Question

Use Hund's rules to derive the ground state of the \(\mathrm{Ce}^{3+}\) ion, and calculate its magnetic moment. (The spin-orbit coupling constant for \(\mathrm{Ce}^{3+}\) is \(1000 \mathrm{cm}^{-1}\) and so the population of states other than the ground state can be neglected at \(298 \mathrm{K} .\) )

Short answer

The ground state of \(\ce{Ce^{3+}}\) is \(^2F_{5/2}\) and its magnetic moment is approximately \(2.96 \mu_B\).

Step-by-step solution

Determine Electron Configuration

First, identify the electron configuration of the \(\ce{Ce^{3+}}\) ion. Neutral cerium (\(\ce{Z=58}\)) has the electron configuration \([\ce{Xe}] 4f^1 5d^1 6s^2\). When 3 electrons are removed to form \(\ce{Ce^{3+}}\), the configuration is \([\ce{Xe}] 4f^1\).

Apply Hund's First Rule

Hund's first rule states that for a given electron configuration, the term with the maximum multiplicity (\(2S + 1\)) has the lowest energy. The single \(4f\) electron contributes a spin multiplicity of \(S = 1/2\), leading to a doublet state (\(2S + 1 = 2\)). Therefore, \(S = 1/2\).

Apply Hund's Second Rule

Hund's second rule indicates that, for terms with the same multiplicity, the term with the largest orbital angular momentum \(L\) is lower in energy. For one \(f\) electron, \(L=3\) (\(f\) orbitals correspond to \(l=3\)). This gives the term symbol \(^2F_{5/2}\) after considering the \(J = L + S = 3 + 1/2 = 5/2\).

Calculate the Magnetic Moment

The magnetic moment \(\mu\) in Bohr magnetons (\(\mu_B\)) for a free ion with total angular momentum \(J\) is given by \[ \mu = \sqrt{J(J+1)} \mu_B. \] For \(J=5/2\), the magnetic moment is \[ \mu = \sqrt{\frac{5}{2} \left(\frac{5}{2} + 1\right)} \mu_B = \sqrt{\frac{35}{4}} \mu_B = \sqrt{8.75} \mu_B \approx 2.96 \mu_B. \]

Confirm Ground State

Considering the spin-orbit coupling constant is large (\(1000 \text{ cm}^{-1}\)) and the temperature is standard \((298 \text{ K})\), thermal excitation to higher states other than the ground state (\(^2F_{5/2}\)) can be neglected. The ground state remains as \(^2F_{5/2}\).

Key concepts

Electron Configuration

The electron configuration is a description of the distribution of electrons in an atom or ion. Each electron resides in a specific energy level or orbitals around the nucleus. In this context, we want to find the electron configuration for the cerium ion, \(\ce{Ce^{3+}}\).
Neutral cerium, with an atomic number of 58, has an electron configuration denoted as \([\ce{Xe}]\ 4f^1\ 5d^1\ 6s^2\). This notation indicates cerium's similarity to the element xenon (Xe) rounded out with three additional orbitals.
To form the positively charged ion \(\ce{Ce^{3+}}\), we remove three electrons. Electrons are typically removed from the outermost orbitals first: two from the 6s orbital and one from the 5d orbital. Reconfiguring, we find the electron configuration \([\ce{Xe}]\ 4f^1\). This configuration highlights a single 4f electron, which is essential for determining the ion's chemical and physical properties, like magnetic characteristics.

Spin Multiplicity

Spin multiplicity is crucial in determining the energy level of an atomic state. It refers to the number of possible orientations of the spin for a given electron configuration. The formula used is \(2S + 1\), where \(S\) is the total spin angular momentum.
In the case of \(\ce{Ce^{3+}}\), the lone 4f electron possesses a spin quantum number \(s = \frac{1}{2}\), making the total spin \(S = \frac{1}{2}\). This results in a spin multiplicity of \(2 \times \frac{1}{2} + 1 = 2\), signifying a doublet state.

• A doublet state indicates two possible orientations for the electron's spin.
• The multiplicity also gives an insight into the ion's stability and potential interactions with magnetic fields.

This is important because a higher multiplicity state is generally lower in energy, according to Hund's first rule, influencing the ground state configuration.

Magnetic Moment

The magnetic moment is a measure of the strength and direction of magnetism in a particle. In atomic physics, it's related to the angular momentum of electrons in an atom or ion. For a free ion, the magnetic moment can be calculated using a formula based on its total angular momentum \(J\).
For the \(\ce{Ce^{3+}}\) ion, we previously determined \(J\) to be \(\frac{5}{2}\), so the magnetic moment \(\mu\) is calculated using the formula: \[ \mu = \sqrt{J(J+1)} \mu_B \]
This results in:
\[ \mu = \sqrt{\frac{5}{2} \left(\frac{5}{2} + 1\right)} \mu_B = \sqrt{8.75} \mu_B \approx 2.96 \mu_B \]
Where \(\mu_B\) is the Bohr magneton, a physical constant used to describe magnetic moment. This calculated value indicates that the \(\ce{Ce^{3+}}\) ion has a moderate magnetic strength as expected for ions with unpaired electrons.

Orbital Angular Momentum

Orbital angular momentum is an integral part of understanding atomic structure and properties. It determines the magnetic and dynamical behavior of electrons in their respective orbitals.
For the \(\ce{Ce^{3+}}\) ion, the single unpaired electron is located in a 4f orbital. Orbitals are characterized by quantum number \(l\), which, for f orbitals, is 3. This means the orbital angular momentum \(L\) is also 3.

• With \(L = 3\), the electron has specific energy related properties, and it follows rules as per Hund's rules to determine energy level occupation.
• Understanding \(L\)'s value helps determine the term symbol, in our case, \(^2F\). With the election added spin of \(\frac{1}{2}\), you deduce the total angular momentum \(J\), forming part of the final term \(^2F_{5/2}\).

The application of Hund's second rule helps us determine that states with larger \(L\) hold lower energy, emphasizing the importance of orbital angular momentum in chemistry and physics.