Q.Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
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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 number of unpaired electrons in 3d orbitals determines the magnetic moment, and in an octahedral field the five d‑orbitals split into t2g (lower energy) and eg (higher energy) sets.
Step 1: Determine the 3d electron count for each ion.
Remove electrons from the 4s orbital first, then from 3d.
- Ti2+: [Ar]3d2
- V2+: [Ar]3d3
- Cr3+: [Ar]3d3
- Mn2+: [Ar]3d5
- Fe2+: [Ar]3d6
- Fe3+: [Ar]3d5
- Co2+: [Ar]3d7
- Ni2+: [Ar]3d8
- Cu2+: [Ar]3d9
Step 2: For hydrated ions (weak field, octahedral), fill t2g first with one electron each, then pair, then fill eg.
Hund’s rule applies — each orbital gets one electron before pairing.
| Ion | 3dn | t2g occupancy | eg occupancy | Unpaired e− |
|---|---|---|---|---|
| Ti2+ | 2 | ↑ ↑ | — | 2 |
| V2+ | 3 | ↑ ↑ ↑ | — | 3 |
The number of 3d electrons in each ion is found by subtracting the ion charge from the neutral atom’s atomic number, then removing 4s electrons first. For hydrated octahedral complexes, the five 3d orbitals split into t2g (lower energy) and eg (higher energy) sets; electrons fill according to Hund’s rule and the ligand field strength (here, weak-field/high-spin for most first-row transition metal aqua ions).
Concept and Intuition
To find the 3d electron count for a transition metal ion, you must remember the Aufbau principle for neutral atoms: for elements in the 3d series, the 4s orbital fills before 3d (e.g., [Ar]4s23dx). When forming a positive ion, electrons are removed first from the 4s orbital, not the 3d — this is a common mistake. So for Ti2+, the neutral Ti has [Ar]4s23d2; removing two electrons takes both 4s electrons, leaving 3d2.
Once we know the 3d count, we consider the hydrated ion in an octahedral crystal field. Water is a weak-field ligand, so the splitting energy Δo is small. This means electrons fill all five orbitals singly before pairing (Hund’s rule) — the high-spin configuration. The five d orbitals split into a lower-energy triplet (dxy,dxz,dyz — called t2g) and a higher-energy doublet (dz2,dx2−y2 — called eg). For weak fields, electrons occupy t2g first, then eg, all with parallel spins as far as possible.
A classic error: for Fe3+, students often write 3d5 but then pair electrons in t2g because they think of the free ion. In a weak octahedral field, Fe3+ has all five orbitals singly occupied — a half-filled t2g3eg2 configuration. Do not pair unless the ligand is strong (like CN⁻).
Let’s work through each ion step by step.
1. Ti2+ (Titanium, Z = 22)
Neutral Ti: [Ar]4s23d2. Remove 2 electrons → both from 4s.
3d electrons = 2.
In octahedral field: two electrons go into t2g (lower energy), both unpaired.
Configuration: t2g2eg0 (2 unpaired electrons).
2. V2+ (Vanadium, Z = 23)
Neutral V: [Ar]4s23d3. Remove 2 electrons → both from 4s.
3d electrons = 3.
Three electrons: all occupy t2g singly (Hund’s rule).
Configuration: t2g3eg0 (3 unpaired).
3. Cr3+ (Chromium, Z = 24)
Neutral Cr: [Ar]4s13d5 (exception: half-filled d gives stability). Remove 3 electrons → first the 4s electron, then two from 3d.
3d electrons = 3.
Same as V²⁺: t2g3eg0 (3 unpaired).
Cr has a special ground state: 4s13d5, not 4s23d4. Always check the periodic table for these exceptions (Cr and Cu). For ions, the 4s is always emptied first, so Cr³⁺ ends up 3d3.
4. Mn2+ (Manganese, Z = 25)
Neutral Mn: [Ar]4s23d5. Remove 2 electrons → both from 4s.
3d electrons = 5.
Five electrons: fill all five orbitals singly — t2g3eg2 (5 unpaired). This is a half-filled d shell, extra stable.
5. Fe2+ (Iron, Z = 26)
Neutral Fe: [Ar]4s23d6. Remove 2 electrons → both from 4s.
3d electrons = 6.
Six electrons: first five singly occupy all orbitals, the sixth pairs in a t2g orbital.
