Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 56

The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion is diamagnetic, but \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is para-magnetic. Explain.

Short Answer

Expert verified
The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion is diamagnetic because all its electrons in the 3d orbitals are paired, resulting in 0 unpaired electrons. On the other hand, the \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ion is paramagnetic because it has 4 unpaired electrons in its 3d orbitals.

Step by step solution

01

Determine the oxidation states of the metal ions

To find the electron configuration, first, we need to know the oxidation state for Cobalt (Co) in \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) and Iron (Fe) in \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\). The overall charge of the \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) complex is +3. Ammonia (NH3) is a neutral ligand, so the oxidation state of Co is +3. Similarly, the overall charge of the \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex is +2. Water (H2O) is also a neutral ligand, so the oxidation state of Fe is +2.
02

Determine the electron configuration of the metal ions

Next, let's find the electron configuration of these metal ions in their respective oxidation states. Cobalt has an atomic number of 27, and its ground state electron configuration is [Ar] 3d7 4s2. In a +3 oxidation state, Co would lose 3 electrons. The 4s electrons are lost first, followed by one of the 3d electrons: [Ar] 3d6. Iron has an atomic number of 26, and its ground state electron configuration is [Ar] 3d6 4s2. In a +2 oxidation state, Fe would lose 2 electrons. Both 4s electrons are lost, giving an electron configuration of [Ar] 3d6.
03

Apply Crystal Field Theory to determine the number of unpaired electrons

To determine the number of unpaired electrons, we will apply Crystal Field Theory. Without going into detail, both \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2}\mathrm{O}\) are considered weak-field ligands. Therefore, no electron pairing will occur in the 3d orbitals for these complexes. For \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) with an electron configuration of [Ar] 3d6, the 3d orbitals will have all paired electrons, resulting in 0 unpaired electrons. For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) with an electron configuration of [Ar] 3d6, there are 4 unpaired electrons.
04

Determine the magnetic properties based on the number of unpaired electrons

A complex ion is diamagnetic if there are no unpaired electrons and paramagnetic if there are unpaired electrons present. The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion has 0 unpaired electrons, so it is diamagnetic. The \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ion has 4 unpaired electrons, so it is paramagnetic. The magnetic properties of each complex ion can now be explained. The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion is diamagnetic due to the absence of unpaired electrons in its 3d orbitals. The \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ion is paramagnetic due to the presence of 4 unpaired electrons in its 3d orbitals.

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 compound related to acetylacetone is 1,1,1-trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise 49.) Both Be \(^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa - having the formula \(\mathrm{M}(\mathrm{tfa})_{2}\) . Two isomers are formed for each metal complex. a. The Be \(^{2+}\) complexes are tetrahedral. Draw the two isomers of \(\mathrm{Be}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Cu}(\mathrm{tfa})_{2} ?\)

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.

Would it be better to use octahedral \(\mathrm{Ni}^{2+}\) complexes or octahe- dral \(\mathrm{Cr}^{2+}\) complexes to determine whether a given ligand is a strong-field or weak-field ligand by measuring the number of unpaired electrons? How else could the relative ligand field strengths be determined?

When an aqueous solution of KCN is added to a solution containing \(\mathrm{Ni}^{2+}\) ions, a precipitate forms, which redissolves on addition of more \(\mathrm{KCN}\) solution. Write reactions describing what happens in this solution. [Hint: \(\mathrm{CN}^{-}\) is a Bronsted-Lowry base \(\left(K_{\mathrm{b}} \approx 10^{-5}\right)\) and a Lewis base.]

Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

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