Q.Use Hund's rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.
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🔒 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.
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 (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 μavg, then:
M=Nμavg
The SI unit of 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.
M=volumetotal magnetic dipole moment
How Magnetization Connects to the Magnetic Field
When a material gets magnetized, it produces its own magnetic field. The total magnetic field B inside the material is the sum of:
- The external applied field H (caused by free currents, like the current in a solenoid)
- The material's response — the magnetization M
The fundamental relation is:
B=μ0(H+M)
where μ0=4π×10−7T⋅m/A is the permeability of free space.
Do not confuse H (magnetic field intensity, or "magnetizing field") with B (magnetic flux density). H is what you apply; M is what the material does; 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 M is proportional to H for most materials (at least for small fields):
M=χmH
where χm is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.
| Material Type | χm | Behaviour | Example |
|---|---|---|---|
| Diamagnetic | Small and negative (≈−10−5) | Weakly repelled by a magnet; M opposes H | Water, copper, bismuth |
| Paramagnetic | Small and positive (≈10−5 to 10−3) | Weakly attracted; M aligns with H | Aluminium, oxygen gas |
| Ferromagnetic | Large and positive (≫1) | Strongly attracted; M can be huge and persists even after 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 (μ) 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 μ=IA?
Step 1: Force on a moving charge
A charge q moving with velocity v in a magnetic field B experiences:
F=q(v×B)
Step 2: Torque on a current loop
For a rectangular loop of sides a and b (A=ab), placed in a uniform B:
- Current I means charge flows. On side of length a, the force magnitude is F=IaB (since I=tq and v=ta).
- These forces on opposite sides form a couple (equal, opposite, not collinear).
- Torque τ=force×perpendicular distance=(IaB)×(bsinθ)
Step 3: Recognize the pattern
τ=I(ab)Bsinθ=IABsinθ
This looks exactly like:
τ=μBsinθ
Comparing, we identify:
μ=IA
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 μ=2evr?
Step 1: Treat orbit as a current loop
- Time for one revolution: T=v2πr
- Current (charge per unit time): I=Te=2πrev
Step 2: Apply μ=IA
- Area of orbit: A=πr2
- So: μ=(2πrev)(πr2)=2evr
Step 3: Express in terms of angular momentum
- Orbital angular momentum: L=mvr
- Therefore: μ=2meL
Why this matters: The magnetic moment is directly proportional to angular momentum. The factor 2me 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 μ=NIA? …
Concept: Magnetic Moment Calculation – The spin-only formula μs=n(n+2) gives the magnetic moment in Bohr magnetons (μB), 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]4f15d16s2. Removing three electrons (the two 6s electrons and the one 5d electron) gives Ce³⁺: [Xe]4f1.
Step 2: Apply Hund’s rule …
Ce3+ has the configuration [Xe]4f1 — one unpaired electron. Its Hund's-rule ground-state term is 2F5/2, and the spin-only magnetic moment is μ=n(n+2)=1(1+2)=3≈1.73 BM.
1. Configuration of Ce3+.
Cerium (Z=58) has the ground-state configuration [Xe]4f15d16s2. Removing three electrons (the two 6s and the one 5d) gives
Ce3+: [Xe]4f1
so there is a single 4f electron, i.e. one unpaired electron.
2. Ground-state term by Hund's rules (4f1).
- Spin: S=21, 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:
[Xe]4f15d16s2
Step 2 — Determine the configuration of the Ce3+ ion
For Ce3+, remove three electrons from the neutral configuration [Xe]4f15d16s2.
The removal order is:
- First: the two 6s electrons
- Then: the single 5d electron
So:
Ce3+:[Xe]4f1
Only one electron remains in the 4f subshell.
Step 3 — Apply Hund’s rules to find the number of unpaired electrons
For a 4f1 configuration:
- Hund’s Rule 1: Maximize total spin. With one electron, S=21.
- Hund’s Rule 2: Maximize orbital angular momentum. For f orbitals (l=3), the single electron occupies the orbital with ml=+3 (highest value). So L=3.
- Hund’s Rule 3: For less than half-filled subshell, J=∣L−S∣=3−21=25.