Configuration: t2g4eg2 (4 unpaired electrons).
6. Fe3+ (Iron, Z = 26)
Neutral Fe: [Ar]4s23d6. Remove 3 electrons → both 4s and one 3d.
3d electrons = 5.
Same as Mn²⁺: t2g3eg2 (5 unpaired).
7. Co2+ (Cobalt, Z = 27)
Neutral Co: [Ar]4s23d7. Remove 2 electrons → both from 4s.
3d electrons = 7.
Seven electrons: fill t2g with three, then eg with two (all singly), then the remaining two pair in t2g.
Configuration: t2g5eg2 (3 unpaired electrons).
8. Ni2+ (Nickel, Z = 28)
Neutral Ni: [Ar]4s23d8. Remove 2 electrons → both from 4s. …
Method: Crystal Field Theory + Spin-Only Magnetic Moment Calculation
Method Name: Spin-Only Magnetic Moment using Crystal Field Splitting (for octahedral dn ions)
Step 1: Determine the number of 3d electrons for each ion
For transition metal ions, remove electrons from the 4s orbital first, then from 3d.
| Ion | Atomic No. | Neutral configuration | Ion configuration | 3d electrons (n) |
|---|---|---|---|---|
| Ti2+ | 22 | [Ar]3d24s2 | [Ar]3d2 | 2 |
| V2+ | 23 | [Ar]3d34s2 | [Ar]3d3 | 3 |
| Cr3+ | 24 | [Ar]3d54s1 | [Ar]3d3 | 3 |
| Mn2+ | 25 | [Ar]3d54s2 | [Ar]3d5 | 5 |
| Fe2+ | 26 | [Ar]3d64s2 | [Ar]3d6 | 6 |
| Fe3+ | 26 | [Ar]3d64s2 | [Ar]3d5 | 5 |
| Co2+ | 27 | [Ar]3d74s2 | [Ar]3d7 | 7 |
| Ni2+ | 28 | [Ar]3d84s2 | [Ar]3d8 | 8 |
| Cu2+ | 29 | [Ar]3d104s1 | [Ar]3d9 | 9 |
Key exam point: For Cr and Cu, the neutral atom has a half-filled or fully-filled d-subshell (exception to Aufbau), but ions follow normal removal order.
Step 2: Occupancy of five 3d orbitals in octahedral field
In an octahedral crystal field, the five d-orbitals split into:
- t2g (lower energy): dxy, dyz, dzx
- eg (higher energy): dx2−y2, dz2
Hund's rule applies: electrons fill degenerate orbitals singly before pairing.
| Ion | n | t2g occupancy | eg occupancy | Unpaired electrons |
|---|---|---|---|---|
| Ti2+ | 2 | ↑↑ | — | 2 |
| V2+ | 3 | ↑↑↑ | — | 3 |
| Cr3+ | 3 | ↑↑↑ | — | 3 |
| Mn2+ | 5 | ↑↑↑ | ↑↑ | 5 |
| Fe2+ | 6 | ↑↓↑↑ | ↑↑ | 4 |
| Fe3+ | 5 | ↑↑↑ | ↑↑ | 5 |
| Co2+ | 7 | ↑↓↑↓↑ | ↑↑ | 3 |
| Ni2+ | 8 | ↑↓↑↓↑↓ | ↑↑ | 2 |
| Cu2+ | 9 | ↑↓↑↓↑↓ | ↑↓↑ | 1 |
Note: H2O is a weak field ligand, so every hydrated ion here is high-spin — each keeps the maximum number of unpaired electrons Hund's rule allows for its d-electron count. (For d8 (Ni2+) and d9 (Cu2+) the octahedral occupancy is the same whatever the field strength.)
Step 3: Calculate spin-only magnetic moment …
Here are the most common mistakes students make when calculating magnetic moments and determining 3d orbital occupancy for hydrated transition metal ions, along with how to avoid each.
1. Mistake: Forgetting to Account for the Charge When Writing Electronic Configuration
The Error: Students often write the configuration for the neutral atom (e.g., Fe: [Ar]3d64s2) and then directly use the same dn count for the ion without removing electrons from the correct subshell.
Example of Mistake: For Fe2+, writing 3d6 but thinking it comes from simply removing two electrons from the 4s orbital after the 3d is filled — which is wrong because in ions, the 4s empties first.
