The d- and f-Block Elements — Class 12 Chemistry Notes & Practice

Take a breath. If you have already flipped through this chapter once and thought “this is just a wall of facts I have to memorise”, you are not alone — almost everyone reacts that way the first time. But here is the secret that changes everything: the d- and f-block chapter is not a list. It is one idea with many symptoms. A single structural fact about where the d and f electrons sit inside the atom quietly produces every single property in this unit — variable oxidation states, colour, magnetism, catalysis, complex formation, and the famous lanthanoid contraction. Learn the cause once, and the symptoms stop being things you memorise and start being things you can derive in the exam hall.

This page is your complete Unit 4 companion: d and f block elements class 12 notes, the full d and f block class 12 formula and oxidation state list, lanthanoid contraction explained with consequences, and a set of d and f block elements class 12 important questions with worked model answers that show exactly where the marks sit. Unit 4 carries 7 marks in the 70-mark CBSE Class 12 Chemistry theory paper for 2026-27, and it is one of the most reliably scoring units in the whole syllabus — because the questions repeat, and because almost all of them can be answered from one root idea.

We will go slowly. No step is skipped. Every configuration, every oxidation state, every balanced equation on this page has been checked against NCERT Class 12 Chemistry Part I, and every magnetic moment has been calculated rather than copied. Ready? Let us begin.

Meet Your Tutor

The d- and f-Block Elements can look like a catalogue of colours, oxidation states and exceptions. I will help you build the chapter from electronic configuration and recurring trends first, then attach each compound and reaction to that framework. Keep a small comparison table nearby and update it as the patterns become visible.

What You’ll Learn

Your Game Plan

1. Day 1 — learn the cause. Read the root-idea section and the position/configuration sections. Write the 3d series configurations from memory until you get all ten right, including Cr and Cu.
2. Day 2 — derive the symptoms. Radii, ionisation enthalpy, oxidation states, colour, magnetism. Do not memorise these as separate facts; each time, ask “how does the poorly-shielding d electron explain this?”
3. Day 3 — the two flagship compounds. K₂Cr₂O₇ and KMnO₄. Write out the preparation and the oxidation equations by hand three times. These are guaranteed marks.
4. Day 4 — lanthanoids and actinoids. Lanthanoid contraction, its consequences, the comparison table, applications.
5. Day 5 — practice. Do the worksheet at the bottom of this page with your book closed, then mark yourself honestly against the model answers.

Study Notes

One Cause, Many Symptoms: The Root Idea Behind This Whole Chapter

Imagine a school where the senior students sit in the outermost row of the hall and a group of newer students keeps getting seated in a row behind them — between the seniors and the stage. Every new student added to that inner row should, in principle, block a bit of the stage light from reaching the seniors. But these particular students are oddly shaped and sit in awkward gaps, so they block the light very badly. The seniors keep feeling the stage lights getting brighter and brighter even though the hall is filling up.

That is exactly what happens in the d- and f-block. As you move across a transition series, each new electron goes into the inner (n−1)d sub-shell, not the outer ns sub-shell. In the f-block, each new electron goes into the even more deeply buried (n−2)f sub-shell. Because d orbitals and especially f orbitals are diffuse and poorly directed, they shield the outer electrons from the nucleus very badly. Meanwhile the nuclear charge goes up by one proton at every step.

Two consequences follow immediately, and they are the whole chapter:

• Consequence A — the effective nuclear charge creeps up. Sizes shrink slowly across a series; this is the seed of the lanthanoid contraction and of the near-identical sizes of the 4d and 5d metals.
• Consequence B — the (n−1)d and ns orbitals end up almost equal in energy. Because they are so close, both sets of electrons are available for bonding. That single fact gives you variable oxidation states, partly-filled d sub-shells in ions, d–d transitions (colour), unpaired electrons (paramagnetism), empty low-lying d orbitals (complex formation) and easy switching between oxidation states (catalysis).

