The Transition Metals: Why This Chapter Matters
The d and f block elements sit in the middle of the periodic table and they are the reason chemistry feels “alive” — coloured solutions, magnetic materials, catalysts that run industrial processes. For Class 12, this chapter carries significant weightage in both CBSE boards (5-6 marks consistently) and JEE Main (1-2 questions every session).
What makes this topic tricky is that students try to memorise everything individually. That’s the wrong approach. Once you understand the pattern — how electronic configuration drives colour, magnetism, and oxidation states — almost everything follows logically.
We’ll build that pattern here, starting from electronic configuration and working up to the specific reactions CBSE and JEE love to test.
Key Terms and Definitions
d Block elements: Elements in groups 3-12 of the periodic table where the last electron enters a d orbital. They form the transition metals — though strictly, transition metals must have at least one ion with an incompletely filled d subshell.
Zinc (Zn), Cadmium (Cd), and Mercury (Hg) are d block but NOT transition metals. Their ions have fully filled d¹⁰ configuration. CBSE loves asking this distinction.
f Block elements: Elements where the last electron enters an f orbital. Split into:
- Lanthanoids (Ce to Lu, 4f series) — the “rare earths”
- Actinoids (Th to Lr, 5f series) — mostly radioactive
Oxidation state: The charge an element would have if all bonds were ionic. d block elements show variable oxidation states because d and s electrons are close in energy and both can be lost.
Paramagnetism: Attraction to a magnetic field due to unpaired electrons. The more unpaired electrons, the stronger the paramagnetism.
Magnetic moment formula:
where = number of unpaired electrons, BM = Bohr Magneton
Core Concepts
1. Electronic Configuration
The general configuration for d block: [noble gas] (n-1)d¹⁻¹⁰ ns⁰⁻²
Two anomalies you must memorise — they appear in PYQs constantly:
| Element | Expected | Actual | Reason |
|---|---|---|---|
| Cr (Z=24) | [Ar] 3d⁴ 4s² | [Ar] 3d⁵ 4s¹ | Half-filled d is extra stable |
| Cu (Z=29) | [Ar] 3d⁹ 4s² | [Ar] 3d¹⁰ 4s¹ | Fully filled d is extra stable |
When writing ionic configurations, remove 4s electrons before 3d. Cu²⁺ is [Ar] 3d⁹, not [Ar] 3d⁷ 4s². Many students forget this and lose a mark in CBSE.
2. Variable Oxidation States
Transition metals show variable oxidation states because 3d and 4s electrons have similar energies. Both can participate in bonding.
Key oxidation states to know cold:
- Mn: +2, +3, +4, +6, +7 (MnO₄⁻ is the most oxidised, deep purple)
- Fe: +2 (Fe²⁺, pale green) and +3 (Fe³⁺, yellow-brown)
- Cr: +3 (most stable) and +6 (CrO₄²⁻ yellow, Cr₂O₇²⁻ orange)
- Cu: +1 (Cu₂O, brick red) and +2 (most common, blue solutions)
The highest oxidation state equals the group number up to group 7 (Mn = +7). After that, lower oxidation states become more stable.
3. Why Do Transition Metal Compounds Show Colour?
This is a JEE favourite and CBSE 2-mark staple. The reason has one root cause: d-d transitions.
When white light hits a transition metal compound, photons of specific wavelengths are absorbed to promote an electron from a lower d orbital to a higher d orbital. The complementary colour is what we see.
This only works when the d subshell is partially filled. That’s why:
- Sc³⁺ (d⁰) → colourless
- Zn²⁺ (d¹⁰) → colourless
- Cu²⁺ (d⁹) → blue ✓
- Fe³⁺ (d⁵) → yellow-brown ✓
CBSE 2023 asked: “Why is Cu²⁺ solution blue but Cu⁺ is colourless?” Answer: Cu²⁺ has d⁹ (incomplete d, d-d transition possible). Cu⁺ has d¹⁰ (complete d, no d-d transition possible).
4. Magnetic Properties
Directly tied to unpaired electrons. Use BM.
| Ion | Config | Unpaired e⁻ | μ (BM) |
|---|---|---|---|
| Fe³⁺ | 3d⁵ | 5 | 5.92 |
| Ni²⁺ | 3d⁸ | 2 | 2.83 |
| Cu²⁺ | 3d⁹ | 1 | 1.73 |
| Zn²⁺ | 3d¹⁰ | 0 | 0 (diamagnetic) |
5. Catalytic Property
Transition metals are exceptional catalysts for two reasons:
- Variable oxidation states — they can easily gain/lose electrons, acting as electron shuttles in reactions
- Ability to adsorb reactants on their surface (the d orbitals form weak bonds with reactant molecules)
Key examples for exams:
- Fe in Haber process (N₂ + 3H₂ → 2NH₃)
- V₂O₅ in Contact process (2SO₂ + O₂ → 2SO₃)
- Ni in hydrogenation of oils
- MnO₂ as catalyst in decomposition of KClO₃
6. Interstitial Compounds
Transition metals have large atomic radii with gaps (interstices) in their crystal lattice. Small atoms like H, C, N can fit into these gaps forming interstitial compounds.