How to Avoid:
- Rule: For transition metal ions, remove electrons from the 4s orbital before the 3d orbital.
- Correct method:
- Write neutral atom: Fe = [Ar]3d64s2
- Remove 2 electrons: first from 4s → [Ar]3d6
- So Fe2+ has 6 3d electrons.
Quick check for all ions asked:
| Ion | Neutral config | Ion config | Number of 3d electrons |
|---|---|---|---|
| Ti2+ | [Ar]3d24s2 | [Ar]3d2 | 2 |
| V2+ | [Ar]3d34s2 | [Ar]3d3 | 3 |
| Cr3+ | [Ar]3d54s1 | [Ar]3d3 | 3 |
| Mn2+ | [Ar]3d54s2 | [Ar]3d5 | 5 |
| Fe2+ | [Ar]3d64s2 | [Ar]3d6 | 6 |
| Fe3+ | [Ar]3d64s2 | [Ar]3d5 | 5 |
| Co2+ | [Ar]3d74s2 | [Ar]3d7 | 7 |
| Ni2+ | [Ar]3d84s2 | [Ar]3d8 | 8 |
| Cu2+ | [Ar]3d104s1 | [Ar]3d9 | 9 |
2. Mistake: Ignoring Hund’s Rule When Filling the Five 3d Orbitals
The Error: Students fill orbitals in pairs before all five are singly occupied, especially for d4, d5, d6, and d7 configurations.
Example of Mistake: For Mn2+ (d5), writing ↑↓ in one orbital and three singles — instead of all five orbitals singly occupied.
How to Avoid:
- Hund’s Rule: Electrons occupy degenerate orbitals singly first, with parallel spins, before pairing.
- For free ions (no ligand field): Fill all five orbitals with one electron each before pairing.
- For hydrated ions (octahedral field): The same rule applies for high-spin complexes (which is the case for all these hydrated ions because water is a weak field ligand).
Correct occupancy for each (high-spin octahedral):
| dn | Occupancy (five orbitals: dxy,dyz,dxz,dx2−y2,dz2) |
|---|---|
| d2 | ↑ ↑ _ _ _ |
| d3 | ↑ ↑ ↑ _ _ |
| d4 | ↑ ↑ ↑ ↑ _ |
| d5 | ↑ ↑ ↑ ↑ ↑ |
| d6 | ↑↓ ↑ ↑ ↑ ↑ |
| d7 | ↑↓ ↑↓ ↑ ↑ ↑ |
| d8 | ↑↓ ↑↓ ↑↓ ↑ ↑ |
| d9 | ↑↓ ↑↓ ↑↓ ↑↓ ↑ |
3. Mistake: Confusing High-Spin vs Low-Spin for Hydrated Ions
The Error: Students assume all octahedral complexes are low-spin, or they apply strong-field rules (like for CN−) to water complexes.
Example of Mistake: For Fe2+ (d6) in water, writing ↑↓ ↑↓ ↑↓ _ _ (low-spin, 0 unpaired electrons) instead of the correct high-spin arrangement with 4 unpaired electrons.
How to Avoid:
- Remember: Water (H2O) is a weak field ligand — it causes a small crystal field splitting (Δo).
- Weak field → high-spin (electrons prefer to occupy all orbitals singly before pairing).
- Strong field ligands (like CN−, CO) cause low-spin.
- For all ions listed, hydrated = high-spin — including Co2+, which is high-spin with water like the rest.
Unpaired electron count for each (high-spin):
| Ion | dn | Unpaired electrons |
|---|---|---|
| Ti2+ | d2 | 2 |
| V2+ | d3 | 3 |
| Cr3+ | d3 | 3 |
| Mn2+ | d5 | 5 |
| Fe2+ | d6 | 4 |
| Fe3+ | d5 | 5 |
| Co2+ | d7 | 3 |
| Ni2+ | d8 | 2 |
| Cu2+ | d9 | 1 |
4. Mistake: Using the Wrong Formula for Magnetic Moment
The Error: Students use μ=n(n+2) but forget that n = number of unpaired electrons, not total d electrons.
Example of Mistake: For Fe2+ (d6), using n=6 → μ=6×8=48≈6.93 BM (wrong). Correct: n=4 → μ=4×6=24≈4.90 BM.
How to Avoid:
- Formula: μ=n(n+2) BM, where n = number of unpaired electrons.