Throughout this page, every time we cash out a symptom of that root cause we will tag it with a navy Key Idea box. Collect those boxes and you have the chapter.

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Position in the Periodic Table

The d-block is the wide central slab of the periodic table — groups 3 to 12. It contains four series, each named after the d sub-shell being filled:

• 3d series (first transition series): Sc (Z = 21) to Zn (Z = 30) — period 4. This is the series your board paper cares about.
• 4d series (second): Y (39) to Cd (48) — period 5.
• 5d series (third): La (57), then Hf (72) to Hg (80) — period 6, skipping over the lanthanoids which are pulled out below.
• 6d series (fourth): Ac (89), then Rf (104) onwards — period 7, mostly synthetic and radioactive, and incomplete.

The f-block sits inside the d-block conceptually — that is why its members are called inner transition elements. There are two series: the lanthanoids, Ce (58) to Lu (71), filling 4f; and the actinoids, Th (90) to Lr (103), filling 5f. Lanthanum (57) and actinium (89) are studied along with them because they sit immediately before each series.

Apply the definition to zinc. Zn is [Ar] 3d¹⁰ 4s²; its only common ion, Zn²⁺, is [Ar] 3d¹⁰. Both the atom and the ion have a completely filled d sub-shell, so strictly Zn, Cd and Hg are not transition elements. We still study them with the d-block because they close each series and their chemistry is discussed alongside. Notice the contrast with Cu: copper’s atom is 3d¹⁰ 4s¹ (full d), but its common Cu²⁺ ion is 3d⁹ — incompletely filled — so copper is a genuine transition element.

Before you go further, it is worth refreshing the Class 10 picture of how metals behave — reactivity order, alloys, corrosion and electrolytic refining all reappear in this chapter in a more sophisticated form. If any of that feels hazy, ten minutes on our Metals and Non-metals chapter for Class 10 will pay for itself.

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Electronic Configuration of the d-Block Elements (Cr and Cu Anomalies)

The general outer configuration of a d-block element is

For the 3d series that reads [Ar] 3d^1–10 4s^1–2. Here is the complete, verified list. Fill it in from memory every morning for a week and it will never leave you.

Why do Cr and Cu break the pattern? Because 3d and 4s are almost equal in energy (that is Consequence B of our root idea), an electron can slide from 4s into 3d if doing so buys enough stability. Two arrangements are unusually stable: a half-filled d⁵ and a completely filled d¹⁰. Both maximise the number of parallel-spin electron pairs, which maximises the exchange energy — a quantum-mechanical stabilisation with no everyday analogue, but you can picture it as electrons of the same spin being able to swap places freely, and more swapping options meaning lower energy.

So chromium prefers [Ar] 3d⁵ 4s¹ over the “expected” 3d⁴ 4s², and copper prefers [Ar] 3d¹⁰ 4s¹ over 3d⁹ 4s².

Why do the anomalies not repeat cleanly lower down? In the 4d and 5d series the energy gaps are different, and several elements (Nb, Mo, Ru, Rh, Pd, Ag) show their own irregularities — palladium is the extreme case at 4d¹⁰ 5s⁰. For the board paper you only need the 3d series and the two anomalies; just do not write “only Cr and Cu are exceptions in the whole d-block”, because that is not true.

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Atomic and Ionic Radii Across a Series

Here is our first symptom of the root cause. Move left to right across the 3d series: nuclear charge rises by one unit each step, but the added electron goes into the inner 3d sub-shell where it shields the 4s electrons poorly. Net effect — the effective nuclear charge felt by the outer electrons creeps upward and the atom shrinks.

But the shrinking is gentle, not dramatic like in the p-block, and it does not continue all the way. The pattern across the 3d series is:

• Sc to Cr: a noticeable decrease, as the added d electrons shield poorly.
• Fe, Co, Ni: the radii become almost constant. By now there are enough d electrons that the extra electron–electron repulsion roughly cancels the extra nuclear pull.
• Cu and Zn: the radii rise again. The d sub-shell is now full (d¹⁰), repulsion dominates, and the atom expands slightly.

