The d- and f-Block Elements Class 12: Trends, Reactions and Board Answers

What The d- and f-Block Elements is about

Weightage: This chapter is Unit IV, d- and f-Block Elements, which carries 7 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit IV d- and f-Block Elements

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper. See the questions

Key concepts

Position, electronic configuration and the definition of a transition element

NCERT section 4.1 Position in the Periodic Table; 4.2 Electronic Configurations of the d-Block Elements

The d-block has four series: 3d (Sc to Zn), 4d (Y to Cd), 5d (La, Hf to Hg) and 6d (Ac, Rf onwards). The general configuration is $(n-1)d^{1-10}n{s}^{1-2}$ (Pd is an exception, $4d^{10}5s^0$).

The 3d and 4s orbitals have very similar energies, so a half-filled or completely filled d set is sometimes reached at the expense of the s orbital:

• Cr: $3d^{5}4s^1$, not $3d^{4}4s^2$
• Cu: $3d^{10}4s^1$, not $3d^{9}4s^2$

A transition element is one that has an incompletely filled d subshell in its ground state or in any of its common oxidation states. By this definition:

• Zn, Cd and Hg ($(n-1)d^{10}n{s}^2$ in the atom and $d^{10}$ in their common +2 ions) are not regarded as transition elements, though they are studied with them.
• Cu, Ag and Au are transition elements, because Cu²⁺ is $3d^9$ and Ag and Au have partly filled d orbitals in their +2 and +3 states.

The partly filled d orbitals are the root of the special properties: variable oxidation states, coloured ions, paramagnetism, complex formation and catalytic activity.

Worked example

Scandium (Z = 21) is counted as a transition element, but zinc (Z = 30) is not. Use their electronic configurations to justify this. (2 marks)

Try it first, then show the solutionSolution

1. Sc: $[Ar]3d^{1}4s^2$. In the atom the 3d subshell is incompletely filled (one electron), so Sc fits the definition. (1)
2. Zn: $[Ar]3d^{10}4s^2$, and Zn²⁺ is $3d^{10}$. Its d subshell is full in the atom and in its only common oxidation state, so it is not regarded as a transition element. (1)

Answer: Sc has an incompletely filled 3d subshell (3d¹) in its ground state; Zn has a completely filled 3d¹⁰ subshell both as the atom and as Zn²⁺.

Physical properties, atomic size and enthalpy of atomisation

NCERT section 4.3.1 Physical Properties; 4.3.2 Variation in Atomic and Ionic Sizes of Transition Metals

Transition elements are typical metals: hard, with high tensile strength, ductility and malleability, high thermal and electrical conductivity and metallic lustre. Apart from Zn, Cd, Hg and Mn, each has at least one of the usual metallic crystal structures at room temperature.

Melting points and enthalpies of atomisation are high. Besides the ns electrons, (n−1)d electrons take part in metallic bonding. The more unpaired d electrons, the stronger the bonding; so the values rise to a maximum around the middle of each series ($d^5$) and then fall. Metals with very high enthalpy of atomisation tend to be noble in their reactions. The 4d and 5d metals have higher enthalpies of atomisation than the 3d metals, so metal–metal bonding is more frequent in compounds of the heavier metals.

Atomic and ionic radii decrease slowly across a series, because each new electron enters an inner d orbital and shields the outer electrons imperfectly, so the effective nuclear charge rises. Down a group, the 4d member is larger than the 3d one, but the 5d member is almost the same size as the 4d one (Zr 160 pm, Hf 159 pm).

This is because of the lanthanoid contraction: 4f orbitals are filled before the 5d series begins, and 4f electrons shield very poorly, so the steady size decrease along the lanthanoids cancels the expected increase. So Zr and Hf have very similar radii and chemistry, and occur together in nature.

As size falls and mass rises across the series, density increases from Ti to Cu.

Worked example

Explain why (a) transition metals have high enthalpies of atomisation, and (b) zirconium and hafnium have almost identical atomic radii. (2 marks)

Try it first, then show the solutionSolution

1. (a) Both ns and (n−1)d electrons, especially unpaired d electrons, take part in metallic bonding, so the interatomic attraction is strong. (1)
2. (b) Hf comes after the 14 lanthanoids, in which 4f orbitals are filled. The poor shielding of 4f electrons causes the lanthanoid contraction, which cancels the normal increase in size from the 4d to the 5d series. (1)

Answer: (a) Strong metallic bonding involving unpaired d electrons. (b) The lanthanoid contraction makes Hf almost the same size as Zr.

Ionisation enthalpies and variable oxidation states

NCERT section 4.3.3 Ionisation Enthalpies; 4.3.4 Oxidation States

Ionisation enthalpy increases along each series, but less steeply than in the main groups, because the extra d electrons shield the outer electrons. Breaks in the trend come from stable configurations: removing an electron from d⁵ (Mn²⁺) or d¹⁰ (Zn²⁺) is difficult, which is why the third ionisation enthalpy of Mn and Zn is high.

Variable oxidation states are a hallmark of transition metals, because the energies of the (n−1)d and ns electrons are close, so different numbers of them can be lost.

