NCERT Solutions for Class 12 Chemistry Chapter 4: The d- and f-Block Elements (NCERT 2026–27)
These Class 12 Chemistry Chapter 4 solutions cover The d- and f-Block Elements with every NCERT Intext Question and end-of-chapter Exercise reproduced verbatim and solved step by step. You will master the position and electronic configurations of the transition (d-block) and inner-transition (f-block) elements, their variable oxidation states, lanthanoid contraction, magnetic behaviour, colour, catalysis and the preparation and oxidising action of K2Cr2O7 and KMnO4 — everything you need for the CBSE 2026–27 board exam.
Class: 12Subject: ChemistryChapter: 4Topic: The d- and f-Block ElementsExercises: 38 + 10 IntextSession: 2026–27
The d-block comprises Groups 3–12, where the inner (n−1)d orbitals are progressively filled across four long periods (3d, 4d, 5d and 6d series). The f-block — the lanthanoids (Ce to Lu, filling 4f) and the actinoids (Th to Lr, filling 5f) — sits in a separate panel at the bottom of the periodic table. By IUPAC, a transition metal has an incompletely filled d subshell in its atom or in one of its common ions; Zn, Cd, Hg and Cn (d10) are therefore not regarded as transition metals. The presence of partly filled d/f orbitals gives these elements their hallmark properties: variable oxidation states, coloured ions, paramagnetism, catalytic activity, complex formation, interstitial compounds and alloys. The chapter closes with the chemistry of two industrially vital oxoanion salts — potassium dichromate (K2Cr2O7) and potassium permanganate (KMnO4).
Key Concepts & Definitions
Transition element: a metal with an incompletely filled d subshell either in the neutral atom or in one of its common oxidation states.
Lanthanoid contraction: the steady decrease in atomic and ionic radii from La to Lu caused by the poor shielding of one 4f electron by another as nuclear charge rises; it makes 4d and 5d congeners (e.g. Zr 160 pm, Hf 159 pm) almost identical in size.
Spin-only magnetic moment: μ = √[n(n+2)] BM, where n is the number of unpaired electrons; a single unpaired electron gives 1.73 BM.
Disproportionation: a species in one oxidation state changes into two species, one with a higher and one with a lower oxidation state (e.g. 2Cu+ → Cu2+ + Cu).
Interstitial compounds: non-stoichiometric solids formed when small atoms (H, C, N) occupy holes in a metal lattice (e.g. TiC, Fe3H); hard, high-melting, inert, metallic-conducting.
Alloy: a blend of metals with similar atomic radii (within ~15%), e.g. brass (Cu–Zn), bronze (Cu–Sn), steels.
4.1 Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?
ANSWERA transition element is one that has an incompletely filled d subshell either in the neutral atom or in one of its common oxidation states. Silver (Z = 47) commonly exhibits the +2 oxidation state (e.g. in AgO, AgF2); Ag2+ has a 4d9 configuration with an incompletely filled d subshell.Because at least one of its oxidation states has a partly filled d subshell, silver is correctly regarded as a transition element.
4.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e. 126 kJ mol−1. Why?
ANSWEREnthalpy of atomisation depends on the strength of the interatomic metallic bonding, which in turn depends on the number of unpaired (n−1)d electrons available for bonding.Zinc has the configuration 3d104s2 — the d subshell is completely filled, so no d electrons take part in metallic bonding. With the weakest interatomic interaction in the series, zinc has the lowest enthalpy of atomisation (126 kJ mol−1).
4.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
ANSWERManganese (Z = 25, 3d54s2) exhibits the largest number of oxidation states, from +2 to +7.This is because manganese has the maximum number of unpaired electrons (five 3d + the 4s pair), so it can lose a variable number of (3d + 4s) electrons one at a time and show every state from +2 up to +7, which equals the sum of its valence electrons.
4.4 The E°(M2+/M) value for copper is positive (+0.34 V). What is the possible reason for this? (Hint: consider its high ΔaH° and low ΔhydH°)
ANSWERE°(M2+/M) depends on the balance of three energy terms: the enthalpy of atomisation (ΔaH°), the sum of the first two ionisation enthalpies, and the hydration enthalpy of the M2+ ion (ΔhydH°).For copper the high ΔaH° and high ionisation enthalpy needed to convert Cu(s) to Cu2+(aq) are not compensated by its comparatively low (less negative) hydration enthalpy. The overall process is energetically unfavourable, so E° is positive — which is why copper does not liberate H2 from acids and is ‘noble’.
4.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
ANSWERThe irregular variation is mainly due to the varying degrees of stability of different 3dn configurations: the empty (d0), half-filled (d5) and fully-filled (d10) sets are extra stable.For example, the second ionisation enthalpy shows unusually high values for Cr and Cu because the M+ ions have the stable 3d5 and 3d10 configurations, which resist removal of a second electron. Removing one electron also alters the relative energies of the 4s and 3d orbitals, adding to the irregularity.