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 Ce3+
- The Error: Students often write the configuration of neutral Cerium (Ce) as [Xe]4f15d16s2 and then, when removing three electrons to form Ce3+, they incorrectly remove the 4f electron first. This leads to a configuration like [Xe]4f0 or [Xe]5d1.
- 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]4f15d16s2
- Remove 6s2 first (highest n=6).
- Remove 5d1 next (next highest n=5).
- Result: Ce3+=[Xe]4f1
- 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 4f1 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 f1, f2, f3... up to f7, 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 μ=n(n+2) but then plug in the wrong value for n, or they use the orbital contribution formula μ=4S(S+1)+L(L+1) when the question explicitly asks for the 'spin-only' formula. …
Showing the 12 most recent of 17 on this concept.
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.Which of the following ion has the maximum theoretical magnetic moment? [Fe (Z=26), Cr (Z=24), Ti (Z=22), Co (Z=27)](a) Fe3+(b) Cr3+(c) Ti3+(d) Co3+
›Reveal solutionSolution
Magnetic moment mu = sqrt(n(n+2)) BM rises with the number of unpaired electrons n; the d5 configuration allows the maximum possible unpaired electrons.
Find the d-electron configuration of each 3+ ion:
- Fe (Z=26): [Ar]3d⁶4s² → Fe3+ = [Ar]3d⁵ → 5 unpaired electrons (high spin) → μ = √(5×7) = √35 ≈ 5.92 BM
- Cr (Z=24): [Ar]3d⁵4s¹ → Cr3+ = [Ar]3d³ → 3 unpaired electrons → μ = √(3×5) = √15 ≈ 3.87 BM
- Ti (Z=22): [Ar]3d²4s² → Ti3+ = [Ar]3d¹ → 1 unpaired electron → μ = √3 ≈ 1.73 BM …
- GUJCET 2025Set 031 markMCQQ.Identify the metal whose divalent ion has 'spin only' magnetic moment 35 BM. (A) Cr (B) Mn (C) Fe (D) Co
›Reveal solutionSolution
[!TLDR]
35 BM corresponds to 5 unpaired electrons, matching Mn2+ (3d5).
Concept
The spin-only magnetic moment is μ=n(n+2) BM, where n is the number of unpaired electrons.
Solution
Solve for n:
n(n+2)=35 ⇒ n(n+2)=35 ⇒ n=5
Now find the divalent ion with 5 unpaired d electrons:
- Cr2+:3d4 → 4 unpaired …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.What is the value of magnetic moment of divalent ion having atomic number 30 in aqueous solution.(a) 0 BM(b) 2.84 BM(c) 1.73 BM(d) 5.92 BM
›Reveal solutionSolution
Atomic number 30 is Zn; its divalent ion Zn2+ has a fully filled 3d10 configuration with zero unpaired electrons, so its magnetic moment is zero.
Element with Z = 30 is Zinc, electronic configuration [Ar] 3d10 4s2.
Zn2+ is formed by removing the two 4s electrons, giving [Ar] 3d10 - a completely filled d-subshell with no unpaired electrons.
…
- GUJCET 2024Set 131 markMCQQ.Which of the following ion show highest spin only magnetic moment value? (A) Co2+ (B) Mn2+ (C) Ti2+ (D) Fe2+
›Reveal solutionSolution
Spin-only moment μ=n(n+2) BM increases with the number of unpaired electrons n; Mn2+ (d5) has the most.
Concept: Count d-electrons and unpaired electrons:
- Ti2+: d2, n=2.
- Mn2+: d5, n=5 ⇒μ=5×7=5.92 BM.
- Fe2+: d6, n=4. …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.What is the magnetic moment of a divalent ion in aqueous solution if its atomic number is 28?(a) 3.87 BM(b) 2.84 BM(c) 1.73 BM(d) 4.90 BM
›Reveal solutionSolution
Atomic number 28 is Nickel; its divalent ion Ni2+ has electronic configuration 3d8, giving 2 unpaired electrons, and the spin-only magnetic moment formula gives 2.84 BM.