- Always count unpaired electrons from the orbital diagram first.
- Memorize common values:
- n=1 → μ≈1.73 BM
- n=2 → μ≈2.83 BM
- n=3 → μ≈3.87 BM
- n=4 → μ≈4.90 BM
- n=5 → μ≈5.92 BM
5. Mistake: Thinking Cr3+ and V2+ (Both d3) Have Different Magnetic Moments — They Are the Same …
Showing the 12 most recent of 14 on this concept.
- COMEDK 2026Set 2026-M1 markMCQQ.The ion that has a spin only magnetic moment of 5.9 BM is: (A) Fe3+ (B) Mn3+ (C) Ni2+ (D) Co3+
›Reveal solutionSolution
The spin-only magnetic moment is given by μ=n(n+2) BM, where n is the number of unpaired electrons. A value of 5.9 BM corresponds to n=5 unpaired electrons. Among the options, only Fe3+ has 5 unpaired electrons, so the correct option is (A).
The key concept here is the spin-only magnetic moment formula:
μ=n(n+2) BM
where n is the number of unpaired electrons. This formula works well for first-row transition metal ions because orbital angular momentum is often "quenched" by the crystal field. The experimental value 5.9 BM is very close to the theoretical value for n=5, which is 5×7=35≈5.92 BM. So we need to find which ion has exactly 5 unpaired electrons in its ground state.
Let’s check each option step by step:
-
Determine the electronic configuration of each ion
- Fe3+: Atomic number of Fe = 26. Neutral Fe: [Ar]3d64s2. Removing three electrons (first from 4s, then from 3d) gives [Ar]3d5.
- Mn3+: Atomic number of Mn = 25. Neutral Mn: [Ar]3d54s2. Removing three electrons gives [Ar]3d4.
- Ni2+: Atomic number of Ni = 28. Neutral Ni: [Ar]3d84s2. Removing two electrons gives [Ar]3d8.
- Co3+: Atomic number of Co = 27. Neutral Co: [Ar]3d74s2. Removing three electrons gives [Ar]3d6.
-
Count unpaired electrons using Hund’s rule
- For Fe3+ (3d5): All five d-orbitals are half-filled, each with one electron → 5 unpaired electrons.
- For Mn3+ (3d4): Four electrons occupy four orbitals singly (Hund’s rule) → 4 unpaired electrons.
- For Ni2+ (3d8): The d-orbitals fill as: ↑↓ ↑↓ ↑ ↑ ↑ (two paired, two singly, one singly) → 2 unpaired electrons. …
-
- KCET 2026Set D31 markMCQQ.The calculated spin only magnetic moment of Cr2+ ion is (A) 3.87 BM (B) 4.90 BM (C) 5.92 BM (D) 2.84 BM
›Reveal solutionSolution
The spin-only magnetic moment depends on the number of unpaired electrons in the ion's d-configuration.
Step 1 — Electron configuration of Cr²⁺
Neutral Cr (Z = 24) is [Ar]3d54s1. Forming Cr2+ removes the single 4s electron first, then one 3d electron, giving Cr2+=[Ar]3d4.
Step 2 — Counting unpaired electrons …
- COMEDK 2025Set 2025-A1 markMCQQ.Which is the correct order of increasing number of unpaired electrons in the following ions? A=Cr2+(Z=24)B=Cu2+(Z=29)C=Ni2+(Z=28)D=Fe3+(Z=26) (A) B<C<A<D (B) D<C<A<B (C) B<C<D<A (D) C<B<D<A
›Reveal solutionSolution
The number of unpaired electrons depends on the electronic configuration of each ion, considering the d-orbital filling and Hund's rule. The correct order is B (Cu²⁺, 1 unpaired) < C (Ni²⁺, 2 unpaired) < A (Cr²⁺, 4 unpaired) < D (Fe³⁺, 5 unpaired), which corresponds to option (A).
Concept & Intuition
The key is to write the ground-state electron configurations for each ion, focusing on the d-electrons. Transition metal ions lose s-electrons first, then d-electrons if needed. Unpaired electrons arise from half-filled or partially filled d-orbitals following Hund's rule (each orbital gets one electron before pairing). The more unpaired electrons, the higher the magnetic moment. Here, we count them directly.
Step-by-step reasoning
-
Cr²⁺ (Z = 24)
- Neutral Cr: [Ar] 4s¹ 3d⁵ (exception: half-filled d gives stability).
