NCERT Exercises and Solutions: The d- and f-Block Elements
🎓 Class 12ChemistryCBSETheoryCh 4 – The d- and f-Block Elements⏱ ~8 min
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NCERT Exercises and Solutions: The d- and f-Block Elements
Chapter 8 Summary — Key Ideas in One Glance
The d- and f-blocks together comprise more than half the periodic table. They host the metals on which industrial civilisation rests (Fe, Cu, Cr, Mn, Ti, V, Pt) and the radioactive nuclei that power reactors (U, Pu).
d-Block Position
Groups 3–12 in 4 series (3d, 4d, 5d, 6d). General config (n-1)d¹⁻¹⁰ ns⁰⁻². Anomalies: Cr 3d⁵4s¹, Cu 3d¹⁰4s¹, Pd 4d¹⁰5s⁰.
Atomic Radii
Smooth decrease across each series; plateau in middle (Mn–Cu). 4d ≈ 5d (lanthanoid contraction). Density rises across each row.
Ionisation Enthalpies
Slow rise across the series. Anomalies at d⁵ and d¹⁰. ΔᵢH₂ very high for Cr and Cu (loss destroys d⁵/d¹⁰).
Oxidation States
Maximum in middle of series (Mn: +2 to +7). Heavier members of a group favour higher oxidation state (W(VI) > Cr(VI)).
NCERT Exercises — Full Solutions (Q 4.1 to Q 4.38)
Q 4.1 Write down the electronic configuration of: (i) Cr³⁺ (ii) Pm³⁺ (iii) Cu⁺ (iv) Ce⁴⁺ (v) Co²⁺ (vi) Lu²⁺ (vii) Mn²⁺ (viii) Th⁴⁺.
(i) Cr³⁺: [Ar] 3d³
(ii) Pm³⁺: [Xe] 4f⁴
(iii) Cu⁺: [Ar] 3d¹⁰
(iv) Ce⁴⁺: [Xe] 4f⁰ (noble-gas core)
(v) Co²⁺: [Ar] 3d⁷
(vi) Lu²⁺: [Xe] 4f¹⁴ 5d¹
(vii) Mn²⁺: [Ar] 3d⁵
(viii) Th⁴⁺: [Rn] 5f⁰ (noble-gas core)
Q 4.2 Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?
Mn²⁺ has the electronic configuration [Ar] 3d⁵ — a half-filled d-shell with maximum exchange energy. Removing a third electron to form Mn³⁺ would destroy this special stability, requiring extra energy. Fe²⁺ has [Ar] 3d⁶; oxidising it to Fe³⁺ ([Ar] 3d⁵) actually leads to a more stable, half-filled configuration. Hence Mn²⁺ resists oxidation to Mn³⁺ while Fe²⁺ is readily oxidised to Fe³⁺.
Q 4.3 Explain briefly how the +2 state becomes more stable in the first half of the first transition row with increasing atomic number.
Across the first transition series, the third ionisation enthalpy (IE₃) rises sharply. By Mn the IE₃ is so large that the M³⁺ state becomes harder to reach, while M²⁺ becomes more accessible. Mn²⁺ is also stabilised by its d⁵ half-filled configuration. So in the first half of the row, going from Sc → Mn the +2 state becomes increasingly stable relative to higher states.
Q 4.4 To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate.
Electronic configurations strongly influence stability. Half-filled (d⁵) and fully filled (d¹⁰) configurations are extra-stable. Examples: Mn²⁺ (d⁵) is unusually stable; Fe³⁺ (d⁵) is more stable than Fe²⁺ in air. Zn²⁺ (d¹⁰) is the only stable state of Zn. Cu²⁺ (d⁹) is more stable than Cu⁺ (d¹⁰) in aqueous solution because Cu²⁺ has a much larger hydration enthalpy that compensates for its higher ionisation energy. So electronic stability often "wins" in solution chemistry, but lattice and hydration energies matter too.
Q 4.5 What may be the stable oxidation state of the transition element with the following d electron configurations: 3d³, 3d⁵, 3d⁸, 3d⁴?
3d³ → V²⁺ (3d³) and Cr³⁺ (3d³); Cr³⁺ in solution is the most stable.
3d⁵ → Mn²⁺ (d⁵) and Fe³⁺ (d⁵) are both half-filled and very stable.
3d⁸ → Ni²⁺ (d⁸) is the stable form.