• The number of oxidation states is greatest in the middle of the series. Mn shows every state from +2 to +7, its highest being equal to the total of 4s and 3d electrons.
• Elements near the ends show fewer: Sc only +3, Zn only +2, because there are too few electrons to lose or too many d electrons (fewer orbitals available to share).
• Unlike p-block elements, oxidation states of transition metals usually differ by one (for example Fe²⁺ and Fe³⁺, Cu⁺ and Cu²⁺).
• Higher oxidation states are shown in fluorides and oxides, because fluorine and oxygen are small and highly electronegative. Oxygen stabilises higher states better than fluorine because it can form multiple bonds to the metal: Mn reaches +7 in Mn₂O₇ but only +4 in MnF₄.

Cr²⁺ is reducing but Mn³⁺ is oxidising although both are d⁴: Cr²⁺ → Cr³⁺ gives d³ (stable half-filled $t_{2g}$ level), while Mn³⁺ → Mn²⁺ gives the stable d⁵ configuration.

Worked example

Explain why (a) manganese shows the largest number of oxidation states in the 3d series, and (b) the highest oxidation state of Mn is shown in its oxide (Mn₂O₇) and not in its fluoride. (2 marks)

Try it first, then show the solutionSolution

1. (a) Mn ($3d^{5}4s^2$) has the largest number of electrons (seven) that can take part in bonding, so it can show every state from +2 to +7. (1)
2. (b) Oxygen can form multiple bonds with the metal, which stabilises the very high oxidation state; fluorine can form only single bonds, so Mn stops at MnF₄. (1)

Answer: (a) Seven 3d + 4s electrons are available. (b) Oxygen forms multiple (π) bonds with Mn; fluorine cannot.

Electrode potentials, stability of oxidation states and reactivity

NCERT section 4.3.5 Trends in the M²⁺/M Standard Electrode Potentials; 4.3.6 Trends in the M³⁺/M²⁺ Standard Electrode Potentials; 4.3.7 Trends in Stability of Higher Oxidation States; 4.3.8 Chemical Reactivity and E° Values

The value of $E^{\circ }(M^{2+}/M)$ depends on three energy terms: enthalpy of atomisation, ionisation enthalpy and hydration enthalpy. Some key points from the NCERT trends:

• Copper has a positive $E^{\circ }(C{u}^{2+}/Cu)$ (+0.34 V), the only one in the 3d series. The large energy needed to turn Cu(s) into Cu²⁺(aq) (its high enthalpy of atomisation plus the first two ionisation enthalpies) is not balanced by the hydration enthalpy of Cu²⁺, so Cu does not liberate hydrogen from acids. It reacts only with oxidising acids such as nitric acid and hot concentrated sulphuric acid.
• Mn, Ni and Zn have more negative $E^{\circ }$ than the trend predicts: Mn²⁺ is d⁵ (half-filled), Zn²⁺ is d¹⁰ (full), and Ni²⁺ has the most negative hydration enthalpy.
• **$E^{\circ }(M^{3+}/M^{2+})$**: the low value for Sc reflects the stability of Sc³⁺ (noble gas core); the high value for Zn reflects d¹⁰ Zn²⁺. Mn³⁺/Mn²⁺ (+1.57 V) is very positive because Mn²⁺ is d⁵, so Mn³⁺ is a strong oxidising agent. Fe³⁺/Fe²⁺ is low because Fe³⁺ is d⁵. Co³⁺/Co²⁺ (+1.97 V) is very high too.
• Cu⁺ is unstable in water and disproportionates: $2C{u}^{+}(aq)\to C{u}^{2+}(aq)+Cu(s)$. Cu²⁺ is more stable because its much more negative hydration enthalpy more than makes up for the second ionisation enthalpy.
• Ti²⁺, V²⁺ and Cr²⁺ are strong reducing agents and can liberate hydrogen from dilute acid: $2C{r}^{2+}(aq)+2H^{+}(aq)\to 2C{r}^{3+}(aq)+H_2(g)$.

Worked example

(a) Why is the E° value for Cu²⁺/Cu positive while that of the other 3d metals is negative? (b) Why does Cu⁺ not exist in aqueous solution? Write the equation. (3 marks)

Try it first, then show the solutionSolution

1. (a) Converting Cu(s) to Cu²⁺(aq) needs a high enthalpy of atomisation plus the first two ionisation enthalpies, and this is not balanced by its hydration enthalpy, so E° is positive. (1)
2. (b) Cu²⁺(aq) has a much more negative hydration enthalpy than Cu⁺(aq), which more than compensates for the second ionisation enthalpy of Cu, so Cu⁺ is less stable. (1)
3. It disproportionates: $2C{u}^{+}(aq)\to C{u}^{2+}(aq)+Cu(s)$. (1)

Answer: (a) High atomisation and ionisation enthalpies are not balanced by hydration. (b) Cu²⁺ is stabilised by its much larger hydration enthalpy; 2Cu⁺ → Cu²⁺ + Cu.