4.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
ANSWEROxygen and fluorine are the two most electronegative and smallest elements and are strong oxidising agents.Because of their small size and high electronegativity, they can oxidise the metal to its highest oxidation state by forming strong bonds — fluorine through high lattice/bond enthalpies and oxygen through its ability to form multiple (M=O) bonds. Hence the highest states appear as oxides (e.g. Mn2O7, CrO3) and fluorides (e.g. CrF6, VF5).
4.7 Which is a stronger reducing agent, Cr2+ or Fe2+, and why?
ANSWERCr2+ is a stronger reducing agent than Fe2+.On oxidation, Cr2+ (3d4) → Cr3+ (3d3): the product has a stable half-filled t2g3 configuration in an aqueous (octahedral) field, which is favourable, so Cr2+ readily gives up an electron. In contrast, Fe2+ (3d6) → Fe3+ (3d5) destroys an otherwise favourable arrangement; although d5 is stable, the change is less driven. Hence Cr2+ is the stronger reductant.
4.8 Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).
ANSWERStep 1 – Identify the ion. Z = 27 is cobalt; the divalent ion Co2+ has the configuration 3d7 (after losing the two 4s electrons).Step 2 – Count unpaired electrons. 3d7 → ↑↓ ↑↓ ↑ ↑ ↑ gives n = 3 unpaired electrons.Step 3 – Apply the spin-only formula. μ = √[n(n+2)] = √[3(3+2)] = √15 = 3.87 BM.
4.9 Explain why Cu+ ion is not stable in aqueous solutions.
ANSWERIn aqueous solution Cu+ undergoes disproportionation: 2 Cu+(aq) → Cu2+(aq) + Cu(s).Although forming Cu2+ requires a second ionisation, the much more negative hydration enthalpy of the smaller, doubly charged Cu2+(aq) more than compensates for it. The overall process is energetically favourable (positive E°), so Cu+ is unstable in water and converts to Cu2+ plus metallic copper.
4.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
ANSWERBoth contractions arise from the imperfect shielding of one electron by another in the same (f) subshell as nuclear charge increases.5f electrons shield the nuclear charge even more poorly than 4f electrons because the 5f orbitals are more diffuse and penetrate less. With weaker shielding from element to element, the effective nuclear charge felt by the outer electrons rises more sharply, so the size decreases more from one actinoid to the next than in the lanthanoids.
NCERT Exercises (4.1–4.38) — Solved
4.1 Write down the electronic configuration of: (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+ (vii) Mn2+ (viii) Th4+
ANSWER(i) Cr3+ (Z = 24, remove 4s1 + 2 of 3d): [Ar] 3d3(ii) Pm3+ (Z = 61): [Xe] 4f4(iii) Cu+ (Z = 29, remove 4s1): [Ar] 3d10(iv) Ce4+ (Z = 58): [Xe] 4f0 (noble-gas core)(v) Co2+ (Z = 27, remove 4s2): [Ar] 3d7(vi) Lu2+ (Z = 71): [Xe] 4f14 5d1(vii) Mn2+ (Z = 25, remove 4s2): [Ar] 3d5(viii) Th4+ (Z = 90): [Rn] 5f0 (noble-gas core)
4.2 Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
ANSWERMn2+ has the configuration 3d5, which is an exactly half-filled and therefore especially stable arrangement (maximum exchange energy).Oxidising Mn2+ (3d5) to Mn3+ (3d4) destroys this stable half-filled set, so it is difficult — Mn2+ resists oxidation. Conversely, oxidising Fe2+ (3d6) to Fe3+ (3d5) produces the stable half-filled 3d5 configuration, so Fe2+ is easily oxidised. Hence Mn2+ compounds are more stable towards oxidation than Fe2+.
4.3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
ANSWERAcross the first half of the 3d series (Sc → Mn), the nuclear charge increases steadily while electrons are added to the inner 3d orbitals. This raises the effective nuclear charge and the sum of the ionisation enthalpies needed for higher states, so removal of more than two electrons becomes progressively harder.As a result, the +2 state (formed by losing the two 4s electrons) becomes more and more favoured with increasing atomic number, the trend culminating at Mn2+, whose stable 3d5 half-filled configuration gives the +2 state particular stability.
4.4 To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
ANSWERElectronic configuration is a major factor: states that leave the ion with a stable empty (d0), half-filled (d5) or fully-filled (d10) configuration are especially stable.Examples: Sc3+ (3d0) and Zn2+ (3d10) are very stable because of their d0 and d10 configurations. Mn2+ and Fe3+ are stable owing to the half-filled 3d5 arrangement. Conversely, Mn3+ (3d4) is unstable and strongly oxidising. Thus configuration largely controls which oxidation states are favoured, though hydration and lattice energies also contribute.
4.5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4?
