The d and f Block Elements — Complete NCERT Notes, Solved Examples & MCQs
A friendly, exam-focused guide to transition elements, lanthanides, and actinides — with easy explanations, diagrams, memory tricks, solved NCERT questions, and previous year CBSE questions, made for students by Jnaanangkur – The Learning Hub.
Welcome, Dear Students !
Namaskar! If you have ever wondered why blood is red, why old coins turn green, why welding uses a "silvery" flame, or why a pink solution of potassium permanganate is used to purify wounds — you have already met the d and f block elements without knowing it.
This chapter introduces the elements sitting in the middle and bottom rows of the periodic table — the transition metals (iron, copper, zinc, silver, gold, and more) and the rare-earth lanthanides and actinides. These elements are chemically fascinating because of their variable oxidation states, colourful compounds, and role as catalysts in industry and the human body.
By the end of this guide, you will be able to explain why transition metals behave the way they do — not just memorise facts — which is exactly what CBSE and State Board exams (and competitive exams like NEET/JEE/CTET) expect from you.
Learning Objectives & Real-Life Importance
By the end of this chapter, you will be able to:
- Locate d-block and f-block elements in the periodic table and write their general electronic configurations.
- Explain why transition metals show variable oxidation states, coloured ions, catalytic behaviour, and magnetic properties.
- Describe the preparation, structure, and reactions of two important compounds: potassium dichromate (K₂Cr₂O₇) and potassium permanganate (KMnO₄).
- Understand lanthanide contraction and its far-reaching consequences.
- Compare lanthanides and actinides on the basis of oxidation states, radioactivity, and chemistry.
Why does this chapter matter in real life?
- 🩸 Haemoglobin in your blood contains iron (Fe), a transition metal, which carries oxygen.
- 💉 KMnO₄ is used as a disinfectant and water purifier in villages and hospitals.
- 🏭 Catalysts like V₂O₅, Ni, and Pt (transition metals) run the Contact Process, Haber Process, and vehicle catalytic converters.
- 💡 Lanthanides are used in LED screens, camera lenses, MRI contrast agents, and rechargeable batteries.
- ☢️ Actinides like uranium and plutonium power nuclear reactors.
- 🪙 Alloys such as stainless steel (Fe-Cr-Ni) and brass (Cu-Zn) are everywhere around you.
Quick Navigation
Position in the Periodic Table
The d and f block elements occupy the "inner" part of the periodic table, between the s-block and p-block elements.
| Series | Elements | Atomic Numbers |
|---|---|---|
| 3d (First transition series) | Sc to Zn | 21–30 |
| 4d (Second transition series) | Y to Cd | 39–48 |
| 5d (Third transition series) | La, Hf to Hg | 57, 72–80 |
| 6d (Fourth transition series) | Ac, Rf to Cn | 89, 104–112 |
| 4f (Lanthanides) | Ce to Lu | 58–71 |
| 5f (Actinides) | Th to Lr | 90–103 |
Electronic Configuration
This is why Zn, Cd, and Hg (all with configuration (n−1)d10 ns²) are technically part of the d-block but are not called transition elements — their d-orbitals are completely full in both the atom and their common ions (Zn²⁺, Cd²⁺, Hg²⁺).
Anomalous configurations to remember: Cr = [Ar] 3d⁵ 4s¹ and Cu = [Ar] 3d¹⁰ 4s¹ (instead of the expected 3d⁴4s² and 3d⁹4s²). This happens because a half-filled (d⁵) or fully-filled (d¹⁰) d-subshell gives extra stability.
General Properties of Transition Elements
6.1 Metallic Character & Physical Properties
All transition elements are metals — hard, high density, high melting and boiling points — because of strong metallic bonding involving both ns and (n−1)d electrons.
6.2 Atomic and Ionic Radii
Radii decrease slowly from left to right across a series (unlike the sharp decrease in s and p blocks) because the added d-electrons partly shield the increasing nuclear charge. Radii tend to rise slightly again near the end of the series due to electron-electron repulsion in the filled orbitals.
6.3 Ionisation Enthalpy
Ionisation enthalpy generally increases across a period (due to increasing nuclear charge) but with small irregularities, since removing electrons from a partially filled, symmetric d-configuration (like d⁵ or d¹⁰) requires extra energy.