Magnetic properties and colour

NCERT section 4.3.9 Magnetic Properties; 4.3.10 Formation of Coloured Ions

Diamagnetic substances are repelled by a magnetic field; paramagnetic ones are attracted, because of unpaired electrons. Many transition metal ions are paramagnetic.

The magnetic moment comes from the spin and orbital motion of unpaired electrons. For the 3d ions, the orbital part is effectively quenched, so the spin-only formula is used:
$\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons and BM is the Bohr magneton.

• n = 1: 1.73 BM; n = 2: 2.83; n = 3: 3.87; n = 4: 4.90; n = 5: 5.92 BM.

Colour: when light falls on a transition metal ion with a partly filled d subshell, an electron is excited from a lower d orbital to a higher one (a d–d transition). The energy absorbed is in the visible region, so the complementary colour is seen. The exact colour depends on the ligands around the ion.

• Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) configurations cannot have d–d transitions and are colourless.
• Examples of coloured aqueous ions: Ti³⁺ purple, V³⁺ green, Mn²⁺ pale pink, Fe³⁺ yellow, Co²⁺ pink, Ni²⁺ green, Cu²⁺ blue.

Worked example

Calculate the spin-only magnetic moment of Fe²⁺ (Z of Fe = 26), and state whether Sc³⁺ (Z of Sc = 21) solution is coloured, with reason. (2 marks)

Try it first, then show the solutionSolution

1. Fe²⁺: $[Ar]3d^6$; in a hydrated (weak-field) ion the six d electrons are arranged with 4 unpaired.
2. $\mu =\sqrt{4(4+2)}=\sqrt{24}=4.90$ BM. (1)
3. Sc³⁺ is $3d^0$: no d electron is available for a d–d transition, so its solution is colourless. (1)

Answer: Fe²⁺: 4.90 BM. Sc³⁺ is colourless because it has no d electrons (3d⁰).

Complexes, catalysis, interstitial compounds and alloys

NCERT section 4.3.11 Formation of Complex Compounds; 4.3.12 Catalytic Properties; 4.3.13 Formation of Interstitial Compounds; 4.3.14 Alloy Formation

Complex formation: transition metals form many complexes, such as $[Fe(CN{)}_6{]}^{3-}$, $[Cu(N{H}_3{)}_4{]}^{2+}$ and $[PtC{l}_4{]}^{2-}$. This is because their ions are small and highly charged, and they have vacant d orbitals of suitable energy to accept lone pairs from ligands.

Catalytic properties: transition metals and their compounds are good catalysts, because they can show several oxidation states and form complexes. Examples: V₂O₅ in the Contact process, finely divided iron in the Haber process, nickel in hydrogenation. Catalysts at a solid surface form bonds with reactant molecules using their 3d and 4s electrons, which raises the concentration of reactants at the surface and weakens their bonds. Iron(III) catalyses the reaction between iodide and persulphate by changing oxidation state:
$2F{e}^{3+}+2I^{-}\to 2F{e}^{2+}+I_2$, then $2F{e}^{2+}+S_2{O}_8^{2-}\to 2F{e}^{3+}+2S{O}_4^{2-}$.

Interstitial compounds form when small atoms such as H, C or N are trapped in the crystal lattice of the metal: TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇. They are non-stoichiometric, have higher melting points than the pure metals, are very hard, keep metallic conductivity and are chemically inert.

Alloys: transition metals have similar atomic radii (within about 15%), so atoms of one can take the place of another in the lattice. Their alloys are hard and have high melting points. Examples: ferrous alloys with Cr, V, W, Mo and Mn for steels and stainless steel; brass (Cu–Zn) and bronze (Cu–Sn).

Worked example

Give reasons: (a) Transition metals and their compounds act as catalysts. (b) Interstitial compounds have higher melting points than the pure metals. (2 marks)

Try it first, then show the solutionSolution

1. (a) They can show variable oxidation states and form intermediate compounds or complexes with reactants, providing a path of lower activation energy; solid metals also adsorb reactants on their surface using 3d and 4s electrons. (1)
2. (b) Small atoms (H, C, N) fill the spaces in the metal lattice and bond with the metal atoms, making the structure more rigid, so more energy is needed to melt it. (1)

Answer: (a) Variable oxidation states and complex/surface intermediate formation. (b) Trapped small atoms bond within the lattice and make it more rigid.