Properties: harder than pure metal, high melting point, retain metallic conductivity. Example: steel is essentially iron with carbon in interstitial positions.
7. Lanthanoid Contraction
As we move across the lanthanoids (La to Lu), nuclear charge increases by 1 each time but the added electron goes into the 4f subshell. The 4f electrons shield poorly — they’re diffuse and don’t effectively screen outer electrons from the nucleus.
Result: atomic radius decreases steadily across the lanthanoids — this is lanthanoid contraction.
The consequence that always appears in exams: The elements following the lanthanoids (Hf, Ta, W…) have nearly identical atomic radii to the elements above them (Zr, Nb, Mo…). This makes 4d and 5d pairs almost inseparable chemically — for example, Zr and Hf are always found together in nature.
Solved Examples
Example 1 — CBSE Level
Q: Calculate the magnetic moment of Fe²⁺ ion.
Fe: [Ar] 3d⁶ 4s²
Fe²⁺: Remove both 4s electrons → [Ar] 3d⁶
Draw the 3d box diagram: ↑↓ | ↑ | ↑ | ↑ | ↑
Unpaired electrons: 4
Example 2 — CBSE/JEE Main Level
Q: Why does Cr show +6 as its maximum oxidation state?
Cr is in Group 6. Maximum oxidation state = group number = +6.
Actual check: Cr = [Ar] 3d⁵ 4s¹, total valence electrons = 6. In CrO₄²⁻ or Cr₂O₇²⁻, all 6 electrons are used for bonding with oxygen → +6 oxidation state. ✓
Example 3 — JEE Main Level
Q: Among Cr²⁺, Mn²⁺, Fe²⁺, and Cu²⁺, which has the highest magnetic moment?
Work out d configurations for each M²⁺ ion (remove 2 electrons from 4s first, then 3d if needed):
| Ion | 3d config | Unpaired e⁻ | μ (BM) |
|---|---|---|---|
| Cr²⁺ | 3d⁴ | 4 | 4.90 |
| Mn²⁺ | 3d⁵ | 5 | 5.92 |
| Fe²⁺ | 3d⁶ | 4 | 4.90 |
| Cu²⁺ | 3d⁹ | 1 | 1.73 |
Mn²⁺ has the highest magnetic moment — half-filled 3d⁵ means all electrons are unpaired (Hund’s rule maximises unpaired count at exactly half-filled).
Example 4 — JEE Advanced Level
Q: KMnO₄ acts as an oxidising agent in both acidic and alkaline medium, but the product is different. Explain.
In acidic medium (H₂SO₄): MnO₄⁻ → Mn²⁺ (colourless) Mn goes from +7 to +2 — gain of 5 electrons.
In alkaline medium: MnO₄⁻ → MnO₄²⁻ (manganate, green) or MnO₂ (brown precipitate)
In neutral/faintly alkaline: Mn goes from +7 to +4 (MnO₂) — gain of 3 electrons.
The product depends on the medium because H⁺ ions stabilise the fully reduced Mn²⁺ through hydration. Without H⁺, the reduction stops at a higher oxidation state.
The n-factor of KMnO₄ changes with medium: 5 in acidic, 3 in neutral, 1 in strongly alkaline. This directly affects calculations in redox titrations — a very common JEE Main numerical.
Exam-Specific Tips
For CBSE Class 12 Boards
The marking scheme rewards specific terminology. Use the exact phrases:
- “d-d transition” (not just “electron jump”)
- “lanthanoid contraction” (spelled correctly — common spelling error)
- “spin-only magnetic moment” when using the formula
High-probability 2-mark questions: Why are transition metals good catalysts? Why do transition metal compounds show colour? Why is Sc³⁺ colourless?
High-probability 3-mark questions: Explain lanthanoid contraction and its consequences. Compare properties of first-row transition metals with explanation.
For JEE Main
Weightage: 1-2 questions per session, mostly from these areas:
- Magnetic moment calculation (appears almost every session)
- KMnO₄ reactions and n-factor
- Oxidation state identification
- Colour of specific ions
JEE Main 2024 Shift 1 asked the magnetic moment of a specific ion — just plug into the formula after correctly finding unpaired electrons.