ANSWER3d3 (Vanadium, 3d34s2): stable states +2, +3, +4 and +5 (+4 and +5 are the most stable).3d5 (Chromium, 3d54s1): stable states +3, +4 and +6 (+3 and +6 being common).3d5 / 3d54s2 (Manganese): stable states +2, +4, +6 and +7.3d8 (Nickel, 3d84s2): stable state +2 (and +3 in some complexes).3d4: there is no element with a ground-state 3d4 configuration in the first transition series (Cr is 3d54s1), so this case does not correspond to a real ground-state atom.
4.6 Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
ANSWERVanadate VO3− (V in +5, Group 5), chromate CrO42− (Cr in +6, Group 6) and permanganate MnO4− (Mn in +7, Group 7).In each of these oxometal anions, the metal’s oxidation state equals its group number.
4.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
ANSWERLanthanoid contraction is the regular decrease in atomic and ionic radii of the elements from lanthanum to lutetium. It arises because, as the 4f orbitals are progressively filled, one 4f electron shields another from the increasing nuclear charge very imperfectly, so the effective nuclear charge rises and the size shrinks across the series.Consequences: (i) the 4d and 5d congeners have almost identical radii (e.g. Zr 160 pm, Hf 159 pm), so they have very similar properties and are difficult to separate and occur together in nature; (ii) there is a small but steady fall in basic strength of Ln(OH)3 from La to Lu; (iii) the radii of the third (5d) transition series resemble those of the second (4d) series rather than being larger.
4.8 What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
ANSWERCharacteristics: typical metallic properties (high tensile strength, malleability, ductility, high m.p./b.p., high thermal and electrical conductivity); variable oxidation states; coloured ions; paramagnetism; catalytic activity; and tendency to form complexes, interstitial compounds and alloys.Why “transition”: originally because their properties are transitional between the highly reactive s-block metals and the p-block elements; they form a bridge in the periodic table.Exceptions: Zn, Cd and Hg (Group 12) have completely filled d10 configurations in their ground state and in their common +2 ions, so they are not regarded as transition elements.
4.9 In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
ANSWERTransition elements have a general outer configuration (n−1)d1–10 ns1–2 — that is, the differentiating electron enters the inner (n−1)d subshell, which is incompletely filled in the atom or its ions.Non-transition elements are the s-block (ns1–2) and p-block (ns2 np1–6) elements, in which the last electron enters the outermost s or p subshell and the inner d orbitals are either empty or completely filled. This partly filled d (or f) subshell is what gives transition elements their distinctive chemistry.
4.10 What are the different oxidation states exhibited by the lanthanoids?
ANSWERThe characteristic and most stable oxidation state of all the lanthanoids is +3.In addition, a few show +2 and +4 states, favoured when they lead to an empty, half-filled or filled 4f subshell — e.g. Ce4+ (4f0) and Tb4+ (4f7); Eu2+ (4f7) and Yb2+ (4f14). These +2 and +4 ions revert to the common +3 state.
4.11 Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.(ii) The enthalpies of atomisation of the transition metals are high.(iii) The transition metals generally form coloured compounds.(iv) Transition metals and their many compounds act as good catalysts.
ANSWER(i) Paramagnetism arises from unpaired electrons. Transition metals and their ions have partly filled (n−1)d orbitals containing unpaired electrons, which give a net magnetic moment, so they are attracted by a magnetic field.(ii) They have a large number of unpaired (n−1)d electrons available for strong interatomic metallic bonding; greater the number of valence electrons, stronger the bonding, hence high enthalpies of atomisation.(iii) Their ions have partly filled d orbitals, so an electron can be promoted from a lower to a higher d level (d–d transition) by absorbing visible light; the complementary colour is transmitted, making the compounds coloured. Ions with d0 or d10 (e.g. Sc3+, Zn2+) are colourless.(iv) Their catalytic activity is due to (a) variable oxidation states, which allow them to form intermediate compounds and provide an alternative low-energy path, and (b) their ability to adsorb reactants on their surface (using 3d/4s electrons), increasing the reactant concentration and weakening bonds.
4.12 What are interstitial compounds? Why are such compounds well known for transition metals?
ANSWERInterstitial compounds are formed when small atoms such as H, C or N are trapped in the interstices (holes) of a metal lattice. They are usually non-stoichiometric and are neither typically ionic nor covalent, e.g. TiC, Mn4N, Fe3H, VH0.56, TiH1.7.They are well known for transition metals because these metals have large interatomic voids in their lattices and can accommodate small non-metal atoms without major structural change. Such compounds are very hard, have high melting points (higher than the pure metal), retain metallic conductivity and are chemically inert.
4.13 How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
ANSWERIn transition metals the oxidation states differ from one another by one unit, because the variation involves loss of (n−1)d electrons one at a time. For example, manganese shows +2, +3, +4, +5, +6 and +7.In non-transition (p-block) metals the variation is selective and the states usually differ by two units (due to the inert pair effect), e.g. tin shows +2 and +4, lead +2 and +4, and phosphorus +3 and +5.
