1. Chapter at a Glance
The d-block (groups 3–12) contains the transition elements, whose atoms or common ions have partly filled d orbitals. The f-block contains the lanthanoids (4f) and actinoids (5f), placed in two separate rows. This chapter explains their electronic configurations and characteristic properties — variable oxidation states, coloured ions, magnetism, catalytic activity and complex formation — along with two important compounds, KMnO₄ and K₂Cr₂O₇, and the lanthanoid contraction. NEET asks one or two questions every year.
- Configurations of Cr, Cu and transition-metal ions.
- Magnetic moment μ = √(n(n + 2)) BM from unpaired electrons.
- Why transition-metal ions are coloured; which ions are colourless.
- Oxidation states and E° trends (Mn³⁺, Co³⁺, Cr²⁺).
- Preparation and reactions of KMnO₄ and K₂Cr₂O₇; lanthanoid contraction.
2. Position and Electronic Configuration
- General configuration: (n−1)d¹⁻¹⁰ns¹⁻². There are four transition series: 3d (Sc–Zn), 4d (Y–Cd), 5d (La, Hf–Hg) and 6d.
- Exceptions due to the stability of half-filled and full d subshells: Cr = [Ar]3d⁵4s¹, Cu = [Ar]3d¹⁰4s¹.
- When transition metals form ions, the 4s electrons are lost first: Fe²⁺ = [Ar]3d⁶, Fe³⁺ = [Ar]3d⁵.
- Zn, Cd and Hg (d¹⁰ in both the atom and common ions) are not regarded as true transition elements. Sc is a transition element because its atom has 3d¹, although Sc³⁺ is d⁰.
3. General Properties of Transition Elements
| Property | Explanation | NEET Note |
|---|---|---|
| Metallic, hard, high melting points | Strong metallic bonding involving d electrons | Maximum around the middle (Cr, Mo, W); Zn, Cd, Hg low |
| Atomic radii | Decrease slowly across the series, then level off | 4d and 5d radii nearly equal (lanthanoid contraction) |
| Ionisation enthalpy | Increases slowly across the series | Irregularities from d-subshell stability |
| Variable oxidation states | ns and (n−1)d energies are close | Mn shows +2 to +7 (maximum in 3d series) |
| Coloured ions | d–d transitions absorb visible light | d⁰ and d¹⁰ ions are colourless |
| Paramagnetism | Unpaired d electrons | μ = √(n(n + 2)) BM |
| Catalytic activity | Variable oxidation states and surface adsorption | Fe (Haber), V₂O₅ (contact), Ni (hydrogenation) |
| Complex formation | Small, highly charged ions with vacant d orbitals | [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺ |
| Interstitial compounds | Small atoms (H, C, N) trapped in metal lattice | Hard, high melting (TiC, Fe₃H) |
| Alloys | Similar atomic sizes | Brass (Cu–Zn), bronze (Cu–Sn), steel |
4. Colour and Magnetism
- In a complex or hydrated ion, the d orbitals split into two energy levels. An electron absorbs visible light to jump between them, and we see the complementary colour. For example, [Ti(H₂O)₆]³⁺ (d¹) absorbs green-yellow light and looks purple.
- Colourless ions: Sc³⁺, Ti⁴⁺ (d⁰), Cu⁺, Zn²⁺ (d¹⁰).
- Coloured ions: Cu²⁺ (blue), Ni²⁺ (green), Fe³⁺ (yellow-brown), Co²⁺ (pink), Mn²⁺ (very pale pink), Cr³⁺ (green/violet).
| Ion | Configuration | Unpaired electrons | μ (BM) |
|---|---|---|---|
| Sc³⁺, Ti⁴⁺ | 3d⁰ | 0 | 0 |
| Ti³⁺, V⁴⁺ | 3d¹ | 1 | 1.73 |
| V³⁺ | 3d² | 2 | 2.83 |
| Cr³⁺ | 3d³ | 3 | 3.87 |
| Cr²⁺, Mn³⁺, Fe⁴⁺ | 3d⁴ | 4 | 4.90 |
| Mn²⁺, Fe³⁺ | 3d⁵ | 5 | 5.92 |
| Fe²⁺, Co³⁺ | 3d⁶ | 4 | 4.90 |
| Ni²⁺ | 3d⁸ | 2 | 2.83 |
| Cu²⁺ | 3d⁹ | 1 | 1.73 |
| Cu⁺, Zn²⁺ | 3d¹⁰ | 0 | 0 |
5. Stability of Oxidation States
- Mn²⁺ (d⁵) and Fe³⁺ (d⁵) are especially stable because they are half-filled. So Mn³⁺ is a strong oxidising agent (it readily gains an electron to become d⁵), and E°(Mn³⁺/Mn²⁺) is high.
