_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) b. \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]\left(\mathrm{NO}_{3}\right)_{2}\) c. \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}(\mathrm{OH})_{2}\right] \mathrm{Cl}\)

Gadolinium(III) ions are used in contrast agents for MRI because they have unpaired electrons. What is the electron configuration for Gd \(^{3+}\) and how many unpaired electrons does it have?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

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.