- Remove two electrons: first from 4s, then from 3d → Cr²⁺: [Ar] 3d⁴.
- Hund's rule: four d-orbitals each get one electron, one orbital empty → 4 unpaired electrons.
-
Cu²⁺ (Z = 29)
- Neutral Cu: [Ar] 4s¹ 3d¹⁰ (exception: fully filled d).
- Remove two electrons: one from 4s, one from 3d → Cu²⁺: [Ar] 3d⁹.
- One d-orbital has a single electron (the rest are paired) → 1 unpaired electron.
-
Ni²⁺ (Z = 28)
- Neutral Ni: [Ar] 4s² 3d⁸.
- Remove two 4s electrons → Ni²⁺: [Ar] 3d⁸.
- Filling: five orbitals get 2, 2, 2, 1, 1 (two unpaired) → 2 unpaired electrons.
-
Fe³⁺ (Z = 26)
- Neutral Fe: [Ar] 4s² 3d⁶. …
-
- COMEDK 2025Set 2025-A1 markMCQQ.The correct order of spin only magnetic moments among the following is: [Given: Atomic numbers: Mn=25,Fe=26,Co=27 ] (A) [Fe(CN)6]4−>[MnCl4]2−>[CoCl4]2− (B) [MnCl4]2−>[CoCl4]2−>[Fe(CN)6]4− (C) [Fe(CN)6]4−>[CoCl4]2−>[MnCl4]2− (D) [MnCl4]2−>[Fe(CN)6]4−>[CoCl4]2−
›Reveal solutionSolution
The spin-only magnetic moment depends on the number of unpaired electrons, which is determined by the metal’s oxidation state and the ligand field (strong-field CN⁻ vs. weak-field Cl⁻). The correct order is [MnCl4]2−>[CoCl4]2−>[Fe(CN)6]4−, corresponding to option (B).
Concept & Intuition
The spin-only magnetic moment is given by μ=n(n+2) Bohr magnetons, where n is the number of unpaired electrons. So the problem reduces to finding n for each complex. The key twist: ligands like CN⁻ are strong-field (low-spin), causing maximum pairing, while Cl⁻ is weak-field (high-spin), favoring unpaired electrons. Also, geometry matters — here all complexes are either octahedral or tetrahedral, which affects the d-orbital splitting and thus the electron configuration.
Step-by-step reasoning
-
Determine oxidation states and d-electron counts
- [Fe(CN)6]4−: CN⁻ is −1 each, so Fe must be +2 (since 6×(−1) + x = −4 → x = +2). Fe (Z=26): [Ar]3d⁶. Fe²⁺: 3d⁶.
- [MnCl4]2−: Cl⁻ is −1 each, so Mn is +2 (4×(−1) + x = −2 → x = +2). Mn (Z=25): [Ar]3d⁵. Mn²⁺: 3d⁵.
- [CoCl4]2−: Cl⁻ is −1 each, so Co is +2 (4×(−1) + x = −2 → x = +2). Co (Z=27): [Ar]3d⁷. Co²⁺: 3d⁷.
-
Analyze [Fe(CN)6]4− — octahedral, strong-field
CN⁻ is a strong-field ligand, causing large splitting. For Fe²⁺ (d⁶) in an octahedral field, strong-field means low-spin: all electrons pair in the t2g set first. Configuration: t2g6eg0. All 6 electrons are paired → 0 unpaired electrons.
μ=0(0+2)=0 BM.
-
Analyze [MnCl4]2− — tetrahedral, weak-field
Cl⁻ is a weak-field ligand. Tetrahedral splitting is smaller than octahedral, so always high-spin. Mn²⁺ is d⁵. In tetrahedral geometry, the e set is lower in energy than t2. High-spin d⁵: each of the five d-orbitals gets one electron (Hund’s rule) → 5 unpaired electrons.
μ=5(5+2)=35≈5.92 BM.
-
Analyze [CoCl4]2− — tetrahedral, weak-field …
-
- COMEDK 2025Set 2025-E1 markMCQQ.Arrange the complex ions in the increasing order of their magnetic moments A. [Fe(H2O)6]2+ B. [Fe(CN)6]4− C. [Fe(CN)6]3− D. [FeF6]3− (A) D<A<B<C (B) B<C<A<D (C) D<C<B<A (D) A<D<C<B
›Reveal solutionSolution
The magnetic moment depends on the number of unpaired electrons, which is determined by the metal’s oxidation state, ligand field strength (weak vs. strong field), and the resulting high-spin or low-spin configuration. The correct increasing order is B < C < A < D, corresponding to option (B).