4.14 Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
ANSWERStep 1 – Fusion with alkali in air: chromite ore is fused with sodium carbonate in free access of air to give sodium chromate.
4 FeCr2O4 + 8 Na2CO3 + 7 O2 → 8 Na2CrO4 + 2 Fe2O3 + 8 CO2Step 2 – Acidification: the yellow chromate solution is filtered and acidified with H2SO4 to give orange sodium dichromate.
2 Na2CrO4 + 2 H+ → Na2Cr2O7 + 2 Na+ + H2OStep 3 – Conversion to the potassium salt: treating sodium dichromate with KCl gives the less soluble potassium dichromate, which crystallises out as orange crystals.
Na2Cr2O7 + 2 KCl → K2Cr2O7 + 2 NaClEffect of increasing pH: the orange dichromate (Cr2O72−) is in equilibrium with the yellow chromate (CrO42−). Increasing the pH (adding OH−) shifts the equilibrium towards yellow chromate: Cr2O72− + 2 OH− → 2 CrO42− + H2O. The colour changes from orange to yellow.
4.15 Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i) iodide (ii) iron(II) solution and (iii) H2S
ANSWERIn acidic solution K2Cr2O7 is a strong oxidising agent, Cr being reduced from +6 to +3:
Cr2O72− + 14 H+ + 6 e− → 2 Cr3+ + 7 H2O (E° = 1.33 V)(i) With iodide (iodide oxidised to iodine):
Cr2O72− + 14 H+ + 6 I− → 2 Cr3+ + 3 I2 + 7 H2O(ii) With iron(II) (Fe2+ oxidised to Fe3+):
Cr2O72− + 14 H+ + 6 Fe2+ → 2 Cr3+ + 6 Fe3+ + 7 H2O(iii) With H2S (sulphide oxidised to sulphur):
Cr2O72− + 8 H+ + 3 H2S → 2 Cr3+ + 3 S + 7 H2O
4.16 Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions.
ANSWERPreparation: MnO2 (pyrolusite) is fused with KOH in presence of air or an oxidiser such as KNO3 to give the dark-green potassium manganate.
2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O
The manganate(VI) is then oxidised to permanganate(VII), either by disproportionation in acid (3 MnO42− + 4 H+ → 2 MnO4− + MnO2 + 2 H2O) or, commercially, by electrolytic oxidation in alkaline solution.(i) With iron(II) ions:
MnO4− + 8 H+ + 5 Fe2+ → Mn2+ + 4 H2O + 5 Fe3+(ii) With SO2:
2 MnO4− + 5 SO2 + 2 H2O → 2 Mn2+ + 5 SO42− + 4 H+(iii) With oxalic acid (at ~333 K):
2 MnO4− + 5 C2O42− + 16 H+ → 2 Mn2+ + 8 H2O + 10 CO2
4.17 For M2+/M and M3+/M2+ systems the E° values for some metals are as follows: Cr2+/Cr −0.9 V; Cr3+/Cr2+ −0.4 V; Mn2+/Mn −1.2 V; Mn3+/Mn2+ +1.5 V; Fe2+/Fe −0.4 V; Fe3+/Fe2+ +0.8 V. Use this data to comment upon: (i) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and (ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
ANSWER(i) Stability of M3+: A more positive E°(M3+/M2+) means M3+ is more easily reduced, i.e. less stable. The values are Cr3+/Cr2+ −0.4 V, Fe3+/Fe2+ +0.8 V, Mn3+/Mn2+ +1.5 V. So the order of stability of the +3 ion is Cr3+ > Fe3+ > Mn3+. Cr3+ is the most stable (t2g3), Mn3+ is least stable (strongly oxidising), and Fe3+ is intermediate.(ii) Ease of oxidation of the metal: The metal is oxidised in the M2+/M couple; a more negative E°(M2+/M) means the metal is more easily oxidised. The values are Mn2+/Mn −1.2 V, Cr2+/Cr −0.9 V, Fe2+/Fe −0.4 V. So the ease of oxidation of the metal is Mn > Cr > Fe — iron is the most difficult of the three to oxidise.
4.18 Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
ANSWERAn ion is coloured if it has partly filled (1–9) d orbitals that allow d–d transitions; it is colourless if d0 or d10.Coloured: Ti3+ (3d1), V3+ (3d2), Mn2+ (3d5), Fe3+ (3d5) and Co2+ (3d7) — all have unpaired/partly filled d electrons and undergo d–d transitions.Colourless: Sc3+ (3d0) and Cu+ (3d10) — with empty or completely filled d orbitals no d–d transition is possible.