Ni (Z=28): [Ar] 3d8 4s2
Ni2+: loses the two 4s electrons -> [Ar] 3d8
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.How many number of unpaired electrons are there in complex ion [Ni(CN)4]2-?(a) 4(b) 3(c) 2(d) 0
›Reveal solutionSolution
[Ni(CN)4]2- has Ni2+ (d8) with the strong-field ligand CN-, which forces pairing of the d-electrons into a square planar, diamagnetic (dsp2) arrangement.
Ni2+: [Ar] 3d8
CN- is a very strong field ligand (high in the spectrochemical series), causing the 3d8 electrons to pair up, freeing one 3d orbital for dsp2 hybridisation (square planar geometry). …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.Which compound has magnetic moment equal to 4.90 BM?(a) Cr2(SO4)3(b) NiSO4(c) FeSO4(d) MnSO4
›Reveal solutionSolution
mu = 4.90 BM means 4 unpaired electrons; Fe2+ (d6) fits, so FeSO4.
Spin-only moment mu = sqrt(n(n+2)) BM. For mu = 4.90: n(n+2) = 24 -> n = 4 unpaired electrons.
Check the metal ions:
- Cr3+ (Cr2(SO4)3): d3 -> 3 unpaired -> 3.87 BM.
- Ni2+ (NiSO4): d8 -> 2 unpaired -> 2.83 BM. …
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.Among the following, which compound has the highest magnetic moment?(a) MnSO4(b) CrCl3(c) Ni(NO3)2(d) FeSO4
›Reveal solutionSolution
Magnetic moment μ = √[n(n+2)] BM, where n = number of unpaired electrons; more unpaired electrons means a higher moment.
MnSO4: Mn2+ is d5 (high spin) → 5 unpaired electrons → μ = √35 = 5.92 BM.
FeSO4: Fe2+ is d6 → 4 unpaired electrons → μ = 4.90 BM.
CrCl3: Cr3+ is d3 → 3 unpaired electrons → μ = 3.87 BM. …
- GUJCET 2021Set 151 markMCQQ.If atomic number of element is 26, then magnetic moment is ___ BM of its divalent aqueous ion? (A) 1.73 (B) 3.87 (C) 2.83 (D) 4.90
›Reveal solutionSolution
Fe²⁺ (3d6) has 4 unpaired electrons, giving spin-only μ=n(n+2)=24=4.90 BM.
Concept: Atomic number 26 = Fe. The divalent ion Fe²⁺ has configuration [Ar]3d6, with 4 unpaired electrons. …
- GUJCET 2020Set 071 markMCQQ.The divalent ion of which of the following element in aqueous solution has magnetic moment 5.92 BM? (A) Fe (B) Cr (C) Co (D) Mn
›Reveal solutionSolution
μ=n(n+2)=5.92⇒n=5 unpaired; the d5 divalent ion is Mn²⁺.
Concept — spin-only magnetic moment. μ=n(n+2) BM. 5.92=5⋅7=35, so n=5 unpaired electrons. Mn²⁺ is [Ar]3d5 (5 …
- GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQQ.Magnetic moment of a divalent ion in aqueous solution if its atomic number is 25 ______.(a) 4.90 BM(b) 5.92 BM(c) 2.84 BM(d) 3.87 BM
›Reveal solutionSolution
Element 25 is manganese; its divalent ion Mn2+ has a 3d5 configuration with 5 unpaired electrons (a spin-only 'half-filled, maximally paramagnetic' case), giving the largest common magnetic moment among first-row transition ions.
…
- GUJCET 2019Set 131 markMCQQ.Which of the following pair has similar magnetic moment? (A) Ni2+,Co2+ (B) Fe2+,Mn2+ (C) Fe3+,Mn2+ (D) Cr3+,Mn3+
›Reveal solutionSolution
Magnetic moment depends only on the number of unpaired electrons; Fe3+ and Mn2+ are both 3d5.
Concept: Spin-only magnetic moment μ=n(n+2)BM, where n = number of unpaired electrons. Two ions have the same moment if they have the same n.
Counting unpaired d-electrons:
- Ni2+: 3d8 → 2 ; Co2+: 3d7 → 3 (not equal).
- Fe2+: 3d6 → 4 ; Mn2+: 3d5 → 5 (not equal). …
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