Q.Use Hund's rule to derive the electronic configuration of $Ce^{3+}$ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.

Concept: Magnetic Moment Calculation – The spin-only formula $\mu_s = \sqrt{n(n+2)}$ gives the magnetic moment in Bohr magnetons ($\mu_B$), where $n$ is the number of unpaired electrons.

Step 1: Electronic configuration of Ce³⁺

Cerium (Ce, atomic number 58) has the ground-state configuration $[Xe]\,4f^1\,5d^1\,6s^2$. Removing three electrons (the two 6s electrons and the one 5d electron) gives Ce³⁺: $[Xe]\,4f^1$.

Step 2: Apply Hund’s rule …

1. Configuration of $\text{Ce}^{3+}$.

Cerium $(Z=58)$ has the ground-state configuration $[\text{Xe}]\,4f^1\,5d^1\,6s^2$. Removing three electrons (the two $6s$ and the one $5d$) gives

so there is a single $4f$ electron, i.e. one unpaired electron.

2. Ground-state term by Hund's rules $(4f^1)$.

• Spin: $S = \tfrac{1}{2}$, so multiplicity $2S+1 = 2$.
• Orbital: for an $f$ electron $l = 3$, giving $L = 3$, i.e. term letter F. …

Method: Spin-Only Magnetic Moment Calculation Using Hund’s Rules

Step 1 — Identify the electronic configuration of the neutral atom

Cerium (Ce) has atomic number 58.

Its ground state configuration is:

Step 2 — Determine the configuration of the $Ce^{3+}$ ion

For $Ce^{3+}$, remove three electrons from the neutral configuration $[\text{Xe}]\,4f^1\,5d^1\,6s^2$.

The removal order is:

• First: the two $6s$ electrons
• Then: the single $5d$ electron

So:

Only one electron remains in the $4f$ subshell.

Step 3 — Apply Hund’s rules to find the number of unpaired electrons

For a $4f^1$ configuration:

• Hund’s Rule 1: Maximize total spin. With one electron, $S = \frac{1}{2}$.
• Hund’s Rule 2: Maximize orbital angular momentum. For $f$ orbitals ($l = 3$), the single electron occupies the orbital with $m_l = +3$ (highest value). So $L = 3$.
• Hund’s Rule 3: For less than half-filled subshell, $J = |L - S| = 3 - \frac{1}{2} = \frac{5}{2}$.

Number of unpaired electrons, $n = 1$.

Step 4 — Apply the spin-only formula

The spin-only magnetic moment is: …

Here are the most common mistakes students make when tackling this exact problem, along with clear strategies to avoid them.

1. Mistake: Writing the wrong electronic configuration for $Ce^{3+}$

• The Error: Students often write the configuration of neutral Cerium ($Ce$) as $[Xe] 4f^1 5d^1 6s^2$ and then, when removing three electrons to form $Ce^{3+}$, they incorrectly remove the $4f$ electron first. This leads to a configuration like $[Xe] 4f^0$ or $[Xe] 5d^1$.
• Why it happens: Confusion about the order of filling vs. the order of removal. Electrons are removed from the outermost shell first (highest $n$ value), not from the subshell that was filled last.
• How to avoid:
  • Remember the removal rule: For transition and inner-transition elements, electrons are removed from the highest principal quantum number ($n$) shell first.
  • Apply it to Cerium:
    • Neutral $Ce$: $[Xe] 4f^1 5d^1 6s^2$
    • Remove $6s^2$ first (highest $n=6$).
    • Remove $5d^1$ next (next highest $n=5$).
    • Result: $Ce^{3+} = [Xe] 4f^1$
  • Key takeaway: The $4f$ subshell is deep inside the atom ($n=4$), so it is the last to be removed, not the first.

2. Mistake: Misapplying Hund's Rule to find $n$

• The Error: After getting the correct $4f^1$ configuration, students sometimes think the number of unpaired electrons ($n$) is 0 or 2. They might pair the single electron or forget that the $4f$ orbital can hold 14 electrons.
• Why it happens: Not visualizing the $4f$ subshell's seven orbitals. A single electron will occupy one orbital by itself (Hund's first rule: maximize spin multiplicity).
• How to avoid:
  • Visualize the orbitals: The $4f$ subshell has 7 degenerate orbitals.
  • Apply Hund's Rule: Place the one electron in any one of these orbitals. It remains unpaired.
  • Conclusion: Number of unpaired electrons, $n = 1$.
  • Quick check: For $f^1$, $f^2$, $f^3$... up to $f^7$, the electrons will occupy separate orbitals with parallel spins, giving $n = 1, 2, 3...7$ respectively.

3. Mistake: Using the wrong formula for magnetic moment

• The Error: Students use the full formula $\mu = \sqrt{n(n+2)}$ but then plug in the wrong value for $n$, or they use the orbital contribution formula $\mu = \sqrt{4S(S+1) + L(L+1)}$ when the question explicitly asks for the 'spin-only' formula. …