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(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}{ }^{2-}\) ? For those ions that are paramagnetic, determine the number of unpaired electrons.

Short Answer

Expert verified
Paramagnetism is a form of magnetism where certain atoms or ions are attracted into the magnetic field due to the presence of unpaired electrons. Experimentally, the Gouy method can be used to determine if a substance is paramagnetic. Among the given ions, \(\mathrm{O}_{2}^{+}\) and \(\mathrm{Li}_{2}^{+}\) are paramagnetic with one unpaired electron each, while \(\mathrm{N}_{2}^{2-}\) and \(\mathrm{O}_{2}{ }^{2-}\) are not paramagnetic as they have no unpaired electrons.

Step by step solution

01

(Answer to Part A: Define Paramagnetism)

Paramagnetism is a form of magnetism where certain atoms or ions are attracted into the magnetic field due to the presence of unpaired electrons in their atomic or molecular orbitals. When the substance is placed into an external magnetic field, these unpaired electrons align with the magnetic field, and the substance exhibits a net magnetic moment.
02

(Answer to Part B: Determine Experimentally Whether a Substance is Paramagnetic)

One common method to determine if a substance is paramagnetic experimentally is by using the Gouy method. In this method, a sample of the substance in question is placed between the poles of a strong electromagnet. The change in the magnetic field caused by the introduction of the substance into the magnetic field is measured. If there is a positive change in the magnetic field, i.e., an increase, then the substance is considered paramagnetic. If there is a negative change in the magnetic field, i.e., a decrease, then the substance is considered diamagnetic.
03

(Answer to Part C: Determine the Paramagnetism of the Given Ions and Unpaired Electrons)

First, let's find the electronic configurations of each ion: 1. \(\mathrm{O}_{2}^{+}\): - Oxygen has 8 electrons, and the diatomic molecule \(\mathrm{O}_{2}\) has 16 electrons. \(\mathrm{O}_{2}^{+}\) has 15 electrons. - Its molecular orbital configuration is: \(\sigma_{1s}^2, \sigma_{1s*}^2, \sigma_{2s}^2, \sigma_{2s*}^2, \sigma_{2p}^2, \pi_{2p}^4, \pi_{2p*}^1\). - There is one unpaired electron in the \(\pi_{2p*}\) orbital. 2. \(\mathrm{N}_{2}^{2-}\): - Nitrogen has 7 electrons, and the diatomic molecule \(\mathrm{N}_{2}\) has 14 electrons. \(\mathrm{N}_{2}^{2-}\) has 16 electrons. - Its molecular orbital configuration is: \(\sigma_{1s}^2, \sigma_{1s*}^2, \sigma_{2s}^2, \sigma_{2s*}^2, \sigma_{2p}^2, \pi_{2p}^4, \pi_{2p*}^2\). - There are no unpaired electrons. 3. \(\mathrm{Li}_{2}^{+}\): - Lithium has 3 electrons, and the diatomic molecule \(\mathrm{Li}_{2}\) has 6 electrons. \(\mathrm{Li}_{2}^{+}\) has 5 electrons. - Its molecular orbital configuration is: \(\sigma_{1s}^2, \sigma_{1s*}^2, \sigma_{2s}^1\). - There is one unpaired electron in the \(\sigma_{2s}\) orbital. 4. \(\mathrm{O}_{2}^{2-}\): - As mentioned earlier, the diatomic molecule \(\mathrm{O}_{2}\) has 16 electrons. \(\mathrm{O}_{2}^{2-}\) has 18 electrons. - Its molecular orbital configuration is: \(\sigma_{1s}^2, \sigma_{1s*}^2, \sigma_{2s}^2, \sigma_{2s*}^2, \sigma_{2p}^2, \pi_{2p}^4, \pi_{2p*}^4\). - There are no unpaired electrons. Based on the molecular orbital configurations and unpaired electrons, we can conclude that \(\mathrm{O}_{2}^{+}\) and \(\mathrm{Li}_{2}^{+}\) are paramagnetic, while \(\mathrm{N}_{2}^{2-}\) and \(\mathrm{O}_{2}{ }^{2-}\) are not. The number of unpaired electrons in \(\mathrm{O}_{2}^{+}\) is 1, and the number of unpaired electrons in \(\mathrm{Li}_{2}^{+}\) is also 1.