- Cr²⁺ is a strong reducing agent: it loses an electron to become Cr³⁺ (d³, with a stable half-filled t₂g set).
- Co³⁺ is a strong oxidising agent in water.
- E°(M²⁺/M) for Cu is positive (+0.34 V), so copper does not liberate H₂ from dilute acids.
- The highest oxidation states are shown in oxides and fluorides (MnO₄⁻, CrO₄²⁻, VF₅), because O and F are highly electronegative and O can form multiple bonds.
6. Potassium Dichromate and Potassium Permanganate
| Feature | K₂Cr₂O₇ | KMnO₄ |
|---|---|---|
| Ore / starting material | Chromite, FeCr₂O₄ | Pyrolusite, MnO₂ |
| Key steps | Fuse with Na₂CO₃ in air → Na₂CrO₄ → acidify to Na₂Cr₂O₇ → add KCl | Fuse MnO₂ with KOH + oxidant → green K₂MnO₄ → oxidise/disproportionate to KMnO₄ |
| Colour | Orange | Deep purple |
| Structure of anion | Two tetrahedra sharing one O (Cr–O–Cr) | Tetrahedral MnO₄⁻ |
| Magnetic nature | Diamagnetic (Cr⁶⁺, d⁰) | Diamagnetic (Mn⁷⁺, d⁰); colour is from charge transfer |
| Acidic half-reaction | Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O | MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O |
| NEET Note | Chromate (yellow) ⇌ dichromate (orange) depends on pH | Self-indicator in titrations |
Acidified KMnO₄ oxidises Fe²⁺ to Fe³⁺, oxalate to CO₂, I⁻ to I₂, and H₂S to S. In neutral or weakly alkaline solution it oxidises I⁻ to IO₃⁻ and is itself reduced to MnO₂.
7. The f-Block: Lanthanoids and Actinoids
- Lanthanoids (Ce to Lu): general configuration [Xe]4f¹⁻¹⁴5d⁰⁻¹6s². The most common oxidation state is +3. Ce⁴⁺ is a good oxidising agent (it becomes Ce³⁺); Eu²⁺ and Yb²⁺ are reducing agents.
- Lanthanoid contraction: a steady decrease in atomic and ionic radii from La to Lu, because the 4f electrons shield the nuclear charge poorly.
- Consequences: Zr and Hf (and Nb and Ta) have almost identical radii and are very hard to separate; basicity of Ln(OH)₃ decreases from La(OH)₃ to Lu(OH)₃; 5d elements have high densities.
- Actinoids (Th to Lr): all are radioactive; they show a wider range of oxidation states (up to +7 for Np and Pu) because 5f, 6d and 7s energies are close. The actinoid contraction is larger than the lanthanoid contraction (5f shields even more poorly).
- Uses: misch metal (lanthanoid alloy with Fe) in lighter flints; lanthanoid oxides in TV screens and phosphors.
8. Formula and Fact Master Table
| Quantity / fact | Formula or value |
|---|---|
| General configuration (d-block) | (n−1)d¹⁻¹⁰ns¹⁻² |
| Spin-only magnetic moment | μ = √(n(n + 2)) BM |
| Ions formed | Lose ns electrons before (n−1)d |
| Dichromate ⇌ chromate | Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O |
| Permanganate in acid | MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O |
| Dichromate in acid | Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O |
| Lanthanoid configuration | [Xe]4f¹⁻¹⁴5d⁰⁻¹6s² |
| Common lanthanoid state | +3 |
9. Master Summary Table
| Topic | Key fact | NEET Note |
|---|---|---|
| Exceptions | Cr 3d⁵4s¹, Cu 3d¹⁰4s¹ | Stability of half/full d |
| Highest oxidation state (3d) | Mn +7 | In MnO₄⁻ |
| Colour | d–d transitions | d⁰, d¹⁰ colourless |
| Max magnetic moment (3d ions) | d⁵: 5.92 BM | Mn²⁺, Fe³⁺ |
| Strong oxidants | Mn³⁺, Co³⁺ | Strong reductant: Cr²⁺ |
| KMnO₄ | Purple, n = 5 in acid | Self-indicator |
| Lanthanoid contraction | Poor 4f shielding | Zr ≈ Hf in size |
| Actinoids | Radioactive, more states | Up to +7 |
10. Mind Map
- (n−1)d ns
- Cr, Cu exceptions
- Ions lose ns first
- Variable states
- Colour, magnetism
- Catalysis, complexes
- K₂Cr₂O₇
- KMnO₄
- Lanthanoids +3
- Contraction
- Actinoids
11. Tips and Tricks
- Unpaired electrons for d¹ to d⁹ (weak field): 1, 2, 3, 4, 5, 4, 3, 2, 1.
- Magnetic moments to remember: 1.73, 2.83, 3.87, 4.90, 5.92 BM for 1–5 unpaired electrons.