Potassium dichromate, K₂Cr₂O₇

NCERT section 4.4.1 Oxides and Oxoanions of Metals (Potassium dichromate)

Preparation (from chromite ore, FeCr₂O₄):
1. Fuse the ore with sodium carbonate in free access of air to get yellow sodium chromate: $4FeC{r}_2{O}_4+8N{a}_{2}C{O}_3+7O_2\to 8N{a}_{2}Cr{O}_4+2F{e}_2{O}_3+8C{O}_2$
2. Filter and acidify with sulphuric acid to get orange sodium dichromate: $2N{a}_{2}Cr{O}_4+2H^{+}\to N{a}_{2}C{r}_2{O}_7+2N{a}^{+}+H_{2}O$
3. Sodium dichromate is more soluble, so add KCl to crystallise potassium dichromate: $N{a}_{2}C{r}_2{O}_7+2KCl\to K_{2}C{r}_2{O}_7+2NaCl$

Chromate–dichromate change with pH: the oxidation state of Cr is +6 in both.
$2Cr{O}_4^{2-}+2H^{+}\to C{r}_2{O}_7^{2-}+H_{2}O$ (yellow to orange in acid)
$C{r}_2{O}_7^{2-}+2O{H}^{-}\to 2Cr{O}_4^{2-}+H_{2}O$ (orange to yellow in alkali)

Structures: chromate is tetrahedral; dichromate is two tetrahedra sharing one corner oxygen, with a Cr–O–Cr angle of 126°.

Oxidising action in acid: $C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$ ($E^{\circ }$ = 1.33 V). It oxidises iodide to iodine, sulphide (H₂S) to sulphur, tin(II) to tin(IV) and iron(II) to iron(III). Add the half-reactions so that electrons cancel. Potassium dichromate is used as a primary standard in volumetric analysis and in the leather industry; the sodium salt, being more soluble, is used as an oxidising agent in organic chemistry.

Worked example

Write balanced ionic equations for the reaction of acidified potassium dichromate with (a) iron(II) ions and (b) iodide ions. (2 marks)

Try it first, then show the solutionSolution

1. Dichromate half: $C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$.
2. (a) Six Fe²⁺ supply six electrons: $C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}\to 2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$. Charge check: +24 = +24. (1)
3. (b) Six I⁻ give 3I₂ and six electrons: $C{r}_2{O}_7^{2-}+14H^{+}+6I^{-}\to 2C{r}^{3+}+3I_2+7H_{2}O$. Charge check: +6 = +6. (1)

Answer: (a) Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O (b) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O

Potassium permanganate, KMnO₄

NCERT section 4.4.1 Oxides and Oxoanions of Metals (Potassium permanganate)

Preparation: fuse MnO₂ with an alkali metal hydroxide and an oxidising agent such as KNO₃ (or air) to get dark green potassium manganate, which disproportionates in neutral or acidic solution to permanganate:
$2Mn{O}_2+4KOH+O_2\to 2K_{2}Mn{O}_4+2H_{2}O$
$3Mn{O}_4^{2-}+4H^{+}\to 2Mn{O}_4^{-}+Mn{O}_2+2H_{2}O$
Commercially, the manganate is oxidised electrolytically in alkaline solution. In the laboratory, a Mn²⁺ salt is oxidised by peroxodisulphate:
$2M{n}^{2+}+5S_2{O}_8^{2-}+8H_{2}O\to 2Mn{O}_4^{-}+10S{O}_4^{2-}+16H^{+}$

Properties: dark purple (almost black) crystals; on heating to 513 K it decomposes: $2KMn{O}_4\to K_{2}Mn{O}_4+Mn{O}_2+O_2$. Manganate (green, one unpaired electron) is paramagnetic; permanganate is diamagnetic. Both ions are tetrahedral, with π bonding between oxygen p orbitals and manganese d orbitals.

Oxidising action depends on pH:

• Acidic: $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$ ($E^{\circ }$ = 1.52 V). Oxidises Fe²⁺ to Fe³⁺, oxalate to CO₂, nitrite to nitrate, iodide to iodine.
• Neutral or faintly alkaline: $Mn{O}_4^{-}+2H_{2}O+3e^{-}\to Mn{O}_2+4O{H}^{-}$. Oxidises iodide to iodate, thiosulphate to sulphate and Mn²⁺ to MnO₂.

Hydrochloric acid is not used to acidify permanganate titrations, because permanganate would oxidise it to chlorine; dilute sulphuric acid is used.

Worked example

Write balanced ionic equations for the reactions of acidified KMnO₄ with (a) oxalate ions and (b) nitrite ions. (2 marks)

Try it first, then show the solutionSolution

1. Permanganate half: $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$; double it to take 10 electrons.
2. (a) $5C_2{O}_4^{2-}\to 10C{O}_2+10e^{-}$, so $2Mn{O}_4^{-}+16H^{+}+5C_2{O}_4^{2-}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$. Charge: +4 = +4. (1)
3. (b) $5N{O}_2^{-}+5H_{2}O\to 5N{O}_3^{-}+10H^{+}+10e^{-}$; adding and cancelling 10H⁺ and 5H₂O: $2Mn{O}_4^{-}+6H^{+}+5N{O}_2^{-}\to 2M{n}^{2+}+5N{O}_3^{-}+3H_{2}O$. Charge: −1 = −1. (1)

Answer: (a) 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O (b) 2MnO₄⁻ + 6H⁺ + 5NO₂⁻ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O

The lanthanoids

NCERT section 4.5 The Lanthanoids

The lanthanoids are the 14 elements after lanthanum (Ce to Lu), in which the 4f orbitals are filled. La is usually included in discussions, and the general symbol is Ln. Their configurations are $[Xe]4f^{1-14}5d^{0-1}6s^2$.