6.4 Variable Oxidation States
Example: Mn shows oxidation states from +2 to +7 (+2, +3, +4, +5, +6, +7). The +2 state is common for most 3d metals as it arises from loss of the two 4s electrons.
6.5 Colour of Compounds/Ions
Most transition metal ions are coloured in the solid state and in aqueous solution because of d-d transitions — an electron jumps from a lower-energy d-orbital to a higher-energy d-orbital by absorbing light from the visible region, and the complementary colour is transmitted/reflected.
| Ion | Configuration | Colour |
|---|---|---|
| Sc³⁺ | 3d⁰ | Colourless |
| Ti³⁺ | 3d¹ | Purple |
| V³⁺ | 3d² | Green |
| Mn²⁺ | 3d⁵ | Light pink |
| Fe³⁺ | 3d⁵ | Yellow-brown |
| Cu²⁺ | 3d⁹ | Blue |
| Zn²⁺ | 3d¹⁰ | Colourless |
Note: Ions with d⁰ or d¹⁰ configuration (like Sc³⁺, Zn²⁺, Ti⁴⁺) are colourless because no d-d transition is possible.
6.6 Magnetic Properties
Most transition metal compounds are paramagnetic due to unpaired d-electrons. The magnetic moment can be calculated using the "spin-only" formula.
6.7 Catalytic Properties
Transition metals and their compounds act as good catalysts because of their variable oxidation states (they can form intermediate compounds and provide a new low-energy reaction pathway) and their ability to adsorb reactants on their surface (using vacant d-orbitals).
Examples: Fe (Haber process for NH₃), V₂O₅ (Contact process for H₂SO₄), Ni (hydrogenation of oils), Pt/Pd/Rh (catalytic converters).
6.8 Formation of Coloured Complex Compounds
Transition metal ions have small size, high charge, and vacant d-orbitals, which lets them accept lone pairs of electrons from ligands (like NH₃, CN⁻, H₂O) to form stable complex/coordination compounds, e.g., [Cu(NH₃)₄]²⁺, [Fe(CN)₆]⁴⁻.
6.9 Alloy Formation
Since transition metals have similar atomic sizes, atoms of one metal can easily replace atoms of another in a crystal lattice, forming alloys — e.g., stainless steel (Fe + Cr + Ni), bronze (Cu + Sn), brass (Cu + Zn).
6.10 Interstitial Compounds
Important Compounds: K₂Cr₂O₇ and KMnO₄
7.1 Potassium Dichromate (K₂Cr₂O₇)
Preparation: Prepared from chromite ore (FeCr₂O₄) in three steps:
- Chromite ore is fused with sodium/potassium carbonate in the presence of air to give sodium chromate (Na₂CrO₄).
- The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to convert it into orange sodium dichromate (Na₂Cr₂O₇).
- Sodium dichromate is treated with potassium chloride; the less soluble potassium dichromate (K₂Cr₂O₇) crystallises out.
Chromate–Dichromate equilibrium: In alkaline medium, dichromate converts to chromate (yellow); in acidic medium, chromate converts back to dichromate (orange).
Oxidising property: K₂Cr₂O₇ is a strong oxidising agent in acidic medium.
Uses: Oxidising agent in organic chemistry, in leather tanning, as a primary standard in volumetric analysis, and (historically) in the "breathalyser" test for alcohol.
7.2 Potassium Permanganate (KMnO₄)
Preparation: Prepared from pyrolusite ore (MnO₂):
- MnO₂ is fused with KOH in the presence of an oxidising agent (like KNO₃ or air) to form potassium manganate (K₂MnO₄), which is green.
- K₂MnO₄ is oxidised (chemically using Cl₂/O₃, or electrolytically) in alkaline/neutral medium to give purple potassium permanganate (KMnO₄).
Oxidising property: KMnO₄ acts as a powerful oxidising agent, and its behaviour depends on the medium.
| Medium | Half Reaction | Product |
|---|---|---|
| Acidic | MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | Mn²⁺ (almost colourless) |
| Neutral / faintly alkaline | MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ | MnO₂ (brown precipitate) |
| Strongly alkaline | MnO₄⁻ + e⁻ → MnO₄²⁻ | MnO₄²⁻ (green manganate) |
Uses: As an oxidising agent in volumetric analysis (redox titrations), for water purification, as a mild disinfectant/antiseptic, and for bleaching wool, cotton, and silk.