4.19 Compare the stability of +2 oxidation state for the elements of the first transition series.
ANSWERThe +2 state is formed by losing the two 4s electrons. Across the series the E°(M2+/M) values become generally less negative (Sc/Ti most reducing → Cu positive), reflecting an increasing tendency for the +2 ion to be reduced to the metal — i.e. broadly increasing stability of M2+ from left to right, related to the rising sum of the first two ionisation enthalpies.There are notable exceptions: Mn2+ (3d5) and Zn2+ (3d10) are extra stable because of their half-filled and fully-filled configurations, while at the start Sc2+ is virtually unknown and Ti2+/V2+ are strongly reducing. Cu2+ is the only stable ion of copper in aqueous solution.
4.20 Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (ii) atomic and ionic sizes and (iii) oxidation state (iv) chemical reactivity.
ANSWER(i) Electronic configuration: lanthanoids fill 4f (general 4f1–14 5d0–1 6s2); actinoids fill 5f (5f1–14 6d0–1 7s2). The 5f orbitals are less buried than 4f, so they can take part in bonding more readily.(ii) Atomic and ionic sizes: both show a steady contraction across the series (lanthanoid and actinoid contraction), but the actinoid contraction is greater from element to element because of the poorer shielding by 5f electrons.(iii) Oxidation state: lanthanoids are almost restricted to +3 (with occasional +2 and +4); actinoids show a much wider range (+3 to +7, e.g. up to +7 for Np) because the 5f, 6d and 7s levels are close in energy.(iv) Chemical reactivity: lanthanoids are highly reactive (like Ca/Al) but their reactions are fairly regular; actinoids are also highly reactive and, in addition, radioactive, which makes their chemistry more complicated and harder to study.
4.21 How would you account for the following:
(i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising.(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.(iii) The d1 configuration is very unstable in ions.
ANSWER(i) Both Cr2+ and Mn3+ are d4. Cr2+ (d4) is readily oxidised to Cr3+ (d3, stable t2g3), so Cr2+ is strongly reducing. Mn3+ (d4) is readily reduced to Mn2+ (d5, stable half-filled), so Mn(III) is strongly oxidising.(ii) In water Co2+ is stable, but strong-field complexing ligands greatly stabilise the d6 Co3+ state through large crystal-field stabilisation energy, which more than offsets the high third ionisation energy. Hence Co(II) is easily oxidised to Co(III) in their presence.(iii) A d1 ion readily loses its single d electron to give the very stable noble-gas-like d0 configuration; the hydration/lattice energy released more than compensates for the ionisation energy, so the d1 state is unstable.
4.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
ANSWERDisproportionation occurs when a species in an intermediate oxidation state becomes unstable and changes simultaneously into a higher and a lower oxidation state.Example 1: 2 Cu+(aq) → Cu2+(aq) + Cu(s).Example 2: 3 MnO42− + 4 H+ → 2 MnO4− + MnO2 + 2 H2O (Mn(VI) → Mn(VII) + Mn(IV)).
4.23 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
ANSWERCopper exhibits the +1 oxidation state most frequently.In the +1 state copper attains the extra-stable, completely filled 3d10 configuration (Cu+: [Ar] 3d10), which makes the +1 state favourable.
4.24 Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution?
ANSWERMn3+ (3d4) → 4 unpaired electrons.Cr3+ (3d3) → 3 unpaired electrons.V3+ (3d2) → 2 unpaired electrons.Ti3+ (3d1) → 1 unpaired electron.Most stable in aqueous solution: Cr3+, owing to its stable half-filled t2g3 configuration in an octahedral field.
4.25 Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.(iii) The highest oxidation state is exhibited in oxoanions of a metal.
ANSWER(i) In a low oxidation state the metal has few positive charges, so the oxide is largely ionic and basic (e.g. MnO is basic). In a high oxidation state the metal is small and highly charged, the bonding becomes more covalent and the oxide is acidic (e.g. Mn2O7 is acidic); intermediate oxides such as Mn2O3/Cr2O3 are amphoteric.(ii) Oxygen and fluorine are small, highly electronegative and strong oxidisers, and oxygen can form multiple bonds; they can therefore oxidise the metal to its highest state, e.g. Mn in Mn2O7 (+7) and Cr in CrF6 (+6).(iii) In oxoanions the metal is surrounded by several oxygen atoms that can stabilise a very high charge through multiple M=O bonding, e.g. MnO4− (Mn +7), CrO42−/Cr2O72− (Cr +6).
4.26 Indicate the steps in the preparation of: (i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore.
ANSWER(i) K2Cr2O7 from chromite (FeCr2O4):
1. Fuse with Na2CO3 in air: 4 FeCr2O4 + 8 Na2CO3 + 7 O2 → 8 Na2CrO4 + 2 Fe2O3 + 8 CO2
2. Acidify the chromate: 2 Na2CrO4 + 2 H+ → Na2Cr2O7 + 2 Na+ + H2O
3. Treat with KCl: Na2Cr2O7 + 2 KCl → K2Cr2O7 + 2 NaCl (orange crystals).(ii) KMnO4 from pyrolusite (MnO2):
1. Fuse MnO2 with KOH and air/KNO3: 2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O
2. Oxidise the green manganate(VI) to purple permanganate(VII), either by disproportionation in acid (3 MnO42− + 4 H+ → 2 MnO4− + MnO2 + 2 H2O) or by electrolytic oxidation in alkaline solution.