Lanthanoid contraction: the atomic and ionic radii decrease steadily from La to Lu. The shielding of one 4f electron by another is less than that by a d electron, so as nuclear charge rises the effective nuclear charge felt by the 4f electrons rises and the size shrinks. Its consequences:

• the 4d and 5d transition series have almost the same radii (Zr and Hf);
• lanthanoids are very similar to each other and hard to separate;
• basic character of the hydroxides falls from La(OH)₃ to Lu(OH)₃.

Oxidation states: +3 is the principal state. Some show +2 or +4 when this gives an empty, half-filled or full f subshell: Ce⁴⁺ ($4f^0$, a strong oxidant, $E^{\circ }$ Ce⁴⁺/Ce³⁺ = +1.74 V, used in analysis), Tb⁴⁺ ($4f^7$), Eu²⁺ ($4f^7$, a strong reducing agent) and Yb²⁺ ($4f^{14}$).

General characteristics: silvery white, soft metals that tarnish in air; similar to calcium in reactivity. They burn in oxygen to give Ln₂O₃, react with dilute acids to give hydrogen, and form hydroxides that are basic like alkaline earth hydroxides. Many trivalent ions are coloured in solid and solution because of f–f transitions (La³⁺ and Lu³⁺, $4f^0$ and $4f^{14}$, are colourless).

Uses: alloy steels, and mischmetall (a lanthanoid metal about 95%, iron about 5% and traces of S, C, Ca and Al), used in Mg-based alloys for bullets, shells and lighter flints. Mixed lanthanoid oxides are used as catalysts in petroleum cracking.

Worked example

What is the lanthanoid contraction? Give its cause and two consequences. (3 marks)

Try it first, then show the solutionSolution

1. Definition: the steady decrease in atomic and ionic radii from La to Lu across the lanthanoids. (1)
2. Cause: poor shielding of one 4f electron by another, so the effective nuclear charge increases as electrons are added. (1)
3. Consequences: Zr and Hf (4d and 5d) have nearly equal radii and similar properties; lanthanoids are hard to separate; basicity of Ln(OH)₃ decreases from La to Lu (any two). (1)

Answer: A steady decrease in size from La to Lu caused by poor 4f shielding; it makes Zr/Hf alike, lanthanoids hard to separate and their hydroxides less basic along the series.

The actinoids and comparison with the lanthanoids

NCERT section 4.6 The Actinoids; 4.7 Some Applications of d- and f-Block Elements

The actinoids are the 14 elements from Th to Lr, in which 5f orbitals are filled. Their configurations are roughly $[Rn]5f^{1-14}6d^{0-1}7s^2$. All are radioactive; the earlier ones have long half-lives, but the later ones have half-lives of days to minutes, so they can be made only in tiny amounts. That makes their chemistry harder to study.

Actinoid contraction: the size decrease from element to element is greater than in the lanthanoids, because 5f electrons shield even more poorly than 4f electrons.

Oxidation states: a much wider range than lanthanoids, because 5f, 6d and 7s energies are close. +3 is common, but Th shows +4, Pa +5, U +6 and Np and Pu up to +7. The highest states are found in the first half of the series.

Comparison:

• Lanthanoids: mostly +3; less tendency to form complexes; only Pm is radioactive; ions often coloured.
• Actinoids: many oxidation states; greater tendency to form complexes; all radioactive; contraction is larger.
• Both are highly reactive metals, especially when finely divided.

The actinoid metals are silvery in appearance and show a range of structures, because of irregular metallic radii.

Worked example

Give reasons: (a) Actinoids show a wider range of oxidation states than lanthanoids. (b) The chemistry of the actinoids is not as smoothly graded or well studied as that of the lanthanoids. (2 marks)

Try it first, then show the solutionSolution

1. (a) The 5f, 6d and 7s levels are of comparable energies, so a variable number of electrons can take part in bonding. In lanthanoids the 4f electrons are more deeply buried. (1)
2. (b) Actinoids are radioactive, and many have short half-lives and can be prepared only in very small amounts, so their study is difficult; they also show many oxidation states. (1)

Answer: (a) Comparable 5f, 6d and 7s energies. (b) Radioactivity and short half-lives make them hard to study; many oxidation states complicate their chemistry.

Chemistry in real life: The d- and f-Block Elements

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Oxidation States and Electrode Potentials of Transition Metals
• Magnetic Properties and Colour of Transition Metal Ions
• Potassium Dichromate and Potassium Permanganate

The d- and f-Block Elements formula sheet

Spin-only magnetic moment

$\mu =\sqrt{n(n+2)}\text{BM}$

n = unpaired electrons: 1 → 1.73, 2 → 2.83, 3 → 3.87, 4 → 4.90, 5 → 5.92 BM.

Sodium chromate from chromite

$4FeC{r}_2{O}_4+8N{a}_{2}C{O}_3+7O_2\to 8N{a}_{2}Cr{O}_4+2F{e}_2{O}_3+8C{O}_2$

Fusion in free access of air.