| Property | K₂Cr₂O₇ | KMnO₄ |
|---|---|---|
| Ore used | Chromite (FeCr₂O₄) | Pyrolusite (MnO₂) |
| Colour | Orange | Purple/Deep pink |
| Oxidation state of metal | Cr = +6 | Mn = +7 |
| Geometry of central atom | Tetrahedral (each Cr) | Tetrahedral |
| Common oxidising medium | Acidic | Acidic, neutral, and alkaline |
Lanthanides (4f Block)
Oxidation states: +3 is the most common and most stable oxidation state. Some lanthanides also show +2 or +4 when it gives them a stable f⁰, f⁷, or f¹⁴ configuration (e.g., Ce⁴⁺ has f⁰, Eu²⁺ has f⁷).
8.1 Lanthanide Contraction
Cause: As we move from La to Lu, each added electron goes into the inner 4f orbital, which has very poor shielding effect (poorer than d or p orbitals) on the outer electrons. So the effective nuclear charge felt by outer electrons increases steadily, pulling them in and shrinking the radius.
Consequences of lanthanide contraction:
- Radii of 4d and 5d transition series elements of the same group become very close (e.g., Zr and Hf), making them difficult to separate.
- Basic character of lanthanide hydroxides decreases steadily from La(OH)₃ to Lu(OH)₃.
- The lanthanides resemble each other closely in properties, making their separation from each other very difficult.
8.2 Other Properties
- Lanthanides are silvery-white soft metals that tarnish in air.
- They form coloured ions mainly due to f-f transitions (though generally less intense than d-d transitions).
- Most lanthanide ions are paramagnetic (except La³⁺ and Lu³⁺, which have f⁰ and f¹⁴ configurations respectively, hence diamagnetic).
Uses: Mischmetal (lanthanide + iron alloy) is used in cigarette lighter flints and jet-engine alloys; lanthanide oxides are used in television and camera lens glass, LED phosphors, and MRI contrast agents.
Actinides (5f Block)
All actinides are radioactive. Elements after uranium (Z=92) are called transuranium elements and are mostly synthetic (man-made).
Oxidation states: +3 is common (like lanthanides), but actinides show a much wider range of oxidation states (+3 to +7) because 5f, 6d, and 7s orbitals have comparable energies, so more electrons can take part in bonding.
9.1 Actinide Contraction
Similar to lanthanide contraction, there is a steady decrease in the size of atoms/ions across the actinide series, called actinide contraction. It is greater than lanthanide contraction because the 5f orbitals shield the nuclear charge even more poorly than 4f orbitals.
9.2 Lanthanides vs Actinides
| Property | Lanthanides | Actinides |
|---|---|---|
| Orbital being filled | 4f | 5f |
| Radioactivity | Mostly non-radioactive (except Pm) | All radioactive |
| Common oxidation state | +3 (dominant) | +3, but wide range +3 to +7 |
| Complex-forming tendency | Lesser tendency to form complexes | Greater tendency to form complexes |
| Availability | Naturally abundant (except Pm) | Many are man-made/synthetic |
| Energy of 4f/5f, 5d/6d, 6s/7s | Clear energy difference | Comparable energies (more complex chemistry) |
Uses: Uranium and plutonium are used as nuclear fuel in reactors and in nuclear weapons; thorium is used in gas mantles and as an alternative nuclear fuel.
Mind Map: The Whole Chapter at a Glance
Solved NCERT Examples
Example 1
Q: Calculate the number of unpaired electrons in Mn²⁺ and hence find its magnetic moment.
Show Step-by-Step Solution
Step 1: Write the configuration of Mn (Z=25): [Ar] 3d⁵ 4s².
Step 2: To form Mn²⁺, remove 2 electrons from the 4s orbital first (not from 3d): Mn²⁺ = [Ar] 3d⁵.
Step 3: A d⁵ configuration in separate orbitals (Hund's rule) gives 5 unpaired electrons (n = 5).
Step 4: Apply μ = √n(n+2) BM = √5(5+2) = √35 ≈ 5.92 BM.
Example 2
Q: Why is Zn not considered a transition element although it belongs to the d-block?
Show Solution
Zn (Z=30) has the configuration [Ar] 3d¹⁰ 4s². Its d-orbital is completely filled (d¹⁰), and in its only common oxidation state, Zn²⁺ ([Ar] 3d¹⁰), the d-orbital remains completely filled. Since a transition element must have a partially filled d-orbital (d¹–d⁹) in the atom or a common ion, Zn does not qualify as a transition element even though it is a d-block element.