Concept & Intuition
Magnetic moment (μ) for a transition metal complex is given by the spin-only formula:
μ=n(n+2) BM
where n is the number of unpaired electrons. So the order of magnetic moments is simply the order of number of unpaired electrons.
To find n, we need:
- The oxidation state of iron in each complex.
- Whether the ligand is weak-field (high-spin) or strong-field (low-spin).
- The d-electron count and how electrons fill the t2g and eg orbitals.
Let’s work through each complex step by step.
1. Determine oxidation states and d-electron counts
- A: [Fe(H2O)6]2+ — Water is neutral, so Fe is +2. Fe²⁺ has electron configuration [Ar] 3d⁶.
- B: [Fe(CN)6]4− — CN⁻ is -1 each, total −6 from ligands, so Fe must be +2 to give overall −4. Again, Fe²⁺, 3d⁶.
- C: [Fe(CN)6]3− — CN⁻ gives −6, so Fe is +3. Fe³⁺ is 3d⁵.
- D: [FeF6]3− — F⁻ is -1 each, total −6, so Fe is +3. Also 3d⁵.
So we have two Fe²⁺ (d⁶) and two Fe³⁺ (d⁵) complexes.
2. Identify ligand field strength
- H₂O is a weak-field ligand (causes small splitting).
- F⁻ is also a weak-field ligand.
- CN⁻ is a strong-field ligand (causes large splitting).
For weak-field ligands, electrons fill according to Hund’s rule (high-spin). For strong-field ligands, pairing occurs before occupying higher orbitals (low-spin).
3. Determine unpaired electrons for each
-
A: Fe²⁺ (d⁶), weak field (H₂O) → high-spin.
In an octahedral field: t2g4eg2 → 4 unpaired electrons.
n=4.
-
B: Fe²⁺ (d⁶), strong field (CN⁻) → low-spin.
All 6 electrons pair in t2g: t2g6eg0 → 0 unpaired electrons.
n=0.
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C: Fe³⁺ (d⁵), strong field (CN⁻) → low-spin. …
- COMEDK 2025Set 2025-M1 markMCQQ.Match the coordination compounds in Column I having the given type of hybridisation of Mn+ ion and magnetic moment as given in Column II. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Coordination compounds Hybridisation & Magnetic nature A. Ni(CO)4 P. sp3,μ=5.92BM B. [Ni(CN)4]2− Q. sp3,μ=2.84BM C. [Ni(Cl)4]2− R. sp3,μ=0 D. [MnBr4]2− S. dsp2,μ=0 (A) A=RB=SC=QD=P (B) A=SB=RC=PD=Q (C) A=SB=PC=RD=Q (D) A=QB=PC=SD=R
›Reveal solutionSolution
The key is to determine the oxidation state, geometry, and number of unpaired electrons for each complex, then match the hybridisation and magnetic moment. The correct matches are A→R, B→S, C→Q, D→P, so option (A) is correct.
We need to match each coordination compound with its hybridisation and magnetic moment. The magnetic moment (in Bohr magnetons, BM) is given by μ=n(n+2) where n is the number of unpaired electrons. Let’s analyse each complex step by step.
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Complex A: Ni(CO)4
- Nickel is in zero oxidation state (CO is neutral).
- CO is a strong field ligand, causing pairing of electrons.
- Ni(0) has atomic number 28: [Ar]3d84s2. In the presence of strong field CO, the 4s electrons are used for bonding, and the 3d electrons pair up.
- The complex is tetrahedral (4 ligands), so hybridisation is sp3.
- All electrons are paired → μ=0 BM.
- So A matches R (sp3,μ=0).
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Complex B: [Ni(CN)4]2−
- CN⁻ is a strong field ligand.
- Oxidation state: Ni + 4(−1) = −2 → Ni is +2.
- Ni²⁺: [Ar]3d8. Strong field CN⁻ causes pairing of 3d electrons, leaving one d-orbital empty.
- Geometry is square planar (4 ligands), hybridisation is dsp2.
- All electrons paired → μ=0 BM.
- So B matches S (dsp2,μ=0).