Chromate to dichromate (acid)

$2N{a}_{2}Cr{O}_4+2H^{+}\to N{a}_{2}C{r}_2{O}_7+2N{a}^{+}+H_{2}O$

Yellow to orange.

Potassium dichromate

$N{a}_{2}C{r}_2{O}_7+2KCl\to K_{2}C{r}_2{O}_7+2NaCl$

K₂Cr₂O₇ is less soluble and crystallises out.

Chromate–dichromate equilibrium

$2Cr{O}_4^{2-}+2H^{+}\to C{r}_2{O}_7^{2-}+H_{2}O$

Reverse in alkali: $C{r}_2{O}_7^{2-}+2O{H}^{-}\to 2Cr{O}_4^{2-}+H_{2}O$. Cr is +6 in both.

Dichromate as oxidant

$C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}\to 2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$

Half-reaction gains 6e⁻; E° = 1.33 V. With H₂S: $C{r}_2{O}_7^{2-}+8H^{+}+3H_{2}S\to 2C{r}^{3+}+3S+7H_{2}O$.

Potassium manganate

$2Mn{O}_2+4KOH+O_2\to 2K_{2}Mn{O}_4+2H_{2}O$

Dark green; fusion with KOH and an oxidant.

Manganate disproportionation

$3Mn{O}_4^{2-}+4H^{+}\to 2Mn{O}_4^{-}+Mn{O}_2+2H_{2}O$

In neutral or acidic solution.

Laboratory preparation of permanganate

$2M{n}^{2+}+5S_2{O}_8^{2-}+8H_{2}O\to 2Mn{O}_4^{-}+10S{O}_4^{2-}+16H^{+}$

Oxidation by peroxodisulphate.

Heating KMnO₄

$2KMn{O}_4\to K_{2}Mn{O}_4+Mn{O}_2+O_2$

At 513 K.

Permanganate in acid

$Mn{O}_4^{-}+8H^{+}+5F{e}^{2+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$

Oxalate: $2Mn{O}_4^{-}+16H^{+}+5C_2{O}_4^{2-}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$.

Permanganate in neutral or faintly alkaline solution

$2Mn{O}_4^{-}+H_{2}O+I^{-}\to 2Mn{O}_2+2O{H}^{-}+I{O}_3^{-}$

Iodide is oxidised to iodate; MnO₂ is formed.

Disproportionation of Cu⁺

$2C{u}^{+}\to C{u}^{2+}+Cu$

Cu²⁺(aq) is stabilised by its high hydration enthalpy.

Cr²⁺ as a reducing agent

$2C{r}^{2+}+2H^{+}\to 2C{r}^{3+}+H_2$

Cr²⁺ (d⁴) → Cr³⁺ (d³, half-filled t₂g).

The d- and f-Block Elements: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks The d- and f-Block Elements

The question patterns to expect, and the method that works for each.

Tricks to remember The d- and f-Block Elements

Short memory aids. Say each one aloud once, then use it on a question the same day.

• 0, 5, 10 are the happy numbers

  d⁰, d⁵ and d¹⁰ (and f⁰, f⁷, f¹⁴) are extra stable. Most 'why' questions about E° values, oxidation states or colour come back to reaching or leaving one of these.

  Use it when: Mn²⁺ stability, Fe³⁺ stability, Ce⁴⁺, Eu²⁺, Yb²⁺, colourless Sc³⁺ and Zn²⁺.

• Spin-only ladder

  1.73, 2.83, 3.87, 4.90, 5.92 for 1 to 5 unpaired electrons. Each value is just under n + 1.

  Use it when: Quick magnetic moment answers and identifying n from a given moment.

• Orange in Acid, Yellow in Alkali

  OA-YA: dichromate (Orange) in Acid, chromate (Yellow) in Alkali. Cr stays +6 in both.

  Use it when: Chromate–dichromate equilibrium questions.

• Permanganate 5-3-1

  Acid: 5 electrons, Mn²⁺. Neutral or faintly alkaline: 3 electrons, MnO₂. Strongly alkaline: 1 electron, MnO₄²⁻.

  Use it when: Choosing the product and electrons when balancing.

• Zr–Hf twins

  Think 'the lanthanoids squeezed hafnium': 14 poorly shielding 4f electrons pull Hf in to Zr's size.

  Use it when: Lanthanoid contraction consequences.

The d- and f-Block Elements: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in The d- and f-Block Elements

The slips examiners see most often, and what to write instead.

• ✗Mistake: Removing 3d electrons before 4s when writing ion configurations

  Wrong working

  Fe²⁺: [Ar]3d⁴4s²

  Right working

  Fe²⁺: [Ar]3d⁶ (the 4s electrons are lost first)

  ✓Do this: Always remove the outer 4s electrons first, then 3d.

  Usually costs: The magnetic moment mark that depends on it.

• ✗Mistake: Unbalanced charge in permanganate and dichromate equations

  Atoms are balanced but charges are not checked.