Down a group the story is stranger, and it is one of the most-asked questions in this unit. Going 3d → 4d, radii increase as you would expect. But going 4d → 5d the radii barely change at all. Zirconium has an atomic radius of about 160 pm and hafnium, one full period below it, about 159 pm. The reason is the lanthanoid contraction, which we unpack in detail later — fourteen 4f electrons are inserted before the 5d series begins, and their catastrophically poor shielding eats up the size increase you would otherwise get from an extra shell.

One more radius fact worth carrying: density increases steadily across the 3d series, from roughly 3.4 g cm⁻³ for scandium to about 8.9 g cm⁻³ for copper. Same reason — atomic mass climbs while the metallic radius shrinks or stays flat, so you are packing more mass into the same volume.

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Ionisation Enthalpy and Standard Electrode Potentials

Ionisation enthalpy generally increases across the 3d series, but the rise is much gentler than in the s- or p-block, and it is bumpy rather than smooth. The gentleness is our root cause again: the added d electrons partly shield the 4s electrons, so the outer electron does not feel the full weight of the extra protons. The bumps come from exchange-energy stabilisation — whenever removing an electron would destroy a d⁵ or d¹⁰ arrangement, that ionisation costs extra.

Two classic bumps worth knowing:

• The second ionisation enthalpies of Cr and Cu are unusually high, because Cr⁺ is 3d⁵ and Cu⁺ is 3d¹⁰ — you are being asked to break an extra-stable shell.
• The third ionisation enthalpy of Mn is unusually high and that of Fe is comparatively low, because Mn²⁺ (3d⁵) is stable and resists further ionisation, whereas Fe²⁺ (3d⁶) actually gains the half-filled configuration on losing an electron to become Fe³⁺ (3d⁵).

Now to standard electrode potentials. Three energy terms decide E°(M²⁺/M): the enthalpy of atomisation (breaking the metal apart), the first two ionisation enthalpies (stripping two electrons), and the hydration enthalpy (the reward for surrounding M²⁺ with water). Across the series the E° values generally become less negative, with three famous outliers.

The other couple you must know is E°(M³⁺/M²⁺), because it tells you at a glance which ion is an oxidising agent and which is a reducing agent. A very positive value means M³⁺ badly wants to become M²⁺, so M³⁺ is a strong oxidant. A very negative value means M²⁺ badly wants to become M³⁺, so M²⁺ is a strong reductant.

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Variable Oxidation States: The d and f Block Class 12 Oxidation State List

Variable oxidation state is the defining habit of the transition metals, and it is Consequence B of our root idea in its purest form. Because 3d and 4s sit at almost the same energy, both sets of electrons are available for bonding — and the atom can hand over two, three, four or even seven of them depending on what the other reagent offers in return.

Three rules organise everything:

1. The states differ by 1, not by 2. Vanadium shows +2, +3, +4 and +5 — a continuous ladder. Contrast the p-block, where the inert pair effect gives jumps of two (Sn is +2 or +4, never +3).
2. From Sc to Mn, the maximum oxidation state equals the total number of 4s + 3d electrons. Sc has 3 such electrons and reaches +3; Mn has 5 + 2 = 7 and reaches +7. That is the rising half of the fan chart.
3. After Mn the maximum falls away sharply. Once the d sub-shell is more than half full the electrons start pairing, and paired d electrons are far harder to remove. Fe reaches +6 only with difficulty; Cu manages +2; Zn is stuck at +2.

Why do the highest states only ever appear with oxygen and fluorine? Look at where +7 actually turns up: Mn₂O₇ and MnO₄⁻, never MnCl₇. Oxygen can form multiple bonds and fluorine is tiny with a very high bond enthalpy; both are able to stabilise a heavily positive metal centre. A big soft halide simply cannot.