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Complex C: [Ni(Cl)4]2−
- Cl⁻ is a weak field ligand.
- Ni is again +2 (same as B).
- Weak field → no pairing; Ni²⁺ has 8 d-electrons, which in a tetrahedral field give 2 unpaired electrons (since t2 orbitals are higher energy, electrons occupy them singly first).
- Geometry is tetrahedral → hybridisation sp3.
- μ=2(2+2)=8≈2.84 BM.
- So C matches Q (sp3,μ=2.84 BM).
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Complex D: [MnBr4]2−
- Br⁻ is a weak field ligand. …
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- COMEDK 2025Set 2025-M1 markMCQQ.What is the spin only magnetic moment of the metal ion in P and the oxidation number of Sulphur in the oxidised product Z ? (A) μ=0 Oxidation number =+4 (B) μ=5.00BM Oxidation number =0 (C) μ=5.92BM Oxidation number =+2 (D) μ=1.732BM Oxidation number =+6
›Reveal solutionSolution
Pyrolusite (MnO2) fused with KOH and KNO3 gives the manganate(VI) ion in P=K2MnO4. The metal ion Mn6+ is d1, so μ=1(1+2)=1.732 BM. Permanganate then oxidises thiosulphate in neutral medium to sulphate, in which sulphur is +6. The correct option is (D).
Concept
In an oxidising alkaline fusion, Mn(IV) in pyrolusite is raised to Mn(VI), giving the green manganate ion MnO42−. The spin-only magnetic moment depends only on the number of unpaired electrons through μ=n(n+2) BM. Separately, the oxidation number of an atom is read from the formula of the species it ends up in.
Solution
- Identify P. MnO2+2KOH+KNO3→K2MnO4+KNO2+H2O. So P=K2MnO4 and the metal ion is Mn6+.
- Magnetic moment of Mn6+. Configuration [Ar]3d1 gives n=1, so μ=1(1+2)=3=1.732 BM.
- Fate in acid. 3MnO42−+4H+→2MnO4−+MnO2+2H2O (disproportionation), giving permanganate and MnO2. …
- KCET 2024Set B-21 markMCQQ.Match the following: I. Zn2+ II. Cu2+ III. Ni2+ i. d8 configuration ii. colourless iii. μ=1.73BM I II III (A) i ii iii (B) ii iii i (C) ii i iii (D) i iii ii
›Reveal solutionSolution
Write the dn configuration of each ion, then read off colour (needs a partly-filled d shell) and magnetic moment from the spin-only formula.
Step 1 — Get the d-configurations.
For a transition-metal cation, remove the 4s electrons first, then the 3d electrons.
Ion Atomic no. Neutral atom M2+ configuration Zn2+ 30 [Ar]3d104s2 [Ar]3d10 Cu2+ 29 [Ar]3d104s1 [Ar]3d9 Ni2+ 28 [Ar]3d84s2 [Ar]3d8 Step 2 — Match III: Ni2+ → (i) d8.
Read straight off the table above. This is the most direct match, and it already eliminates options (A) and (C), which both send d8 to Zn2+.
Step 3 — Match I: Zn2+ → (ii) colourless.
Colour in transition-metal ions arises from d–d electronic transitions: an electron absorbs visible light and is promoted from the lower (t2g) to the higher (eg) set. This demands a partially filled d subshell. Zn2+ is d10 — completely full, no vacancy to promote into — so no d–d transition is possible and the ion is colourless.
Step 4 — Match II: Cu2+ → (iii) μ=1.73 BM.
Use the spin-only formula, where n = number of unpaired electrons: …
- COMEDK 2024Set 2024-A1 markMCQQ.The metallic ions that have almost same spin only magnetic moment are :(i) Co2+(ii) Mn2+(iii) Cr2+(iv) Cr3+ (A) $$ \text {(iii) and(iv) } (B) \text {(i) and(ii) } (C) \text {(i) and(iv) } (D) \text {(i) and(iii) } $$
›Reveal solutionSolution
Counting unpaired electrons: Co2+ (d7) = 3, Mn2+ (d5) = 5, Cr2+ (d4) = 4, Cr3+ (d3) = 3. Co2+ and Cr3+ share n=3, hence the same μ=3.87 BM → (i) and (iv).
