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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 refers to the attraction of a material to an external magnetic field due to unpaired electrons in its atomic or molecular structure. Experimentally, one can determine if a substance is paramagnetic using a SQUID magnetometer or the Gouy balance technique. Among the given ions, \(\mathrm{O}_{2}{ }^{+}\) and \(\mathrm{Li}_{2}{ }^{+}\) are paramagnetic, each having 1 unpaired electron.

Step by step solution

01

(a) Definition of Paramagnetism

Paramagnetism is the phenomenon in which a material is attracted to an external magnetic field due to the presence of unpaired electrons in its atomic or molecular structure. These unpaired electrons have magnetic moments, and when the material is subjected to a magnetic field, they tend to align with the field, producing a net magnetic attraction.
02

(b) Experimental Determination of Paramagnetism

Experimentally, one can determine if a substance is paramagnetic by using a device called a "SQUID magnetometer" (Superconducting Quantum Interference Device). This device measures the substance's magnetic response when subjected to an external magnetic field. If the material exhibits a positive magnetic response (i.e., it is attracted to the magnetic field), then it is considered paramagnetic. Another experimental method for determining paramagnetism is the "Gouy balance" technique. In this method, a sample of the substance is placed between the poles of an electromagnet, and the force exerted on the sample by the magnetic field is measured. If the force is positive (attraction), then the substance is paramagnetic.
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(c) Identification of Paramagnetic Ions and Calculation of Unpaired Electrons

To identify if the given ions are paramagnetic, we need to check if they have unpaired electrons in their molecular orbitals. We will look at the electronic configuration of each ion and then determine the number of unpaired electrons accordingly: 1. \(\mathrm{O}_{2}{ }^{+}\): This ion has a total of 15 valence electrons. Using molecular orbital theory, the electronic configuration is: \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\Pi_{2p}^{4}\sigma_{2p}^2\sigma_{2p}^{*1}\) Number of unpaired electrons = 1. 2. \(\mathrm{N}_{2}{ }^{2-}\): This ion has a total of 12 valence electrons. Using molecular orbital theory, the electronic configuration is: \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\Pi_{2p}^{4}\sigma_{2p}^0\sigma_{2p}^{*0}\) Number of unpaired electrons = 0. 3. \(\mathrm{Li}_{2}{ }^{+}\): This ion has a total of 5 valence electrons. Using molecular orbital theory, the electronic configuration is: \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^1\sigma_{2s}^{*0}\) Number of unpaired electrons = 1. 4. \(\mathrm{O}_{2}{ }^{2-}\): This ion has a total of 18 valence electrons. Using molecular orbital theory, the electronic configuration is: \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\Pi_{2p}^{4}\sigma_{2p}^2\sigma_{2p}^{*2}\) Number of unpaired electrons = 0. Based on this analysis, \(\mathrm{O}_{2}{ }^{+}\) and \(\mathrm{Li}_{2}{ }^{+}\) are paramagnetic, with 1 unpaired electron each.

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Key Concepts

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

Unpaired Electrons
Unpaired electrons are crucial when discussing paramagnetism. These are electrons that do not have a pair within an atom's electron cloud. When electrons are paired, their magnetic moments cancel each other out, resulting in no net magnetism.
However, when electrons are unpaired, their magnetic moments remain, allowing the atom or molecule to be attracted to external magnetic fields. This unpaired electron's presence leads to the term *paramagnetism*. When under the influence of a magnetic field, these unpaired electrons align themselves with the field lines.
This alignment, albeit temporary and weak in most cases, causes a net attraction which is the fundamental principle behind paramagnetic substances. Identifying unpaired electrons can be performed using electron configurations in molecular orbital theory or observing direct interactions with magnetic fields.
Molecular Orbital Theory
Molecular Orbital Theory provides a framework for understanding the electronic structure of molecules. It expands beyond simpler atomic orbital concepts by considering that atomic orbitals combine to form molecular orbitals, which can be occupied by electrons shared by the entire molecule.
In assessing paramagnetism, we utilize this theory to determine if a molecule or ion has unpaired electrons. For example, the given ion \(\text{O}_2^+\) has 15 electrons occupying molecular orbitals like \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\Pi_{2p}^{4}\sigma_{2p}^2\sigma_{2p}^{*1}\).
Here, the presence of one unpaired electron in the antiparallel \(\sigma_{2p}^{*}\) orbital results in paramagnetism. Thus, molecular orbital theory not only explains the distribution of electrons but also predicts magnetic properties of molecules by analyzing electron pairing in these orbitals.
SQUID Magnetometer
The SQUID magnetometer, short for Superconducting Quantum Interference Device, is a sensitive instrument used to measure extremely subtle magnetic fields, making it perfect for detecting paramagnetic materials.
The device operates based on the quantum interference of superconducting loops containing electron pairs. When a paramagnetic sample is placed in its magnetic field, the unpaired electrons align along the field lines.
This alignment causes a change in magnetic flux that the SQUID detects with high sensitivity. The device then provides researchers or students with a clear indication that a substance is indeed paramagnetic. Due to its precise measurements, SQUID magnetometers are often utilized in advanced research labs studying magnetic materials.
Gouy Balance
The Gouy balance is a more traditional method to determine if a substance exhibits paramagnetism. This technique doesn’t require highly sophisticated equipment but rather uses fundamental principles of magnetism.
A sample is placed on a balance between the poles of an electromagnet. As the electromagnet is activated, paramagnetic materials are drawn towards it due to their unpaired electrons' orientation aligning with the field.
The change in weight measured by the balance reflects the attraction force, indicating the presence of paramagnetism. The Gouy balance method, while less modern compared to devices like the SQUID magnetometer, remains an effective and practical approach for educational settings and basic experimental demonstrations.

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Most popular questions from this chapter

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \mathrm{~A}\) (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5\), is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-} ?\) (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-} .(\mathrm{e}) \mathrm{It}\) is often observed that \(\sigma\) bonds that involve an sp hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{PF}_{3}\), (b) \(\mathrm{CH}_{3}{ }^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{4}^{-}(\mathrm{e}) \mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\), is a fragrant substance used bothas a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\). A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2} .\) The Lewis structures of these two substances are c1ccc(N=Nc2ccccc2)cc1 \(\begin{array}{ll}\text { Azobenzene } & \text { Hydrazobenzene }\end{array}\) (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(\mathrm{C}\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(S\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electrondomain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

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