A useful contrast with the p-block. Going down a d-block group, the higher oxidation state becomes more stable — Cr(VI) in Cr₂O₇²⁻ is a fierce oxidant, but Mo(VI) and W(VI) in MoO₃ and WO₃ are mild. That is the exact opposite of the inert pair effect you learned for the p-block.

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Formation of Coloured Ions

Pour copper sulfate into water and you get that unmistakable blue. Nickel salts give green, cobalt gives pink, potassium dichromate gives orange. Why does one whole region of the periodic table behave like a paint box while sodium and magnesium salts are stubbornly colourless?

When a transition metal ion sits in water, the surrounding water molecules act as ligands. Their electric field splits the five d orbitals, which were degenerate in the free ion, into two groups with a small energy gap between them. That gap happens to correspond to the energy of visible light. An electron in a lower d orbital absorbs a photon of exactly the right wavelength and jumps to a higher d orbital — a d–d transition. The light that survives is the original white light minus the absorbed colour, and what reaches your eye is the complementary colour.

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Magnetic Properties and the Spin-Only Formula

An unpaired electron is a tiny spinning magnet. Pair it up with another electron of opposite spin and the two magnets cancel. So the magnetic behaviour of a transition metal ion is a direct read-out of how many unpaired d electrons it has.

• Diamagnetic — all electrons paired; the substance is weakly repelled by a magnetic field.
• Paramagnetic — one or more unpaired electrons; weakly attracted.
• Ferromagnetic — an extreme case where spins align across whole domains of the solid; very strongly attracted. Iron, cobalt and nickel metals are the classic examples.

Here are the five values, each computed rather than copied. Learn these and you can answer any magnetic-moment question in seconds.

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Catalytic Behaviour

A catalyst is a chemical middleman: it takes something from one reactant, hands it to another, and walks away unchanged. To do that job repeatedly, a species has to be able to change its state and then change back. Transition metals are built for exactly this, because switching between oxidation states costs them almost nothing — Consequence B again.

The textbook illustration is iron catalysing the otherwise sluggish reaction between iodide ions and persulfate ions. Both reactants are negatively charged, so they repel each other and rarely meet. Iron solves the problem by splitting the job in two:

Transition metals also catalyse heterogeneously, by adsorbing reactants onto their surface using partly filled d orbitals. This holds the molecules in place, weakens their bonds and lowers the activation energy. Industrially important examples worth memorising:

• V₂O₅ — Contact process, 2 SO₂ + O₂ → 2 SO₃.
• Finely divided Fe — Haber process, N₂ + 3 H₂ → 2 NH₃.
• Ni — hydrogenation of unsaturated vegetable oils into vanaspati.
• TiCl₄ with Al(C₂H₅)₃ — the Ziegler–Natta catalyst for polythene.
• PdCl₂ — Wacker process, converting ethene to ethanal.
• MnO₂ — decomposition of potassium chlorate to oxygen.

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Complex Formation

Transition metal ions are unusually good at gathering ligands around themselves — [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺, [Co(NH₃)₆]³⁺ and so on. Three features make this possible, and an examiner wants all three:

• Small size — the ion is compact, so its electric field at the surface is intense.
• High charge density — a +2 or +3 charge packed into that small volume strongly attracts electron-pair donors.
• Vacant d orbitals of the right energy — empty, low-lying orbitals that can accept lone pairs from the ligands.

Compare with sodium: Na⁺ is larger, carries only +1, and has no accessible vacant d orbitals, so it forms hardly any stable complexes. The contrast makes the point sharply.

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Interstitial Compounds and Alloy Formation

Picture a crate of oranges. However tightly you pack them, there are gaps — small triangular holes between the fruit. Now imagine dropping peas into those gaps. The crate does not get bigger, but it does get much harder to squash. That is exactly an interstitial compound.