Example 3
Q: Write the ionic equation for the reaction of acidified potassium dichromate with ferrous ions (Fe²⁺).
Show Solution
Step 1: Dichromate is reduced: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Step 2: Fe²⁺ is oxidised: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 6 to balance electrons)
Step 3: Add both half-reactions:
Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
Example 4
Q: Explain why the atomic radii of the third (5d) transition series are very close to those of the second (4d) transition series, unlike the usual increase in size down a group.
Show Solution
This happens because of lanthanide contraction. Before the 5d series begins, the 4f orbitals (lanthanides) are filled. The poor shielding of 4f electrons causes a steady contraction in size across the lanthanides. By the time we reach the 5d series, this contraction almost exactly cancels out the expected size increase from adding an extra shell, so 4d and 5d elements of the same group end up with nearly identical atomic radii (e.g., Zr ≈ Hf).
Important NCERT Questions with Answers
Q1. Why are Mn²⁺ compounds more stable than Fe²⁺ compounds towards oxidation to +3 state?
Mn²⁺ has configuration 3d⁵, which is a stable, exactly half-filled d-subshell. Removing another electron to form Mn³⁺ (3d⁴) disturbs this extra stability, so oxidation is difficult and Mn²⁺ is more stable. Fe²⁺ (3d⁶) does not have this extra stability, so it is more easily oxidised to Fe³⁺ (3d⁵), which itself gains the stable half-filled configuration.
Q2. Explain why transition metals form a large number of complex compounds.
Transition metal ions are comparatively small in size, carry a high positive charge, and have several vacant d-orbitals of suitable energy. These features let them act as strong Lewis acids, accepting electron pairs (lone pairs) donated by ligands such as NH₃, CN⁻, Cl⁻, and H₂O, resulting in the formation of stable complex/coordination compounds.
Q3. What is meant by "disproportionation"? Give an example from Mn chemistry.
Disproportionation is a reaction in which an element in one oxidation state is simultaneously oxidised and reduced, forming two different oxidation states of the same element as products.
Example: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
Here, manganese in the +6 state (MnO₄²⁻) disproportionates into +7 (MnO₄⁻) and +4 (MnO₂) states in acidic medium.
Q4. Why do transition elements show variable oxidation states differing generally by one, unlike p-block elements?
In transition metals, both the (n−1)d and ns electrons have very close energies, so electrons can be removed one at a time from the d-orbital without a big jump in energy. This allows a whole range of oxidation states differing by just 1 unit (e.g., Fe²⁺, Fe³⁺). In contrast, p-block elements usually lose or share electrons in pairs due to the inert pair effect, so their oxidation states typically differ by 2.
Q5. Give reasons: (a) Transition metals and many of their compounds act as catalysts. (b) Interstitial compounds are hard.
(a) Transition metals show variable oxidation states, which allows them to form unstable intermediate compounds with reactants, providing a new pathway of lower activation energy. Their large surface area and vacant d-orbitals also let them adsorb reacting molecules, weakening bonds within the reactants.
(b) In interstitial compounds, small atoms (H, C, N) occupy the empty interstitial spaces within the metal lattice without disturbing the metallic bonding. This increases the rigidity of the lattice, making these compounds extremely hard.
Previous Year CBSE / State Board Questions
CBSE 2023: Assign reasons for the following: (i) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺. (ii) Cr²⁺ is a strong reducing agent.
(i) Mn³⁺ (3d⁴) gains extra stability on being reduced to Mn²⁺ (3d⁵, half-filled), so Mn³⁺ is a strong oxidising agent and the E° value for Mn³⁺/Mn²⁺ is highly positive. Cr³⁺ (3d³) is already a stable configuration, so it does not have the same strong drive to get reduced.
(ii) Cr²⁺ (3d⁴) readily loses an electron to become Cr³⁺ (3d³), which is a more stable, exactly half-filled t2g configuration. This strong tendency to get oxidised makes Cr²⁺ a good reducing agent.
CBSE 2022: Complete the following equation: MnO₄⁻ + 8H⁺ + 5e⁻ → ? What colour change is observed?
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
The deep purple/pink colour of permanganate (MnO₄⁻) fades to almost colourless (very pale pink) as it is reduced to Mn²⁺ in acidic medium.