  Wrong working

  MnO₄⁻ + 8H⁺ + Fe²⁺ → Mn²⁺ + Fe³⁺ + 4H₂O

  Right working

  MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O (charge +17 on each side)

  ✓Do this: Balance electrons between half-reactions, then add up charges on both sides.

  Usually costs: The full equation mark.

• ✗Mistake: Saying the oxidation state of Cr changes between chromate and dichromate

  ✓Do this: Cr is +6 in both CrO₄²⁻ and Cr₂O₇²⁻; the change is an acid–base reaction, not redox.

  Usually costs: 1 mark in an assertion-reason item or a reason question.

• ✗Mistake: Calling zinc a transition element

  ✓Do this: Zn has 3d¹⁰ both in the atom and in Zn²⁺, so by definition it is not a transition element, though it is placed in the d-block.

  Usually costs: 1 mark.

• ✗Mistake: Explaining Zr/Hf similarity by 'same group'

  Being in the same group does not make sizes equal (Ti is much smaller than Zr).

  ✓Do this: Name the lanthanoid contraction and the poor shielding by 4f electrons.

  Usually costs: 1 mark in a 2-mark reason.

NCERT exercise guide: The d- and f-Block Elements

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

The d- and f-Block Elements: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper; 28 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from The d- and f-Block Elements

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every The d- and f-Block Elements question, by year

Practice questions on The d- and f-Block Elements

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which of the following has the highest number of unpaired electrons?
   (a) Fe³⁺ (b) Cr³⁺ (c) Cu²⁺ (d) Ti³⁺

   Hint

   Write the d configuration of each ion.

   Answer

   (a) Fe³⁺ (3d⁵, five unpaired electrons).

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): The oxidation state of chromium is the same in CrO₄²⁻ and Cr₂O₇²⁻.
   Reason (R): Chromate changes to dichromate by a redox reaction in acid.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Is any element oxidised or reduced?

   Answer

   (c) Cr is +6 in both, so the change in acid is not a redox reaction.

3. Q3 · 2 marks · Very short answerBoard level

   Why do transition elements show variable oxidation states? Why does scandium show only +3? (2 marks)

   Hint

   Compare the energies of (n−1)d and ns electrons.

   Answer

   The (n−1)d and ns electrons have similar energies, so different numbers of them can be lost or shared (1). Sc ([Ar]3d¹4s²) loses all three to reach the stable noble gas configuration of Sc³⁺ (1).

4. Q4 · 3 marks · Short answerHarder board level

   Write the configurations and calculate the spin-only magnetic moments of Ti³⁺, Mn²⁺ and Ni²⁺. (Z: Ti 22, Mn 25, Ni 28) (3 marks)

   Hint

   Remove 4s electrons first.

   Answer

   Ti³⁺: [Ar]3d¹, n = 1, 1.73 BM (1). Mn²⁺: [Ar]3d⁵, n = 5, 5.92 BM (1). Ni²⁺: [Ar]3d⁸, n = 2, 2.83 BM (1).

5. Q5 · 3 marks · Short answerHarder board level

   Complete and balance: (a) MnO₄⁻ + H⁺ + I⁻ → (b) Cr₂O₇²⁻ + H⁺ + Sn²⁺ → (c) MnO₄²⁻ + H⁺ → (3 marks)

   Hint

   Equalise electrons before adding half-reactions.

   Answer

   (a) 2MnO₄⁻ + 16H⁺ + 10I⁻ → 2Mn²⁺ + 5I₂ + 8H₂O (1). (b) Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O (1). (c) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (1).

6. Q6 · 2 marks · Very short answerBoard level

   Explain why (a) transition metals form a large number of complexes, and (b) the 4d and 5d elements of the same group have very similar atomic radii. (2 marks)

   Hint

   Think size, charge, vacant orbitals; then the 4f electrons.

   Answer

   (a) Their ions are small and highly charged and have vacant d orbitals of suitable energy to accept lone pairs from ligands (1). (b) The lanthanoid contraction, caused by poor shielding of 4f electrons, cancels the expected increase in size from 4d to 5d (1).

7. Q7 · 3 marks · Short answerHarder board level

   Write the electronic configurations of Ce⁴⁺, Eu²⁺ and Yb²⁺ (Z: Ce 58, Eu 63, Yb 70) and use them to explain why these unusual oxidation states exist. Which of them is a strong oxidising agent and which is a strong reducing agent? (3 marks)

   Hint

   Look for f⁰, f⁷ and f¹⁴.

   Answer

   Ce⁴⁺ [Xe]4f⁰, Eu²⁺ [Xe]4f⁷, Yb²⁺ [Xe]4f¹⁴: empty, half-filled and full f subshells are especially stable (2). Ce⁴⁺ tends to return to +3, so it is a strong oxidant; Eu²⁺ tends to become Eu³⁺, so it is a strong reducing agent (1).

8. Q8 · 5 marks · Long answerHarder board level

   (a) Compare the lanthanoids and actinoids with respect to electronic configuration, oxidation states and chemical reactivity. (3) (b) What is mischmetall? Give one use. (1) (c) Why is the separation of lanthanoids difficult? (1) (5 marks)

   Hint

   Make a small table for part (a).