Spin-only moment μ=n(n+2) BM, where n = number of unpaired electrons:
- (i) Co2+: 3d7 → 3 unpaired → μ=15=3.87 BM
- (ii) Mn2+: 3d5 → 5 unpaired → μ=35=5.92 BM …
- COMEDK 2024Set 2024-E1 markMCQQ.Based on Valence Bond Theory, match the complexes listed in Column I with the number of unpaired electrons on the central metal ion, given in Column II .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} No. Complex ions No. Number of unpaired electrons (A) [FeF6]3− (P) 0 (B) [Fe(CN)6]4− (P) 1 (C) [Fe(H2O)6]2+ (R) 5 (D) [Fe(CN)6]3− (S) 4 (A) A=SB=QC=RD=P (B) A=SB=QC=PD=R (C) A=RB=PC=SD=Q (D) A=RB=SC=QD=P
›Reveal solutionSolution
[!TLDR]
Using d-electron counts and strong/weak-field ligands, the unpaired electrons are 5, 0, 4, 1 for A-D, matching option (C).
Concept
In Valence Bond Theory a strong-field ligand (like CN-) forces electron pairing, giving a low-spin inner-orbital complex, while a weak-field ligand (F-, H2O) leaves the electrons unpaired (high spin). Always count the d-electrons on the metal ion first (CBSE/NCERT Class 12, Coordination Compounds).
Solution
- (A) [FeF6]3−: Fe3+ is 3d5; F− is weak-field ⇒ high spin ⇒ 5 unpaired ⇒ R.
- (B) [Fe(CN)6]4−: Fe2+ is 3d6; CN− strong-field ⇒ low spin (t2g6) ⇒ 0 unpaired ⇒ P. …
- COMEDK 2024Set 2024-M1 markMCQQ.A d - block metal X(Z=26) forms a compound [X(CN)2(CO)4]+. Calculate its spin magnetic moment value. (A) 2.83 BM (B) 1.73 BM (C) 3.87 BM (D) 5.92 BM
›Reveal solutionSolution
The metal is iron (Z=26); in [Fe(CN)2(CO)4]+ it is Fe3+ (d5), and since CN− and CO are both strong-field ligands, the complex is low-spin with 1 unpaired electron, giving μ=1.73 BM — option (B).
Step-by-step reasoning
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Identify the metal. Z=26 is iron (Fe).
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Find the oxidation state.
CN− contributes −1 each (×2), CO is neutral (×4), overall complex charge =+1:
x+2(−1)+4(0)=+1⇒x=+3
So Fe is in the +3 state.
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Find the d-electron count.
Neutral Fe is [Ar]3d64s2; Fe3+ is [Ar]3d5.
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Apply the ligand field. …
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- COMEDK 2024Set 2024-M1 markMCQQ.The permanganate ion in acid medium acts as an oxidant and gets converted to its lower oxidation state. What would be the spin only magnetic moment of such reduced manganese ion? (A) 2.84 BM (B) 4.90 BM (C) 3.87 BM (D) 5.92 BM
›Reveal solutionSolution
In acidic medium, permanganate (MnO4−) is reduced to Mn2+, which has five unpaired 3d electrons; the spin-only magnetic moment is 5(5+2)=35≈5.92 BM, so the correct option is (D).
The key concept here is the relationship between oxidation state, electronic configuration, and magnetic moment. The permanganate ion (MnO4−) contains manganese in the +7 oxidation state. In acidic medium, it is a powerful oxidant and gets reduced to a lower oxidation state — specifically, to the manganous ion (Mn2+). The question asks for the spin-only magnetic moment of this reduced ion. The spin-only formula is μ=n(n+2) BM, where n is the number of unpaired electrons. So we need to find n for Mn2+.
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Determine the electronic configuration of the reduced manganese ion.
Manganese (atomic number 25) has the ground-state configuration [Ar]3d54s2.
In the Mn2+ ion, two electrons are removed — first from the 4s orbital (as is standard for transition metals), giving [Ar]3d5.
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Find the number of unpaired electrons in Mn2+.
The 3d subshell has five orbitals. For a d5 configuration, Hund’s rule tells us that electrons occupy all five orbitals singly with parallel spins before any pairing occurs. Thus, all five 3d electrons are unpaired.
So n=5.
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Apply the spin-only magnetic moment formula.
The formula is:
μ=n(n+2) BM
Substituting n=5:
μ=5×7=35 BM …
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