4.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
ANSWERAn alloy is a homogeneous blend of two or more metals (or a metal with a non-metal) prepared by mixing the molten components; the atoms of one metal are distributed among those of another, usually when their radii are within ~15% of each other.An important lanthanoid-containing alloy is mischmetall, which is about 95% lanthanoid metal and ~5% iron, with traces of S, C, Ca and Al. It is used in Mg-based alloys to make bullets, shells and lighter flints (it is pyrophoric and produces sparks).
4.28 What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.
ANSWERInner transition elements are the f-block elements — the lanthanoids (Z = 58–71) and actinoids (Z = 90–103) — in which the inner 4f or 5f orbitals are progressively filled.Among the given atomic numbers, the inner transition elements are 59 (Pr), 95 (Am) and 102 (No). The others are: 29 (Cu) and 74 (W) are d-block transition metals, and 104 (Rf) is a d-block element.
4.29 The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
ANSWERThe lanthanoids are dominated by a single, regular +3 state (with only occasional +2/+4), making their chemistry smooth and predictable.The actinoids show a much wider and irregular range of oxidation states because the 5f, 6d and 7s levels are close in energy: the maximum state rises from +4 (Th) to +5 (Pa), +6 (U) and +7 (Np) and then falls in later elements. This variability, together with their radioactivity, makes actinoid chemistry far more complicated than that of the lanthanoids.
4.30 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
ANSWERThe last element of the actinoid series is lawrencium (Lr, Z = 103).Its electronic configuration is [Rn] 5f14 6d1 7s2.The expected and common oxidation state of lawrencium is +3, like the other actinoids.
4.31 Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.
ANSWERConfiguration: Ce (Z = 58) is [Xe] 4f1 5d1 6s2; on forming Ce3+ three electrons (6s2 and 5d1) are removed, giving Ce3+ : [Xe] 4f1 — one unpaired electron in the 4f subshell (Hund’s rule).Magnetic moment: n = 1, so μ = √[n(n+2)] = √[1(1+2)] = √3 = 1.73 BM.
4.32 Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
ANSWER+4 state: Ce, Pr, Nd, Tb and Dy. Ce4+ is favoured because it gives the empty 4f0 (noble-gas) configuration, and Tb4+ because it gives the half-filled 4f7 configuration.+2 state: Sm, Eu and Yb. Eu2+ is favoured because it gives the half-filled 4f7 configuration and Yb2+ because it gives the fully-filled 4f14 configuration.Thus the +2 and +4 states appear where they lead to a stable empty (4f0), half-filled (4f7) or fully-filled (4f14) f subshell.
4.33 Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity.
ANSWER(i) Electronic configuration: lanthanoids progressively fill the 4f orbitals (4f1–14 5d0–1 6s2); actinoids fill the 5f orbitals (5f1–14 6d0–1 7s2). The 5f orbitals are more exposed and can participate in bonding.(ii) Oxidation states: lanthanoids show mainly +3 (occasionally +2, +4); actinoids show a wider range, +3 to +7, because the 5f, 6d and 7s levels are close in energy.(iii) Chemical reactivity: both are highly reactive and electropositive; the lanthanoids react in a regular way, whereas the actinoids are radioactive and react in a more complex, less regular manner.
4.34 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
4.35 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: (i) electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes.
ANSWER(i) Electronic configurations: the three series fill 3d, 4d and 5d respectively (general (n−1)d1–10 ns1–2), but the heavier series show more exceptions because of the smaller energy gaps between (n−1)d and ns.(ii) Oxidation states: higher oxidation states are more stable in the 4d and 5d series than in the 3d series; e.g. in Group 6, Mo(VI) and W(VI) are more stable than Cr(VI), so Cr(VI) is a strong oxidiser whereas MoO3/WO3 are not.(iii) Ionisation enthalpies: generally the 5d elements have higher ionisation enthalpies than the 3d/4d elements, owing to lanthanoid contraction and poor shielding by intervening f electrons.(iv) Atomic sizes: sizes increase from the 3d to the 4d series, but the 5d series has almost the same radii as the 4d series (e.g. Zr 160 pm, Hf 159 pm) because of lanthanoid contraction; hence 4d and 5d congeners are chemically very similar.
4.36 Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
ANSWERIn a weak (water) octahedral field the 3d electrons fill the five d orbitals according to Hund’s rule (high spin) — singly first, then pairing.
