Calculating the Effective Magnetic Moment for Lanthanide f¹⁰ Ions

The magnetic properties of lanthanide ions are primarily determined by the unpaired electrons in the 4f subshell. Due to the shielded nature of the 4f electrons, the crystal field effects from the surrounding ligands are relatively weak. Therefore, the magnetic moment is well described by considering the coupling of the orbital angular momentum (L) and the spin angular momentum (S) of the electrons, which is known as Russell-Saunders coupling or L-S coupling.

The effective magnetic moment ($\mu_{eff}$) for a lanthanide ion in its ground state, exhibiting L-S coupling, is given by the formula:

\[ \mu_{eff} = g_J \sqrt{J(J+1)} \text{ BM} \]

where $g_J$ is the Landé g-factor and J is the total angular momentum quantum number of the ground state. The Landé g-factor is calculated using:

\[ g_J = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \]

Here, L is the total orbital angular momentum quantum number and S is the total spin angular momentum quantum number for the ground state term symbol.

Step-by-Step Calculation for f¹⁰ Configuration

We need to determine the ground state term symbol (^2S+1L_J) for an f¹⁰ configuration.

Step 1: Determine S and L for the f¹⁰ configuration.

An f subshell has 7 orbitals with $m_l$ values ranging from +3 to -3. It can hold a maximum of 14 electrons. An f¹⁰ configuration can be treated as having 4 "holes" in a filled f¹⁴ subshell. The properties of the ground state term symbol for an n-electron configuration are the same as those for an (N-n)-hole configuration (where N is the maximum capacity of the subshell, i.e., 14 for f orbitals).

Consider 4 holes in the f subshell. To find the term symbol, we fill these 4 holes into the f orbitals (+3, +2, +1, 0, -1, -2, -3) following Hund's rules for holes (which are similar to electrons, but often treated with reversed spins for simplicity, although the end result for L and S is the same). To maximize S, put holes in different orbitals with the same spin (say, spin down, $m_s = -1/2$).

• Hole 1: $m_l = +3$, $m_s = -1/2$
• Hole 2: $m_l = +2$, $m_s = -1/2$
• Hole 3: $m_l = +1$, $m_s = -1/2$
• Hole 4: $m_l = 0$, $m_s = -1/2$

Total spin S for holes: $S = \sum m_s = 4 \times (1/2) = 2$. So the total spin quantum number S for the f¹⁰ electrons is 2.

Total orbital angular momentum L for holes: $L = |\sum m_l| = |+3 +2 +1 +0| = 6$. So the total orbital angular momentum quantum number L for the f¹⁰ electrons is 6.

An L value of 6 corresponds to the term symbol letter 'I' (S=0, P=1, D=2, F=3, G=4, H=5, I=6, K=7...).

The spin multiplicity is $2S+1 = 2(2) + 1 = 5$.

So the ground state term symbol is ⁵I.

Step 2: Determine J for the ground state.

For a subshell that is more than half-filled (f¹⁰ is more than half-filled as f⁷ is half-filled), the ground state J value is given by J = L + S.

J = L + S = 6 + 2 = 8.

The ground state term symbol is ⁵I₈.

Step 3: Calculate the Landé g-factor ($g_J$).

Using the formula for $g_J$ with S=2, L=6, and J=8:

\[ g_J = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \]

\[ g_J = 1 + \frac{8(8+1) + 2(2+1) - 6(6+1)}{2 \times 8(8+1)} \]

\[ g_J = 1 + \frac{8 \times 9 + 2 \times 3 - 6 \times 7}{2 \times 8 \times 9} \]

\[ g_J = 1 + \frac{72 + 6 - 42}{144} \]

\[ g_J = 1 + \frac{78 - 42}{144} \]

\[ g_J = 1 + \frac{36}{144} \]

\[ g_J = 1 + \frac{1}{4} = 1.25 \]

The Landé g-factor is 1.25.

Step 4: Calculate the effective magnetic moment ($\mu_{eff}$).

Using the formula $\mu_{eff} = g_J \sqrt{J(J+1)}$ with $g_J = 1.25$ and J=8:

\[ \mu_{eff} = 1.25 \sqrt{8(8+1)} \]

\[ \mu_{eff} = 1.25 \sqrt{8 \times 9} \]

\[ \mu_{eff} = 1.25 \sqrt{72} \]

To evaluate $\sqrt{72}$: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$. Using $\sqrt{2} \approx 1.414$, $\sqrt{72} \approx 6 \times 1.414 = 8.484$. A more precise value for $\sqrt{72}$ is approximately 8.485.

\[ \mu_{eff} = 1.25 \times 8.485 \]

\[ \mu_{eff} \approx 10.606 \text{ BM} \]

The effective magnetic moment for a lanthanide f¹⁰ ion is approximately 10.61 BM. Comparing this value with the given options, 10.6 is the closest value.

Revision Table: Key Concepts for Magnetic Moment Calculation

Additional Information: Lanthanide Magnetic Moments

Lanthanide ions exhibit fascinating magnetic properties. Unlike transition metal ions where the spin-only formula ($\mu_{eff} = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons) is often a good approximation because orbital contribution is quenched by crystal fields, for lanthanides, the 4f electrons are buried deep inside the atom and are well shielded by the 5s²5p⁶ shells.

• The orbital angular momentum contribution is significant and is not effectively quenched by the crystal field.
• Therefore, the magnetic moment calculation must include both spin and orbital contributions using the L-S coupling scheme and the $g_J\sqrt{J(J+1)}$ formula.
• The calculated values often agree well with experimental data, except for a few ions like Sm³⁺ and Eu³⁺, where the energy difference between the ground state and the first excited state is small enough that thermal population of the excited state affects the measured magnetic moment. This effect is not considered in the basic ground state calculation.

The f¹⁰ configuration corresponds to ions like Dy³⁺. Experimental values for Dy³⁺ are typically around 10.63 BM, which is very close to the calculated value using the L-S coupling method.