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Chapter 9: Problem 67

(a) What does the term diamagnetism mean? (b) How does a diamagnetic substance respond to a magnetic field? (c) Which of the following ions would you expect to be diamagnetic: \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{2-}, \underline{\mathrm{Be}_{2}}^{2+}, \mathrm{C}_{2}{ }^{-} ?\)

Short Answer

Expert verified
(a) Diamagnetism is a property of materials with zero net magnetic moment, meaning they lack unpaired electrons and don't exhibit magnetic behavior without an external field. (b) When a diamagnetic substance is placed in an external magnetic field, it develops weak magnetization in the opposite direction of the field, resulting in being slightly repelled by the magnetic field. (c) Based on their molecular orbital configurations, \(\mathrm{N}_{2}{ }^{2-}\), \(\mathrm{O}_{2}{ }^{2-}\), and \(\underline{\mathrm{Be}_{2}}^{2+}\) ions are expected to be diamagnetic, as they have no unpaired electrons.

Step by step solution

01

Definition of Diamagnetism

Diamagnetism refers to a property of certain materials in which the net magnetic moment is zero. This means that these materials do not have any unpaired electrons and do not exhibit any magnetic behavior in the absence of an external magnetic field.
02

Diamagnetic Substance's Response to Magnetic Field

When a diamagnetic material is placed in an external magnetic field, it develops a weak magnetization in the opposite direction of the applied field. This means that the material will be slightly repelled by the magnetic field. Diamagnetic materials include substances like bismuth, copper, and water.
03

Determining Diamagnetic Ions

To predict whether an ion is diamagnetic, we need to determine if it has any unpaired electrons in its molecular orbitals. For this, we will have to analyze the molecular orbital configurations of the given ions, \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{2-}, \underline{\mathrm{Be}_{2}}^{2+}, \mathrm{C}_{2}{ }^{-}\). 1. \(\mathrm{N}_{2}{ }^{2-}\): The total number of electrons is 16 (10 from two N atoms and 6 from the 2- charge). The molecular orbital configuration is: \[\sigma_{1s}^2, \sigma_{1s^*}^2, \sigma_{2s}^2, \sigma_{2s^*}^2, \pi_{2p}^4, \sigma_{2p}^2.\] All electrons are paired, so \(\mathrm{N}_{2}{ }^{2-}\) is diamagnetic. 2. \(\mathrm{O}_{2}{ }^{2-}\): The total number of electrons is 18 (12 from two O atoms and 6 from the 2- charge). The molecular orbital configuration is: \[\sigma_{1s}^2, \sigma_{1s^*}^2, \sigma_{2s}^2, \sigma_{2s^*}^2, \sigma_{2p}^2, \pi_{2p}^4, \pi_{2p^*}^4.\] All electrons are paired, therefore \(\mathrm{O}_{2}{ }^{2-}\) is also diamagnetic. 3. \(\underline{\mathrm{Be}_{2}}^{2+}\): The total number of electrons is 4 (8 from two Be atoms and -4 from the 2+ charge). The molecular orbital configuration is: \[\sigma_{1s}^2, \sigma_{1s^*}^2.\] No unpaired electrons are present, so \(\underline{\mathrm{Be}_{2}}^{2+}\) is diamagnetic. 4. \(\mathrm{C}_{2}{ }^{-}\): The total number of electrons is 13 (12 from two C atoms and 1 from the 1- charge). The molecular orbital configuration is: \[\sigma_{1s}^2, \sigma_{1s^*}^2, \sigma_{2s}^2, \sigma_{2s^*}^2, \pi_{2p}^4, \sigma_{2p}^1.\] There is one unpaired electron, making \(\mathrm{C}_{2}{ }^{-}\) paramagnetic. Hence, the following ions are expected to be diamagnetic: \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{2-}, \underline{\mathrm{Be}_{2}}^{2+}\).

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Key Concepts

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

Magnetic Moment
The magnetic moment is a measure of the magnetic strength and orientation of a material or particle. In simple terms, it's like a tiny magnet within atoms and molecules. The presence of unpaired electrons usually gives rise to a net magnetic moment.
  • For materials with all electrons paired, the magnetic moments cancel each other out, resulting in zero net magnetic moment. This leads to diamagnetism, which means the material is not attracted to a magnetic field.
  • Diamagnetic substances develop a weak, opposite magnetization to an external magnetic field, causing slight repulsion.
  • For example, bismuth and copper exhibit this behavior.
Unpaired Electrons
Unpaired electrons are the key to understanding the magnetic properties of a substance. Electrons generally pair up in atoms, creating a stable system with opposite spins that cancel out magnetic properties.
  • If an electron does not have a pair, it contributes to a net magnetic moment. This occurs because unpaired electrons align with an external magnetic field, making substances like oxygen paramagnetic.
  • In a diamagnetic material, all electrons are paired, leading to no net magnetic moment and an opposite reaction to magnetic fields.
  • Checking electron configurations helps determine whether a substance will be diamagnetic or paramagnetic based on the presence of unpaired electrons.
Molecular Orbital Theory
Molecular Orbital (MO) Theory is a method used to determine the electron configuration in molecules. This theory expands upon atomic orbital theory by describing how electrons are distributed in a molecule.
  • Electrons fill molecular orbitals in a similar fashion to atomic orbitals, but with the consideration of the entire molecule.
  • To determine if a molecule or ion is diamagnetic, we use MO theory to find its molecular orbital configuration.
  • If all electrons are paired within these orbitals, the molecule is diamagnetic. Otherwise, it is paramagnetic.
  • The MO diagrams for ( ext{N}_{2}{ }^{2-} ), ( ext{O}_{2}{ }^{2-} ), and ( ext{Be}_{2}{ }^{2+} ) show no unpaired electrons, confirming their diamagnetism.

Studying these configurations also helps predict how molecules will interact with magnetic fields. This insight is valuable for understanding and predicting the behavior of different chemical species in various conditions.

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Most popular questions from this chapter

Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\) - that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state? Explain.

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{4}\) is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \mathrm{~A}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ}\). The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the 2 s orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the z-axis), (c) the \(2 s\) orbital on one atom and the \(2 p_{z}\) orbital on the other atom.

(a) Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{BF}_{3}\). (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) Are there any valence atomic orbitals of \(\mathrm{B}\) that are left unhybridized? If so, how are they oriented relative to the hybrid orbitals?
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