Transition metal lattices have these holes, and small atoms — H, C, N and B — fit into them. The resulting solids have four signature properties:

• Non-stoichiometric — the formulae are odd because the holes need not all be filled: VH_0.56, TiH_1.7, Fe₃H, Mn₄N, TiC.
• Very hard — the trapped small atoms block the metal layers from sliding over one another. Steel is harder than pure iron for exactly this reason.
• Very high melting points — often higher than the pure metal.
• Still metallic in conductivity — the delocalised electron sea survives.

Alloys work on the neighbouring principle. Because transition metals across a series have very similar metallic radii — within about fifteen percent of each other — the atoms of one can simply replace the atoms of another in the lattice without wrecking it. These are called substitutional alloys. Familiar examples: stainless steel (Fe with roughly 12–18 percent Cr plus Ni), brass (Cu–Zn), bronze (Cu–Sn). If you want the Class 10 groundwork on why alloying changes hardness and corrosion resistance, it is set out in the Metals and Non-metals notes.

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Potassium Dichromate: Preparation, Structure and Oxidising Action

This is one of the two compounds the board asks about almost every year. Take it slowly and write every equation out by hand at least three times — these are the easiest marks in the unit. If balancing redox equations feels shaky, spend fifteen minutes first on the Chemical Reactions and Equations foundation from Class 10; everything below is built on it.

Preparation from chromite ore, FeCr₂O₄ — three steps.

The chromate–dichromate equilibrium. This is the colour change every practical student remembers: add acid to a yellow chromate solution and it turns orange; add alkali to an orange dichromate solution and it turns yellow.

Structure. The chromate ion CrO₄²⁻ is a simple tetrahedron with chromium at the centre and four equivalent Cr–O bonds. The dichromate ion Cr₂O₇²⁻ is two CrO₄ tetrahedra joined at a shared corner oxygen, with a Cr–O–Cr bridge angle of 126°. Think of two tents pegged to the same central peg.

Oxidising action. In acidic solution dichromate accepts six electrons and the solution changes from orange to the green of Cr³⁺:

Bolt any six-electron supply onto that half-reaction and you have a complete equation. The four you should be able to write cold:

Uses of K₂Cr₂O₇: a primary standard in volumetric analysis (it is not deliquescent and can be obtained pure), chrome tanning of leather, preparation of azo dyes, and as a laboratory oxidant. Titrations with it are done at a known concentration, so if molarity and normality feel rusty, revisit the Solutions chapter of Class 12 Chemistry.

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Potassium Permanganate: Preparation, Structure and Oxidising Action

The second flagship compound, and the one with the most quotable equations. Permanganate is that deep purple solid you have probably seen used as a disinfectant.

Preparation from pyrolusite, MnO₂ — two steps.

In the laboratory Mn(II) can also be taken all the way to Mn(VII) in one go using peroxodisulfate:

2 Mn²⁺ + 5 S₂O₈²⁻ + 8 H₂O → 2 MnO₄⁻ + 10 SO₄²⁻ + 16 H⁺

Structure. Both MnO₄⁻ and MnO₄²⁻ are tetrahedral. The interesting difference is electronic:

Oxidising action. Unlike dichromate, permanganate has three different behaviours depending on the pH. This is the single most examinable fact about it.

Uses of KMnO₄: oxidising agent in volumetric analysis and in organic synthesis (for example oxidising the side chain of toluene to benzoic acid), bleaching of wool, cotton, silk and oils, and as a topical disinfectant.

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The Lanthanoids: Lanthanoid Contraction Explained With Consequences

The lanthanoids are the fourteen elements from cerium (Z = 58) to lutetium (Z = 71), in which the 4f sub-shell is progressively filled. Lanthanum (57) is always discussed with them. The general configuration is [Xe] 4f^1–14 5d^{0 or 1} 6s², and the common tripositive ion has the tidy form Ln³⁺ = [Xe] 4fⁿ.