   Answer

   (a) Configuration: 4f¹⁻¹⁴5d⁰⁻¹6s² versus 5f¹⁻¹⁴6d⁰⁻¹7s² (1). Oxidation states: lanthanoids mostly +3 (some +2, +4); actinoids show many states, up to +7 (1). Reactivity: both highly reactive metals; actinoids are radioactive and form complexes more readily (1). (b) About 95% lanthanoid metal, 5% iron with traces of S, C, Ca, Al; used in Mg-based alloy for bullets, shells and lighter flints (1). (c) Because of the lanthanoid contraction their sizes and properties are very similar (1).

9. Q9 · 3 marks · Short answerHOTS / competency

   Suppose an unknown hydrated 3d metal ion X²⁺ is green and has a spin-only magnetic moment of 2.83 BM. (a) How many unpaired electrons does it have? (b) Which ion could it be? (c) Explain why it is coloured. (3 marks)

   Hint

   Solve n(n+2) = 8.

   Answer

   (a) n(n+2) = 8, so n = 2 (1). (b) Ni²⁺ (3d⁸, two unpaired electrons, green in water) (1). (c) Its partly filled d subshell allows d–d transitions that absorb part of visible light; the complementary colour, green, is seen (1).

10. Q10 · 3 marks · Short answerHOTS / competency

    Explain why the enthalpy of atomisation of zinc is the lowest in the 3d series, and why the first series metals differ in many properties from the heavier 4d and 5d metals of the same group. (3 marks)

    Hint

    Count unpaired d electrons in Zn; compare metal–metal bonding down a group.

    Answer

    Zn (3d¹⁰4s²) has no unpaired d electrons, so only the 4s electrons take part in metallic bonding, which is weak (1). The 4d and 5d metals have higher enthalpies of atomisation, so metal–metal bonding is more frequent in their compounds (1); they also show higher oxidation states more readily and have similar sizes to each other because of the lanthanoid contraction (1).

11. Q11 · 2 marks · Very short answerHOTS / competency

    Iron(III) catalyses the reaction between iodide ions and persulphate ions. Write the two steps and explain how this shows why transition metal ions are good catalysts. (2 marks)

    Hint

    The catalyst changes oxidation state and is then regenerated.

    Answer

    2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂; 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻ (1). Iron can switch easily between +3 and +2, providing a lower-energy path and being regenerated at the end (1).

12. Q12 · 1 mark · MCQHarder board level

    Which of the following is NOT a property of interstitial compounds?
    (a) They are very hard
    (b) They are chemically very reactive
    (c) They retain metallic conductivity
    (d) They have higher melting points than the pure metals

    Hint

    Recall the four properties listed in NCERT.

    Answer

    (b) Interstitial compounds are chemically inert.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define a transition element.Show answer

   An element that has an incompletely filled d subshell in its ground state or in any of its common oxidation states.

2. 2.Write the configurations of Cr and Cu.Show answer

   Cr: [Ar]3d⁵4s¹; Cu: [Ar]3d¹⁰4s¹.

3. 3.Which 3d element shows the most oxidation states?Show answer

   Manganese, from +2 to +7.

4. 4.Why is Mn³⁺ a strong oxidising agent?Show answer

   Mn³⁺ (d⁴) gains an electron to become Mn²⁺, which has the stable half-filled d⁵ configuration.

5. 5.Give the spin-only formula for magnetic moment.Show answer

   μ = √(n(n+2)) BM, n = number of unpaired electrons.

6. 6.Why is Sc³⁺ colourless?Show answer

   It has no d electrons (3d⁰), so no d–d transition is possible.

7. 7.What is the Cr–O–Cr bond angle in the dichromate ion?Show answer

   126°.

8. 8.What is the product of permanganate reduction in neutral or faintly alkaline solution?Show answer

   MnO₂.

9. 9.Name three lanthanoid ions in unusual oxidation states and why they form.Show answer

   Ce⁴⁺ (f⁰), Eu²⁺ (f⁷) and Yb²⁺ (f¹⁴): empty, half-filled or full f subshells.

10. 10.What is mischmetall?Show answer

    An alloy of about 95% lanthanoid metal and 5% iron with traces of S, C, Ca and Al, used in lighter flints, bullets and shells.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why Zn, Cd and Hg are not regarded as transition elements?
• ☐Can you explain why the enthalpy of atomisation is highest around the middle of the 3d series?
• ☐Can you explain why Cr²⁺ is reducing and Mn³⁺ oxidising?
• ☐Can you explain why Cu⁺ disproportionates in water, with the equation?
• ☐Can you calculate μ for Ti³⁺, Cr³⁺, Mn²⁺, Fe²⁺ and Cu²⁺?
• ☐Can you write the three steps for making K₂Cr₂O₇ from chromite?
• ☐Can you balance permanganate reactions with Fe²⁺, oxalate, nitrite and iodide?
• ☐Can you state the cause and two consequences of the lanthanoid contraction and compare lanthanoids with actinoids?

The d- and f-Block Elements: frequently asked questions