For JEE, memorise the colours of key ions: MnO₄⁻ (purple), Cr₂O₇²⁻ (orange), CrO₄²⁻ (yellow), Cu²⁺ in water (blue), Fe³⁺ (yellow-brown), Fe²⁺ (pale green). These appear in assertion-reasoning and matching questions.
Common Mistakes to Avoid
Mistake 1: Removing 3d electrons before 4s when forming ions.
The filling order (Aufbau) goes 4s before 3d, but when removing electrons to form cations, we always remove 4s first because 4s becomes higher energy once electrons are in 3d. Fe²⁺ = [Ar] 3d⁶ (not [Ar] 3d⁴ 4s²).
Mistake 2: Calling Zn, Cd, Hg “transition metals.”
They’re d block elements but not transition metals. Their stable ions (Zn²⁺, Cd²⁺) have completely filled 3d¹⁰. This distinction earns or loses a mark in CBSE value-based questions.
Mistake 3: Forgetting the Cr and Cu anomalies under exam pressure.
Draw a small box: Cr is group 6, so “expected” 3d⁴ 4s². It “actually” goes to 3d⁵ 4s¹ because half-filled is more stable. Cu: expected 3d⁹ 4s², actual 3d¹⁰ 4s¹ for fully filled stability. Write this on your rough sheet at the start of the exam.
Mistake 4: Using wrong n-factor for KMnO₄ in titration problems.
If a problem doesn’t specify medium, it’s almost always acidic (n-factor = 5) in JEE. CBSE sometimes gives alkaline — read carefully. Using the wrong n-factor gives an entirely wrong answer.
Mistake 5: Confusing lanthanoid contraction consequence.
Students say “lanthanoid contraction makes 4d and 5d elements have similar properties” — partially correct but imprecise. The actual consequence is that 5d elements have nearly the same atomic radius as the 4d elements directly above them in the periodic table. This is why Zr/Hf and Nb/Ta pairs are hard to separate.
Practice Questions
Q1. Write the electronic configuration of Cr³⁺ and calculate its magnetic moment.
Cr: [Ar] 3d⁵ 4s¹
Cr³⁺: Remove 4s¹ first, then 2 from 3d → [Ar] 3d³
3d³: three unpaired electrons (by Hund’s rule, three separate orbitals each with one electron)
BM
Q2. Why does the colour of K₂Cr₂O₇ change from orange to yellow on adding NaOH?
In acidic/neutral medium, Cr₂O₇²⁻ (dichromate, orange) is stable.
On adding NaOH (alkaline conditions):
CrO₄²⁻ (chromate) is yellow. The equilibrium shifts to chromate in alkaline medium because OH⁻ removes H⁺, driving the reaction right.
This is a reversible reaction — adding acid brings back the orange colour.
Q3. Among V²⁺, Cr²⁺, Mn²⁺, and Fe²⁺, which is most stable and why?
Mn²⁺ is most stable.
Mn²⁺ has configuration [Ar] 3d⁵ — a half-filled d subshell. Half-filled subshells have extra stability due to maximum exchange energy (electrons in singly occupied orbitals of the same subshell can exchange positions, lowering energy).
V²⁺ = 3d³, Cr²⁺ = 3d⁴, Fe²⁺ = 3d⁶ — none have the half-filled advantage.
Q4. Balance the following reaction in acidic medium using the ion-electron method:
Oxidation half: Fe²⁺ → Fe³⁺ + e⁻ … (i) × 5
Reduction half: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O … (ii) × 1
Multiply (i) by 5 and add to (ii):
This is exactly the reaction used in the Mohr’s salt titration — a standard CBSE practical question.
Q5. An ion of a 3d transition metal has magnetic moment of 3.87 BM. Identify the ion (two possible answers).
, so unpaired electrons.
We need an ion with 3 unpaired electrons in 3d subshell:
- Cr³⁺: [Ar] 3d³ → 3 unpaired ✓
- Co²⁺ would be [Ar] 3d⁷ → 3 unpaired ✓ (↑↓|↑↓|↑↓|↑|↑ — wait, let’s recount: 3d⁷ = ↑↓|↑↓|↑↓|↑|↑ → actually 3 unpaired) ✓
- V²⁺: [Ar] 3d³ → 3 unpaired ✓
Most commonly expected answer in CBSE: Cr³⁺ or V²⁺.
Q6. What happens when acidified K₂Cr₂O₇ reacts with H₂S? Write the balanced equation.