CBSE 2020: Why do the transition elements exhibit higher enthalpies of atomisation?
Transition elements have a large number of unpaired electrons in their (n−1)d orbitals, which allows for stronger metallic bonding (through greater orbital overlap) compared to s-block metals. Stronger inter-atomic bonding means more energy is required to separate the atoms, giving transition elements higher enthalpies of atomisation.
Assam Board (SEBA)-style: State one similarity and one difference between the properties of lanthanides and actinides.
Similarity: Both series show +3 as their most common and stable oxidation state, and both undergo a steady contraction in atomic/ionic size (lanthanide contraction and actinide contraction respectively).
Difference: Lanthanides are mostly non-radioactive (except promethium), while all actinides are radioactive.
MCQs, Assertion–Reason & Practice Questions
14.1 Multiple Choice Questions
1. Which of the following ions is colourless?
- (a) Ti³⁺
- (b) Cu²⁺
- (c) Sc³⁺
- (d) Mn²⁺
Answer: (c) Sc³⁺ — it has a d⁰ configuration, so no d-d transition is possible.
2. The number of unpaired electrons in Cr³⁺ (Z=24) is:
- (a) 1
- (b) 3
- (c) 5
- (d) 0
Answer: (b) 3. Cr³⁺ = [Ar]3d³, with three unpaired electrons in separate d-orbitals.
3. Lanthanide contraction is caused by:
- (a) Poor shielding by 4f electrons
- (b) Strong shielding by 5d electrons
- (c) Increase in atomic number only
- (d) Radioactivity
Answer: (a) Poor shielding by 4f electrons.
4. Which ore is used for the preparation of potassium permanganate?
- (a) Chromite
- (b) Pyrolusite
- (c) Bauxite
- (d) Haematite
Answer: (b) Pyrolusite (MnO₂).
5. Which of the following elements is NOT a transition element?
- (a) Fe
- (b) Cu
- (c) Zn
- (d) Mn
Answer: (c) Zn — its d-orbitals are completely filled (d¹⁰) in both atom and ion.
14.2 Assertion–Reason Questions
Directions: Choose (a) if both A and R are true and R is the correct explanation of A; (b) if both are true but R is not the correct explanation; (c) if A is true but R is false; (d) if A is false but R is true.
1. Assertion (A): Cu²⁺ (aq) is coloured while Zn²⁺ (aq) is colourless.
Reason (R): Cu²⁺ has an incomplete d-subshell (d⁹), enabling d-d transitions, while Zn²⁺ has a complete d-subshell (d¹⁰).
Answer: (a) Both A and R are true, and R correctly explains A.
2. Assertion (A): Actinides show a greater range of oxidation states than lanthanides.
Reason (R): In actinides, 5f, 6d, and 7s orbitals have comparable energies.
Answer: (a) Both A and R are true, and R correctly explains A.
3. Assertion (A): K₂Cr₂O₇ turns to K₂CrO₄ on adding excess alkali.
Reason (R): Chromate is more stable in acidic medium than dichromate.
Answer: (c) A is true, R is false. Chromate (CrO₄²⁻, yellow) is actually more stable in alkaline medium, not acidic; it is dichromate (Cr₂O₇²⁻, orange) that is stable in acidic medium.
HOTS & Case-Based / Competency Questions
Case-Based Question
Read the passage and answer the questions that follow: "A student was given two coloured aqueous solutions labelled X and Y in the laboratory. Solution X was orange in colour and turned green on adding a reducing agent like FeSO₄. Solution Y was deep purple and turned almost colourless on adding excess dilute H₂SO₄ and a reducing agent."
- Q: Identify solutions X and Y.
- Q: Write the balanced ionic equation for the reaction of solution X with FeSO₄ in acidic medium.
- Q: What is the oxidation state of the metal in solution Y before and after the reaction?
Show Answers
1. X = Potassium dichromate solution (K₂Cr₂O₇), orange, reduced to green Cr³⁺. Y = Potassium permanganate solution (KMnO₄), purple, reduced to almost colourless Mn²⁺.
2. Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
3. Before: Mn is in +7 oxidation state (in MnO₄⁻). After: Mn is reduced to +2 oxidation state (Mn²⁺).
HOTS 1: Why is Sc (Z=21) not treated as a "true" transition element by some textbooks despite having a partially filled d-orbital in its atomic state?
