/*! 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 30 An aqueous solution of iron (II)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An aqueous solution of iron (II) sulfate is paramagnetic. If \(\mathrm{NH}_{3}\) is added, the solution becomes diamagnetic. Why does the magnetism change?

Short Answer

Expert verified
The magnetism changes because ext{NH}_{3} causes d electrons in iron to pair up, shifting from paramagnetic to diamagnetic.

Step by step solution

01

Understanding Paramagnetism

Paramagnetism occurs in substances with unpaired electrons. Iron (II) sulfate, when dissolved in water, forms ext{Fe}^{2+} ions. These ions have unpaired electrons in their d orbitals, which makes the solution paramagnetic.
02

Effect of Ammonia Addition

When ext{NH}_{3} is added to the solution, it acts as a ligand, creating a complex with the ext{Fe}^{2+} ions. This new complex involves the d electrons forming covalent bonds with ammonia's lone pair.
03

Formation of Octahedral Complex

The ext{NH}_{3} ligands surround the ext{Fe}^{2+} ion forming an octahedral complex. In this process, the d orbital electrons pair up because ammonia is a strong field ligand.
04

Role of Ligand Field Theory

According to ligand field theory, a strong field ligand like ext{NH}_{3} causes the splitting of d orbitals. This results in all the d electrons pairing up, leading to no unpaired electrons in the new configuration.
05

Reason for Diamagnetism

With no unpaired electrons in the iron complex, the substance becomes diamagnetic, meaning it is repelled by a magnetic field, rather than attracted as in the paramagnetic state.

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 91Ó°ÊÓ!

Key Concepts

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

Paramagnetism
Magnetism is a fascinating property of matter that refers to how a substance responds to a magnetic field. Paramagnetism is one type of magnetism. This phenomenon occurs in materials that have unpaired electrons. These electrons, spinning on their own axes, create small magnetic fields. In a paramagnetic material, when an external magnetic field is applied, these little magnetic fields align with it.
This alignment causes the substance to be attracted to the magnetic field. An everyday example of a paramagnetic substance is a dissolved solution of iron(II) sulfate, which has one or more unpaired electrons in its orbitals.
These unpaired electrons are the main reason the solution shows paramagnetic behavior.
Diamagnetism
In contrast to paramagnetism, diamagnetism occurs in substances where all of the electrons are paired. Paired electrons have their magnetic fields cancel each other out, leaving no unpaired electrons or net magnetic moment. As a result, diamagnetic materials are not attracted to a magnetic field; rather, they are slightly repelled by it.
This is an intriguing property because it contrasts with the attractive behavior seen in paramagnetic substances. In the case of our exercise, adding ammonia (\(\text{NH}_3\)) to the iron(II) sulfate solution eventually transforms the system into a diamagnetic state. This happens because the resulting complex no longer has any unpaired electrons after completing its reaction.
Ligand Field Theory
Ligand field theory is a vital concept in coordination chemistry, especially in identifying the behavior of metal ions in complexes. It explains how ligands - molecules or ions - interact with metal ions' d orbitals, resulting in the splitting of these d orbitals. The extent of this splitting can greatly depend on the ligand's nature.
Strong field ligands, such as ammonia (\(\text{NH}_3\)), cause a more significant splitting of the d orbitals. They often lead to high energy differences between the split d orbitals and can cause electron pairing within these orbitals.
Understanding these interactions is key to understanding why certain complexes display specific magnetic properties such as paramagnetism or diamagnetism.
Electron Pairing
Electron pairing is an essential process where unpaired electrons in an atom or ion are paired together. This process often results from strong magnetic or electric fields typically introduced by strong field ligands.
Pairing changes the magnetic properties of a substance. Consider an iron(II) ion surrounded by ligands like \(\text{NH}_3\). As the ligand forms covalent bonds with the metal ion, the electrons in the d orbitals are subjected to repulsion and increased energy states, forming pairs, if energetically favorable.
When unpaired electrons pair up, they no longer contribute to magnetism, shifting the substance to be diamagnetic due to an absence of unpaired electrons.
Octahedral Complex
In coordination chemistry, the term octahedral complex describes the spatial arrangement of ligands around a central metal ion when six ligands form bonds with it. This geometery is based on an octahedron, where the central metal ion sits in the middle, and the ligands occupy the vertices.
For instance, when ammonia is added to an aqueous solution of iron(II) sulfate, it forms an octahedral complex by surrounding the \(\text{Fe}^{2+}\) ion.
This octahedral arrangement is significant because it influences the electronic configuration of the central metal ion and often alters its magnetic properties. When strong field ligands, like ammonia, adhere in an octahedral arrangement, significant changes in electronic and magnetic properties can occur, as observed in the transition from paramagnetic to diamagnetic behavior.

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

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: $$\begin{aligned} &\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\\\ &K_{f}=10^{8} \end{aligned}$$ $$\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq}) \quad \quad K_{\mathrm{f}}=10^{18}$$ The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{r} H^{\circ}\) for the NH seaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn},\) and \(\Delta_{r} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}\) -rxn. Is the difference in \(\Delta_{r} H^{\circ}\) sufficient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.

One of the following nitrogen compounds or ions is not capable of serving as a ligand: \(\mathrm{NH}_{4}^{+}, \mathrm{NH}_{3}, \mathrm{NH}_{2}^{-} .\) Identify this species, and explain your answer.

For a tetrahedral complex of a metal in the first transition series, which of the following statements concerning energies of the \(3 d\) orbitals is correct? (a) The five \(d\) orbitals have the same energy. (b) The \(d_{x^{2}-y^{2}}\) and \(d_{z^{2}}\) orbitals are higher in energy than the \(d_{x z}, d_{y z},\) and \(d_{x y}\) orbitals. (c) The \(d_{x z}, d_{y z},\) and \(d_{x y}\) orbitals are higher in energy than the \(d_{x^{2} y^{2}}\) and \(d_{z^{2}}\) orbitals. (d) The \(d\) orbitals all have different energies.

The following are low-spin complexes. Use the ligand field model to find the electron configuration of the central metal ion in each ion. Determine which are diamagnetic. Give the number of unpaired electrons for the paramagnetic complexes. (a) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}\) (c) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (d) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right] \mathrm{SO}_{4}\)

Experiments show that \(\mathrm{K}_{4}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]\) is paramagnetic and has two unpaired electrons. The related complex \(\mathrm{K}_{4}\left[\mathrm{Cr}(\mathrm{SCN})_{6}\right]\) is paramagnetic and has four unpaired electrons. Account for the magnetism of each compound using the ligand field model. Predict where the SCN - ion occurs in the spectrochemical series relative to CN'.

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.