Ion
No. of 3d electrons
Occupation of five 3d orbitals (high-spin)
Ti2+
2
t2g2 eg0 (2 unpaired)
V2+
3
t2g3 eg0 (3 unpaired)
Cr3+
3
t2g3 eg0 (3 unpaired)
Mn2+
5
t2g3 eg2 (5 unpaired)
Fe2+
6
t2g4 eg2 (4 unpaired)
Fe3+
5
t2g3 eg2 (5 unpaired)
Co2+
7
t2g5 eg2 (3 unpaired)
Ni2+
8
t2g6 eg2 (2 unpaired)
Cu2+
9
t2g6 eg3 (1 unpaired)
4.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
ANSWERThe 3d (first series) elements differ from the heavier 4d and 5d elements in several ways: (i) their atomic radii and enthalpies of atomisation are smaller, so metal–metal bonding is less common; (ii) lower oxidation states are more stable for 3d elements, whereas higher states are more stable for 4d/5d elements (e.g. Cr(VI) is a strong oxidiser but Mo(VI)/W(VI) are not).Also, the 3d elements form more high-spin (often paramagnetic) ionic compounds, while the heavier elements more often form low-spin and metal–metal-bonded compounds; and 4d/5d congeners closely resemble each other (lanthanoid contraction) but differ from the 3d member. Hence the statement is justified.
4.38 What can be inferred from the magnetic moment values of the following complex species?
ANSWER
Species
μ (BM)
Unpaired e− (n)
Inference
K4[Mn(CN)6]
2.2
≈ 1
Mn2+ (d5) with strong-field CN− → low-spin, inner-orbital (d2sp3) complex
[Fe(H2O)6]2+
5.3
≈ 4
Fe2+ (d6) with weak-field H2O → high-spin, outer-orbital (sp3d2) complex
K2[MnCl4]
5.9
≈ 5
Mn2+ (d5) with weak-field Cl− → high-spin tetrahedral (sp3) complex
Using μ = √[n(n+2)]: 2.2 BM → n ≈ 1 (low-spin), 5.3 BM → n ≈ 4 (high-spin), 5.9 BM → n ≈ 5 (high-spin). Thus the magnetic moment reveals the number of unpaired electrons, and hence whether the ligand is strong-field (low-spin) or weak-field (high-spin).
Extra Practice Questions
Short Answer Type Questions
Q1. Why do transition metals form a large number of complex compounds?
ANSWERBecause of their comparatively small ionic sizes, high ionic charges and the availability of vacant d orbitals of suitable energy for accepting lone pairs from ligands, transition metal ions readily form coordinate bonds and hence a large number of complexes.
Q2. Why is Zn2+ colourless while Cu2+ is coloured?
ANSWERCu2+ is 3d9 — it has partly filled d orbitals, so d–d transitions absorb visible light and it appears blue. Zn2+ is 3d10 (completely filled), so no d–d transition is possible and it is colourless.
Q3. Why does Cr2+ act as a reducing agent whereas Mn3+ acts as an oxidising agent (both being d4)?
ANSWERCr2+ (d4) is oxidised to the stable Cr3+ (d3, t2g3), so it is reducing; Mn3+ (d4) is reduced to the stable Mn2+ (d5, half-filled), so it is oxidising.
Q4. Why are the melting points of transition metals high, reaching a maximum about the middle of each series?
ANSWERTheir high melting points are due to strong interatomic metallic bonding involving (n−1)d electrons in addition to ns electrons. The maximum near the middle (around d5) corresponds to the greatest number of unpaired d electrons available for bonding.
Q5. Why is the E° value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+?
ANSWERBecause the change Mn3+ → Mn2+ (d4 → d5) produces the very stable half-filled d5 configuration (large third ionisation energy of Mn). This makes Mn3+ strongly oxidising and the +3 state of Mn of little importance.
Long Answer Type Questions
Q1. Account for the catalytic activity of transition metals and their compounds, giving suitable examples.
ANSWERTransition metals are good catalysts for two main reasons. First, they exhibit variable oxidation states, so they can form unstable intermediate compounds with the reactants and provide an alternative reaction path of lower activation energy; for example, Fe3+ catalyses the reaction between I− and S2O82− by cycling between Fe3+ and Fe2+. Second, in heterogeneous catalysis they adsorb reactant molecules on their surface using free valencies of 3d and 4s electrons, increasing the local concentration of reactants and weakening their bonds. Industrially important examples include V2O5 in the Contact process, finely divided iron in the Haber process for ammonia, and nickel in the hydrogenation of oils. Their ability to form complexes and change oxidation state easily underlies all these catalytic actions.
Q2. Discuss the lanthanoid contraction in detail and explain its consequences.