The oxidation state story is refreshingly simple. Almost every lanthanoid shows only +3. The exceptions all come from the same source you have already met — the extra stability of empty, half-filled and completely filled sub-shells:

• Ce⁴⁺ is 4f⁰ (the bare xenon core) — hence cerium’s +4 state, and Ce⁴⁺ is a good oxidising agent because it wants to fall back to Ce³⁺.
• Tb⁴⁺ is 4f⁷ — half filled.
• Eu²⁺ is 4f⁷ — half filled, and Eu²⁺ is a good reducing agent because it wants to rise to Eu³⁺.
• Yb²⁺ is 4f¹⁴ — completely filled.

Now the headline act: the lanthanoid contraction. As you cross the series from La to Lu, the atomic and (much more regularly) the ionic radii steadily decrease. The ionic radius of Ln³⁺ falls from 106 pm for La³⁺ to 85 pm for Lu³⁺ — a total contraction of 21 pm, roughly 1.5 pm at each step.

The three consequences you must be able to state.

1. The second and third transition series become almost the same size. Fourteen f-electron insertions cancel out the size increase you would expect from adding a shell. Zr (about 160 pm) and Hf (about 159 pm) are the standard pair; Nb/Ta and Mo/W behave the same way. Because their radii and configurations are so alike, their chemistry is nearly identical and they are extremely difficult to separate — hafnium was not identified as a distinct element until 1923, well over a century after zirconium.
2. The lanthanoids themselves are hard to separate. Their chemistry is dominated by the +3 state and their sizes differ by only about 1.5 pm per step, so classical chemical separation fails. Modern separation uses ion-exchange chromatography, exploiting the fact that complex stability rises smoothly as the ion gets smaller.
3. The basicity of the hydroxides decreases from La(OH)₃ to Lu(OH)₃. As the cation shrinks, its polarising power rises, the Ln–OH bond becomes more covalent, and the hydroxide becomes progressively less basic.

Other general characteristics worth a line each: lanthanoids are silvery-white, soft, highly electropositive metals that tarnish rapidly in air; most Ln³⁺ ions are coloured because of f–f transitions, with La³⁺ (4f⁰) and Lu³⁺ (4f¹⁴) colourless; most Ln³⁺ ions are paramagnetic, although for the f-block the spin-only formula does not work well because orbital angular momentum is not quenched.

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The Actinoids

The actinoids are the fourteen elements from thorium (Z = 90) to lawrencium (Z = 103), in which the 5f sub-shell fills. Actinium (89) is discussed alongside them. Their general configuration is [Rn] 5f^1–14 6d^0–1 7s².

The single structural difference that drives everything: 5f orbitals are not buried as deeply as 4f orbitals. They extend further out, so 5f electrons are more available for bonding and are more easily ionised. Two things follow at once — actinoid chemistry is richer than lanthanoid chemistry, and the actinoid contraction is sharper per step because 5f shielding is even poorer than 4f shielding.

The other headline fact is that all actinoids are radioactive. Thorium and uranium have very long half-lives and occur naturally in reasonable quantity; the later members are synthetic, short-lived and available only in vanishingly small amounts, which is why their chemistry is patchily known.

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Some Applications of d- and f-Block Elements

This sub-topic is usually worth one or two marks and it is pure recall, so build a short list you can reproduce quickly.

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Memory Devices and the d and f Block Class 12 Formula List

Everything you might need to reproduce under time pressure, on one screen.

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Practice Worksheet

Ten original questions in board style. Close the book, write full answers on paper, and only then open the accordions. Marks are indicated so you can practise pacing.

Score yourself honestly. Anything below 24 out of 30 means you should re-read the two sections that cost you marks rather than re-reading the whole chapter — targeted repair beats broad revision every time.

Kaizen: you do not need to master this chapter today. You need to be one percent clearer about it than you were yesterday. Ten focused minutes tonight — one table rewritten from memory, one equation balanced by hand — compounds into a score you will be proud of in March.

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