- If an ion's colour question appears, first check for d⁰ or d¹⁰ — those are colourless.
- Colour change in dichromate titration: orange → green (Cr³⁺).
- Removing 3d electrons before 4s when writing ion configurations.
- Saying KMnO₄ is coloured because of unpaired electrons (it is d⁰; the colour is from charge transfer).
- Calling Zn a typical transition metal.
- Thinking lanthanoid contraction makes 5d elements much smaller than 4d elements (they are about the same size).
12. Important One-liners
- Manganese shows the largest number of oxidation states in the 3d series.
- Zn²⁺ salts are white because Zn²⁺ is d¹⁰.
- Mercury is the only metal that is liquid at room temperature.
- Iron is the catalyst in the Haber process.
- Chromate is yellow and dichromate is orange.
- Zr and Hf have almost the same atomic radius because of the lanthanoid contraction.
- All actinoids are radioactive.
- Misch metal is used in lighter flints.
13. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
Which ion is colourless in aqueous solution?
- Cu²⁺
- Ni²⁺
- Zn²⁺
- Fe³⁺
Show solution
Zn²⁺ is 3d¹⁰, so no d–d transition is possible and it is colourless. Cu²⁺ (d⁹, blue), Ni²⁺ (d⁸, green) and Fe³⁺ (d⁵, yellow-brown) are all coloured.
Answer: (C)
The electronic configuration of Cr (Z = 24) is:
- [Ar]3d⁴4s²
- [Ar]3d⁵4s¹
- [Ar]3d⁶
- [Ar]3d³4s²4p¹
Show solution
A half-filled 3d⁵ subshell is extra stable, so one 4s electron moves into 3d. (A) is what the Aufbau order would predict, but it is not the observed configuration.
Answer: (B)
Which ion has the highest spin-only magnetic moment?
- Cr³⁺
- Fe²⁺
- Mn²⁺
- Cu²⁺
Show solution
Mn²⁺ (d⁵) has 5 unpaired electrons, giving μ ≈ 5.92 BM. Fe²⁺ (d⁶) has 4, Cr³⁺ (d³) has 3 and Cu²⁺ (d⁹) has 1.
Answer: (C)
The oxidation state of Mn in K₂MnO₄ (manganate) is:
- +7
- +6
- +4
- +2
Show solution
2(+1) + x + 4(−2) = 0, so x = +6. Green manganate(VI) is the intermediate in making KMnO₄, where Mn is +7.
Answer: (B)
The lanthanoid contraction is due to:
- Perfect shielding by 4f electrons
- Imperfect shielding by 4f electrons
- Increase in the number of shells
- Decrease in nuclear charge
Show solution
4f electrons shield the rising nuclear charge poorly. The outer electrons are therefore pulled in more strongly across the series, and the radius shrinks. The number of shells stays the same, and the nuclear charge increases.
Answer: (B)
When acidified K₂Cr₂O₇ oxidises Fe²⁺, the colour changes from:
- Orange to green
- Yellow to orange
- Purple to colourless
- Green to orange
Show solution
Cr₂O₇²⁻ (orange) is reduced to Cr³⁺ (green). Purple to colourless describes KMnO₄. Yellow to orange describes chromate turning into dichromate when acid is added.
Answer: (A)
Mn³⁺ is a strong oxidising agent while Cr²⁺ is a strong reducing agent. The reason is:
- Both change to a d³ configuration
- Mn³⁺ becomes d⁵ and Cr²⁺ becomes d³, both more stable
- Both change to d⁵
- Mn is more electronegative
Show solution
Mn³⁺ (d⁴) gains an electron to become Mn²⁺, a stable half-filled d⁵ ion. Cr²⁺ (d⁴) loses an electron to become Cr³⁺ (d³, a stable half-filled t₂g set). Both reach a more stable configuration, but by opposite routes.
Answer: (B)
The pair of elements that is most difficult to separate is:
- Zr and Hf
- Fe and Co
- Cu and Ag
- Ti and Zr
Show solution
Because of the lanthanoid contraction, Hf (5d) has almost the same radius as Zr (4d). Their chemistry is nearly identical, so they are very hard to separate. The other pairs differ clearly in size or in chemistry.
Answer: (A)
14. Quick Revision Summary
- (n−1)d¹⁻¹⁰ns¹⁻²; Cr 3d⁵4s¹, Cu 3d¹⁰4s¹; ions lose ns first.
- Coloured if d¹–d⁹; μ = √(n(n + 2)); d⁵ has the maximum 5.92 BM.
- Mn³⁺, Co³⁺ oxidising; Cr²⁺ reducing; Mn up to +7.
- K₂Cr₂O₇ (orange → green, n = 6); KMnO₄ (purple, n = 5 in acid).
- Lanthanoids +3; contraction from poor 4f shielding; Zr ≈ Hf; actinoids radioactive.