ANSWERLanthanoid contraction is the steady decrease in the atomic and ionic radii of the lanthanoids from La to Lu. As we move along the series the nuclear charge increases by one unit at each step, but the additional electron enters an inner 4f orbital. Because the 4f orbitals are diffuse and one 4f electron shields another from the nuclear charge very imperfectly, the effective nuclear charge experienced by the outer electrons rises steadily, drawing them inward and shrinking the size. The total contraction across the series is small but significant. Its consequences are far-reaching: (i) the 4d and 5d congeners (e.g. Zr–Hf, Nb–Ta) have almost identical radii and very similar properties, so they occur together and are extremely hard to separate; (ii) the basic strength of the hydroxides Ln(OH)3 decreases steadily from La to Lu; (iii) the contraction allows the lanthanoids themselves to be separated by ion-exchange because of the small, regular change in their ionic radii.
Q3. Describe the structure and oxidising behaviour of acidified KMnO4 with three examples.
ANSWERIn KMnO4 the permanganate ion MnO4− is tetrahedral, with manganese in the +7 oxidation state; it is diamagnetic (no unpaired electrons) and intensely purple due to charge-transfer. In acidic solution it is a powerful oxidising agent, Mn being reduced from +7 to +2: MnO4− + 8 H+ + 5 e− → Mn2+ + 4 H2O (E° = 1.52 V). Examples: (i) it oxidises Fe2+ to Fe3+: 5 Fe2+ + MnO4− + 8 H+ → Mn2+ + 4 H2O + 5 Fe3+; (ii) it oxidises oxalate to CO2: 5 C2O42− + 2 MnO4− + 16 H+ → 2 Mn2+ + 8 H2O + 10 CO2; (iii) it oxidises iodide to iodine: 10 I− + 2 MnO4− + 16 H+ → 2 Mn2+ + 8 H2O + 5 I2. Because of these reactions it is widely used as an oxidant in volumetric analysis and organic chemistry.
MCQs & Assertion–Reason
1. Which of the following is NOT regarded as a transition element?
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: Zinc is not regarded as a transition element.
Reason: Zinc has a completely filled 3d10 configuration in its ground state and in its common +2 ion.
A-R 2. Assertion: Mn2+ is more stable towards oxidation than Fe2+.
Reason: Mn2+ has a stable half-filled 3d5 configuration.
A-R 3. Assertion: Sc3+ ion is colourless in aqueous solution.
Reason: Sc3+ has a 3d0 configuration, so no d–d transition is possible.
A-R 4. Assertion: The actinoid contraction is greater from element to element than the lanthanoid contraction.
Reason: 5f electrons shield the nuclear charge more effectively than 4f electrons.
A-R 5. Assertion: Transition metals are good catalysts.
Reason: They show variable oxidation states and can form intermediate compounds with reactants.
Answer key: 1-(A), 2-(A), 3-(A), 4-(C), 5-(A).
Common Mistakes & Exam Tips
Watch out for these
Forgetting that ns electrons are removed before (n−1)d electrons when writing ion configurations — e.g. Fe2+ is 3d6, not 3d44s2.
Calling Zn, Cd, Hg transition metals — they are d-block but NOT transition elements (d10 in atom and common ion).
Mis-balancing redox equations — always balance charge and atoms (H+, H2O) on both sides.
Using molecular formulas instead of ionic equations when the question asks for ionic equations.
Treating actinoid chemistry like lanthanoid chemistry — actinoids show many more oxidation states and are radioactive.
How to score full marks in this chapter
Learn the spin-only formula μ = √[n(n+2)] and practise counting unpaired electrons for 3d ions. Memorise the three preparation steps and the key oxidising half-reactions of K2Cr2O7 (E° = 1.33 V) and KMnO4 (E° = 1.52 V) with correct subscripts and superscripts. For “account for” questions, always link the answer to the stability of d0, d5 or d10 configurations. State lanthanoid contraction and its consequences (Zr–Hf similarity, decreasing basicity of Ln(OH)3) precisely, as it is a frequent board question.
Frequently Asked Questions
What is Class 12 Chemistry Chapter 4 about?
Chapter 4, The d- and f-Block Elements, deals with the transition (d-block) and inner-transition (f-block) elements — their position in the periodic table, electronic configurations, variable oxidation states, colour, magnetic and catalytic properties, lanthanoid contraction, and the preparation and oxidising action of K2Cr2O7 and KMnO4.
Why are Zn, Cd and Hg not regarded as transition elements?
Because they have completely filled d10 configurations both in their ground state and in their common +2 oxidation states. A transition element must have an incompletely filled d subshell in the atom or in one of its ions, which these elements do not have.
What is lanthanoid contraction and why is it important?
It is the steady decrease in atomic and ionic radii from La to Lu, caused by the poor shielding of one 4f electron by another as nuclear charge increases. It makes the 4d and 5d congeners (e.g. Zr and Hf) almost identical in size and properties, so they are difficult to separate, and it lowers the basicity of Ln(OH)3 across the series.
Are these Class 12 Chemistry Chapter 4 solutions free?
Yes. All ClearStudy NCERT Solutions for Class 12 Chemistry are free and follow the official NCERT textbook for session 2026–27, with every Intext Question and Exercise solved step by step.