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Key Concepts

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

Molecular Orbital Theory
Molecular Orbital (MO) Theory offers a comprehensive explanation of how electrons are distributed in a molecule's atomic orbitals. It plays a critical role in determining the magnetic properties of molecules by demonstrating the presence or absence of unpaired electrons.

MO theory posits that atomic orbitals combine to form molecular orbitals that are spread over the entire molecule. These MOs are of two types: bonding and antibonding orbitals. Bonding orbitals are lower in energy, leading to a stable molecule, while antibonding orbitals are higher in energy and may contain unpaired electrons if occupied.

For example, take the ion \(\mathrm{O}_{2}^{+}\), which has one unpaired electron. MO theory explains this by the filling up of molecular orbitals: the last electron resides in an antibonding orbital that is typically higher in energy than the rest. Hence, molecular orbitals not only explain the structure and bond order of molecules and ions but also their magnetic properties.
Magnetic Properties of Ions
The magnetic properties of ions are fundamentally influenced by the arrangement of their electrons, specifically whether they possess unpaired electrons. Ions with unpaired electrons exhibit paramagnetism because unpaired electrons have intrinsic magnetic moments that align with an external magnetic field, causing attraction.

Contrarily, ions with all paired electrons are diamagnetic because paired electrons have opposite spins that cancel out their magnetic moments, leading to a net zero moment. As elucidated in the solution, ions such as \(\mathrm{O}_{2}^{+}\) and \(\mathrm{Li}_{2}^{+}\) which have unpaired electrons, are paramagnetic. On the other hand, \(\mathrm{N}_{2}^{2-}\) and \(\mathrm{O}_{2}^{2-}\), with all electrons paired, are classified as diamagnetic. This magnetic behavior significantly impacts how ions interact with magnetic fields, thus influencing experiments and applications that utilize magnetic properties.
Gouy Method
The Gouy method is an experimental technique to measure the magnetic susceptibilities of substances, which directly relates to their magnetism. This practical approach helps determine whether a substance is paramagnetic or diamagnetic.

In this method, a substance is suspended between the poles of a magnet, where a change in the weight of the substance is observed. If the weight appears to increase, it indicates that the substance is paramagnetic, as it's pulled towards the magnetic field. Conversely, a decrease in weight suggests diamagnetism, as the substance is repelled by the field.

The beauty of the Gouy method lies in its simplicity and its ability to provide quantitative data about a sample's magnetic properties. It commonly uses a balance to detect small changes in weight, which correspond to changes in the magnetic field caused by the substance’s presence. It's a powerful tool for verifying the magnetic behavior predicted by molecular orbital theory and the electron configurations of ions.

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Most popular questions from this chapter

(a) Starting with the orbital diagram of a sulfur atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{SF}_{2}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4} ?\) Explain.

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$ \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6,}\) you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) Suggest a hybridization scheme for the Xe atom in XeF \(_{6}\). (e) The molecule \(\mathrm{IF}_{7}\) has a pentagonal-bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\)

Dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has three forms (isomers), each of which is a different substance. (a) Draw Lewis structures of the three isomers, all of which have a carbon-carbon double bond. (b) Which of these isomers has a zero dipole moment? (c) How many isomeric forms can chloroethylene, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), have? Would they be expected to have dipole moments?

\((\mathrm{a})\) If the valence atomic orbitals of an atom are \(s p\) 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? Explain.

How would you expect the extent of overlap of atomic orbitals to vary in the series IF, ICl, \(\mathrm{IBr}\), and \(\mathrm{I}_{2}\) ?

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