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.

Use Hund's rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the b

Exercises · 4.31

Ce^{3+}

Tripura TbseTextbookSubjective· 3mImportance★★★★★

39% · 51/132 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Magnetic Moment Calculation

From a Paperclip to a Magnet: The Intuition

You already know that a magnet can pick up iron nails. But what is actually happening inside that nail when it gets near the magnet? And why does a plastic comb, rubbed on hair, pick up tiny bits of paper — but never iron filings?

The answer lies in magnetization — the process by which a material becomes magnetic.

Think of a piece of iron as a chaotic crowd of tiny compass needles. Each needle is an atomic magnetic moment (a tiny magnet, arising from the spin of electrons). In unmagnetized iron, these needles point in random directions. Their magnetic effects cancel out, so the iron as a whole shows no net magnetism.

Now bring a strong magnet close. Its magnetic field acts like a command: "Line up!" The tiny compass needles inside the iron start rotating, aligning themselves with the external field. The more they align, the stronger the iron's own magnetic field becomes. This alignment is magnetization.

Note

Magnetization is not the same as inducing a current. It is a purely magnetic reorientation of atomic dipoles inside a material.

The Precise Definition

Magnetization ( $\vec{M}$ ) is the net magnetic dipole moment per unit volume of a material. It tells you how strongly a material is magnetized — how many tiny atomic magnets are aligned, and in which direction.

If a material has $N$ atoms per unit volume, each with an average magnetic moment $\vec{\mu}_{\text{avg}}$ , then:

\vec{M} = N \vec{\mu}_{\text{avg}}

The SI unit of $\vec{M}$ is amperes per metre (A/m). Why? Because a magnetic dipole moment has units of A·m², and dividing by volume (m³) gives A/m.

\vec{M} = \frac{\text{total magnetic dipole moment}}{\text{volume}}

How Magnetization Connects to the Magnetic Field

When a material gets magnetized, it produces its own magnetic field. The total magnetic field $\vec{B}$ inside the material is the sum of:

The external applied field $\vec{H}$ (caused by free currents, like the current in a solenoid)
The material's response — the magnetization $\vec{M}$

The fundamental relation is:

\vec{B} = \mu_0 (\vec{H} + \vec{M})

where $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space.

Watch out

Do not confuse $\vec{H}$ (magnetic field intensity, or "magnetizing field") with $\vec{B}$ (magnetic flux density). $\vec{H}$ is what you apply; $\vec{M}$ is what the material does; $\vec{B}$ is the total field you measure.

The Three Kinds of Magnetic Materials

Not all materials respond the same way to an external field. The magnetization $\vec{M}$ is proportional to $\vec{H}$ for most materials (at least for small fields):

\vec{M} = \chi_m \vec{H}

where $\chi_m$ is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.

Material Type	$\chi_m$	Behaviour	Example
Diamagnetic	Small and negative ( $\approx -10^{-5}$ )	Weakly repelled by a magnet; $\vec{M}$ opposes $\vec{H}$	Water, copper, bismuth
Paramagnetic	Small and positive ( $\approx 10^{-5}$ to $10^{-3}$ )	Weakly attracted; $\vec{M}$ aligns with $\vec{H}$	Aluminium, oxygen gas
Ferromagnetic	Large and positive ( $\gg 1$ )	Strongly attracted; $\vec{M}$ can be huge and persists even after $\vec{H}$ is removed	Iron, nickel, cobalt

Why this formula?

Magnetic Moment Calculation: Why the Formula Holds

Let's build this from first principles — understanding the why before the formula.

1. What is Magnetic Moment?

A magnetic moment ( $\vec{\mu}$ ) is a measure of the strength and orientation of a magnet or current loop. It tells us how strongly an object will interact with an external magnetic field.

The core idea: any moving charge creates a magnetic field. A loop of current is like a tiny bar magnet — its magnetic moment quantifies this.

2. The Fundamental Formula: Current Loop

The Setup

Consider a planar loop of wire carrying a steady current $I$ , enclosing an area $A$ .

Why $\mu = I A$ ?

Step 1: Force on a moving charge

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences:

\vec{F} = q (\vec{v} \times \vec{B})

Step 2: Torque on a current loop

For a rectangular loop of sides $a$ and $b$ ( $A = ab$ ), placed in a uniform $\vec{B}$ :

Current $I$ means charge flows. On side of length $a$ , the force magnitude is $F = I a B$ (since $I = \frac{q}{t}$ and $v = \frac{a}{t}$ ).
These forces on opposite sides form a couple (equal, opposite, not collinear).
Torque $\tau = \text{force} \times \text{perpendicular distance} = (I a B) \times (b \sin\theta)$

Step 3: Recognize the pattern

\tau = I (ab) B \sin\theta = I A B \sin\theta

This looks exactly like:

\tau = \mu B \sin\theta

Comparing, we identify:

\boxed{\mu = I A}

Why this works: The torque on a current loop is proportional to the current and the area — this product naturally defines the magnetic moment.

3. For a Single Moving Charge (Orbital Magnetic Moment)

The Setup

An electron of charge $-e$ moves in a circular orbit of radius $r$ with speed $v$ .

Why $\mu = \frac{e v r}{2}$ ?

Step 1: Treat orbit as a current loop

Time for one revolution: $T = \frac{2\pi r}{v}$
Current (charge per unit time): $I = \frac{e}{T} = \frac{e v}{2\pi r}$

Step 2: Apply $\mu = I A$

Area of orbit: $A = \pi r^2$
So: $\mu = \left(\frac{e v}{2\pi r}\right) (\pi r^2) = \frac{e v r}{2}$

Step 3: Express in terms of angular momentum

Orbital angular momentum: $L = m v r$
Therefore: $\boxed{\mu = \frac{e}{2m} L}$

Why this matters: The magnetic moment is directly proportional to angular momentum. The factor $\frac{e}{2m}$ is called the gyromagnetic ratio — it links mechanics to magnetism.

4. For a Solenoid (Many Turns)

The Setup

A solenoid of $N$ turns, length $l$ , carrying current $I$ , cross-sectional area $A$ .

Why $\mu = N I A$ ? …

Start your 14-day free trial to unlock the full solution →

$\text{Ce}^{3+}$ has the configuration $[\text{Xe}]\,4f^1$ — one unpaired electron. Its Hund's-rule ground-state term is ${}^2F_{5/2}$ , and the spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM.

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

\text{Ce}^{3+}:\ [\text{Xe}]\,4f^1

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. …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

←Previous question4.30 — Which is the last element in the series of the actinoids? Write the el…→Next question4.32 — Name the members of the lanthanoid series which exhibit +4 oxidation s…

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 understanding — Magnetic Moment Calculation

From a Paperclip to a Magnet: The Intuition

The Precise Definition

How Magnetization Connects to the Magnetic Field

The Three Kinds of Magnetic Materials

Why this formula?

Magnetic Moment Calculation: Why the Formula Holds

1. What is Magnetic Moment?

2. The Fundamental Formula: Current Loop

The Setup

Why $\mu = I A$ ?

3. For a Single Moving Charge (Orbital Magnetic Moment)

The Setup

Why $\mu = \frac{e v r}{2}$ ?

4. For a Solenoid (Many Turns)

The Setup

Why $\mu = N I A$ ? …

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

Step 1 — Identify the electronic configuration of the neutral atom

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

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

Step 4 — Apply the spin-only formula

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

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

3. Mistake: Using the wrong formula for magnetic moment

Q.Use Hund's rule to derive the electronic configuration of Ce3+Ce^{3+}Ce3+ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.

🎓Unlock everything free for 14 days

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.

Unlock everything free for 14 days