Cr₂O₇²⁻ is a strong oxidising agent. It oxidises H₂S to sulphur (S) and itself gets reduced to Cr³⁺ (green).
The orange solution turns green (Cr³⁺). The yellow precipitate of sulphur forms. Colour change is an easy observation-type CBSE question.
Q7. Why do actinoids show a larger number of oxidation states than lanthanoids?
In lanthanoids, 4f, 5d, and 6s subshells have significantly different energies. Only 6s and occasionally 5d electrons participate in bonding → limited oxidation states (mostly +3).
In actinoids, 5f, 6d, and 7s subshells are much closer in energy (smaller energy gap). All three subshells can lose electrons → wider range of oxidation states.
Also, the 5f orbitals in actinoids extend further from the nucleus than 4f do in lanthanoids, so they participate more in bonding.
Q8. Why is the second ionisation enthalpy of Cr much higher than expected?
Cr has configuration [Ar] 3d⁵ 4s¹.
First ionisation removes the 4s¹ electron → Cr⁺ = [Ar] 3d⁵ (half-filled, extra stable)
Second ionisation must remove an electron from the very stable half-filled 3d⁵ → requires much more energy than expected.
Compare: if Cr had the “normal” 3d⁴ 4s² config, the second ionisation would just remove one more 4s electron — easy. But because of the anomalous config, the second IE is abnormally high.
Frequently Asked Questions
Why do transition metals form coloured compounds while s-block metals don’t?
s-block metals (Na, Ca, Mg) have completely filled or empty subshells — no partially filled d orbitals. Without incomplete d subshells, there’s no d-d transition possible, so no selective wavelength absorption, so no colour. Transition metals have partially filled d orbitals that allow electrons to jump between d levels by absorbing specific visible light frequencies.
What is the difference between a d-block element and a transition metal?
All transition metals are d-block elements, but not all d-block elements are transition metals. The definition of a transition metal requires at least one stable ion with an incompletely filled d subshell. Zn, Cd, and Hg are d-block (last electron enters d) but their common ions (Zn²⁺, etc.) have d¹⁰ configuration — complete, so they’re excluded from “transition metals.”
Why does lanthanoid contraction affect the 5d series so strongly?
The 14 lanthanoids are inserted between La (5d¹) and Hf (5d²) in the periodic table. Over these 14 elements, the nuclear charge increases by 14 but the 4f electrons shield poorly. By the time we reach the 5d elements (Hf onwards), the effective nuclear charge is much higher than expected — pulling the outer electrons in. This offsets the expected increase in atomic radius from 4d to 5d, making 4d and 5d pairs nearly the same size.
Can the same element show both paramagnetism and diamagnetism depending on its oxidation state?
Yes. Fe is paramagnetic (it’s a metal with unpaired d electrons). But Fe with specific ligands in certain complex configurations can have all electrons paired → diamagnetic. More simply: Zn metal is paramagnetic (it has unpaired 4s electrons in the metallic band), but Zn²⁺ is diamagnetic (d¹⁰, all paired). This kind of question tests whether you’ve understood which species you’re analysing.
Why is MnO₄⁻ purple but CrO₄²⁻ is yellow — shouldn’t they both show d-d transitions?
This is a subtle and important point. The colours of MnO₄⁻ and CrO₄²⁻ are actually not from d-d transitions — they come from charge transfer transitions (ligand-to-metal electron transfer). Mn in MnO₄⁻ is actually d⁰ (+7 state), so d-d transition is impossible. The intense purple comes from oxygen donating electrons to the Mn — charge transfer absorbs in the green/yellow region, making the solution appear purple. This is a JEE Advanced level distinction.
Why does copper form Cu⁺ and Cu²⁺ but not Cu³⁺ under normal conditions?
Cu⁺ has d¹⁰ config (fully filled, stable). Cu²⁺ has d⁹ (incomplete but manageable — many complexes stabilise it). Cu³⁺ would be d⁸, not particularly stable, and the third ionisation energy of Cu is very high (removing from a d¹⁰ core requires more energy than gained in bond formation under normal conditions). Cu³⁺ exists only in a few specialised complexes.
Which transition metals are most important for JEE Main?
Focus on: Fe (reactions with acids, FeCl₃ colour, Fe₃O₄ formula), Cr (K₂Cr₂O₇ reactions, chromate-dichromate equilibrium), Mn (KMnO₄ in three media, MnO₂ as catalyst), Cu (Cu⁺ vs Cu²⁺, Cu₂O vs CuO), and Zn (why it’s not a transition metal despite being in the d-block). These five cover 80% of exam questions from this chapter.