Scandium has configuration [Ar]3d¹4s², so its atom does have a partially filled d-orbital. However, its only common oxidation state is Sc³⁺, which has a configuration of [Ar]3d⁰ — completely empty. Since a transition element must show a partially filled d-orbital in a common oxidation state (not just in the neutral atom) for its characteristic properties (colour, catalysis, etc.) to appear, Sc³⁺ compounds are colourless and behave more like a typical s/p block ion, making Sc a borderline case.
HOTS 2: Predict which will have a higher melting point: chromium or zinc, and justify using the concept of metallic bonding.
Chromium has a much higher melting point than zinc. Chromium ([Ar]3d⁵4s¹) has 6 unpaired electrons available for metallic bonding, leading to very strong covalent-character metallic bonds involving both d and s electrons. Zinc ([Ar]3d¹⁰4s²) has no unpaired d-electrons (its d-subshell is full), so only the two 4s electrons contribute to weaker metallic bonding, giving zinc a much lower melting point.
Common Mistakes & Exam-Writing Tips
Fix: Always remember Cr and Cu as the two key exceptions where a half-filled or fully-filled d-subshell gives extra stability.
Fix: Always remove ns electrons FIRST when forming a cation, even though d was filled after s. E.g., Fe²⁺ = [Ar]3d⁶, not [Ar]3d⁴4s².
Fix: Remember "Acidic = oraANGe (dichromate)"; alkaline = yellow (chromate).
Fix: Only actinides are radioactive; lanthanides are mostly non-radioactive (promethium, Pm, is the only radioactive lanthanide).
One-Page Quick Revision Sheet
📌 The d and f Block Elements — At a Glance
Frequently Asked Questions (FAQs)
Q1. What is the difference between d-block elements and transition elements?
All transition elements are d-block elements, but not all d-block elements are transition elements. D-block simply refers to position (groups 3–12). Transition elements must additionally have a partially filled d-orbital in the atom or a common ion — this excludes Zn, Cd, and Hg from being called transition elements.
Q2. Why are transition metal compounds generally coloured?
Because of d-d transitions: an electron absorbs energy from visible light and jumps from a lower to a higher d-orbital. The colour we see is the complementary colour of the light that gets absorbed.
Q3. What is lanthanide contraction, in one line?
It is the steady decrease in atomic/ionic size across the lanthanide series (La to Lu) due to poor shielding by 4f electrons.
Q4. Is this chapter important for NEET/JEE along with boards?
Yes. "d and f Block Elements" is a high-weightage inorganic chemistry chapter for both board exams and competitive exams like NEET and JEE, especially topics like KMnO₄/K₂Cr₂O₇ reactions, magnetic moment, and lanthanide contraction.
Q5. How many marks does this chapter usually carry in CBSE Class 12 board exams?
This chapter typically contributes questions worth 4–6 marks in the CBSE board exam, often through short-answer "reasoning" questions, one numerical (magnetic moment), and occasionally a case-based question.
Q6. What is the easiest way to remember the colours of common transition metal ions?
Link the colour to the number of unpaired d-electrons and common lab experience: Cu²⁺ solutions are blue (CuSO₄ used commonly in labs), Mn²⁺ is very pale pink, Fe³⁺ is yellow-brown, and d⁰/d¹⁰ ions (Sc³⁺, Ti⁴⁺, Zn²⁺) are always colourless.
Chapter Summary
The d-block elements (also called transition elements when they have partially filled d-orbitals) sit between the s and p blocks and show unique properties: variable oxidation states, coloured ions, catalytic activity, magnetism, and the ability to form alloys, complexes, and interstitial compounds — all traceable back to their electronic configuration. Two compounds, K₂Cr₂O₇ and KMnO₄, are must-know examples of transition metal oxidising agents. The f-block elements — lanthanides and actinides — sit at the bottom of the periodic table. Lanthanides mostly show a stable +3 oxidation state and undergo lanthanide contraction, which has important consequences for the periodic table. Actinides are all radioactive and show a wider range of oxidation states due to comparable orbital energies. Mastering the "why" behind each property (using electronic configuration) is the key to scoring full marks in this chapter.
Keep the Momentum Going! 🚀
You've just built a rock-solid foundation in d and f Block Elements. Practice a few more numericals on magnetic moment, revise the KMnO₄/K₂Cr₂O₇ reactions once more, and you'll be exam-ready.
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