3d⁴ → no special stability; Cr²⁺ (d⁴) and Mn³⁺ (d⁴) both exist but are easily converted (Cr²⁺ → Cr³⁺ as reductant; Mn³⁺ → Mn²⁺ as oxidant).
Q 4.6 Name the oxometal anions of the first transition series in which the metal exhibits the oxidation state equal to its group number.
Group 5 (V): VO₄³⁻ — V in +5
Group 6 (Cr): CrO₄²⁻ and Cr₂O₇²⁻ — Cr in +6
Group 7 (Mn): MnO₄⁻ — Mn in +7
In each case the oxidation state of the metal equals the group number.
Q 4.7 What is lanthanoid contraction? What are its consequences?
Lanthanoid contraction is the steady decrease in atomic and ionic radii of the lanthanoids from Ce to Lu, caused by imperfect shielding of one 4f electron by another as nuclear charge increases.
Consequences:
Atomic radii of 4d and 5d series elements become similar (Zr 160 pm ≈ Hf 159 pm) — Hf and Zr are very hard to separate.
Basic strength of Ln(OH)₃ decreases from La(OH)₃ to Lu(OH)₃ (smaller, more polarising Ln³⁺).
Increasing covalent character of Ln–O bonds across the series.
Similar chemical behaviour of all lanthanoids — separation requires ion-exchange chromatography.
Q 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 transition elements?
Characteristics: high melting/boiling points; high tensile strength; ductile and malleable; show variable oxidation states; form coloured compounds; form numerous complex ions; many are paramagnetic; act as catalysts; form interstitial compounds and alloys.
Why "transition": their position lies between the strongly electropositive s-block and the largely non-metallic p-block — chemically transitional in character. By IUPAC, an element is a transition element if its atom or any of its common ions has an incomplete d sub-shell.
Not regarded as transition elements: Zn, Cd and Hg (group 12) have completely filled d¹⁰ both as atoms and in their common +2 ions — they fail the IUPAC definition.
Q 4.9 In what way is the electronic configuration of the transition elements different from that of non-transition elements?
Non-transition elements have configurations of the type ns¹⁻² (s-block) or ns² np¹⁻⁶ (p-block) — no electrons in the (n-1)d sub-shell. Transition elements have (n-1)d¹⁻¹⁰ ns⁰⁻² — partly filled inner d-orbitals. Because of this, transition elements show variable oxidation states, paramagnetism, coloured ions, complex formation and catalytic properties not generally seen in non-transition elements.
Q 4.10 What are the different oxidation states exhibited by the lanthanoids?
The most common oxidation state of all lanthanoids is +3. A few elements also show +2 (Eu²⁺ 4f⁷, Yb²⁺ 4f¹⁴, Sm²⁺ 4f⁶) and +4 (Ce⁴⁺ 4f⁰, Pr⁴⁺, Nd⁴⁺, Tb⁴⁺ 4f⁷, Dy⁴⁺), all driven by the extra stability of empty (f⁰), half-filled (f⁷) or fully filled (f¹⁴) f-sub-shells.
Q 4.11 Explain giving reasons: (i) paramagnetism (ii) high enthalpies of atomisation (iii) coloured compounds (iv) catalytic activity.
Paramagnetism — most transition-metal ions have unpaired d-electrons; each unpaired electron contributes a magnetic moment, attracting the substance into a magnetic field.
High ΔₐH — large number of unpaired d/s electrons engage in strong metallic bonding; breaking the lattice requires substantial energy.
Coloured compounds — d-d electronic transitions in the visible region; the colour seen is the complement of the wavelength absorbed.
Catalytic activity — (a) availability of multiple oxidation states allows electron-shuttle catalysis; (b) surface adsorption of reactants on metals via 3d/4s orbitals weakens reactant bonds and lowers activation energy.
Q 4.12 What are interstitial compounds? Why are such compounds well known for transition metals?
Interstitial compounds form when small atoms (H, B, C, N) occupy the gaps (interstices) between metal atoms in a crystal lattice. They are typically non-stoichiometric (e.g. TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇).
Transition metals form them readily because (i) their lattices have suitable-sized voids, and (ii) the metal atoms supply enough valence electrons to bond with the trapped non-metal atoms. Properties: very hard, high melting points, retain metallic conductivity, chemically inert.
Q 4.13 How is the variability in oxidation states of transition metals different from that of non-transition metals? Illustrate.
In transition metals, oxidation states usually differ by 1 unit (e.g. V²⁺, V³⁺, V⁴⁺, V⁵⁺ — successive states by one electron). This is because (n-1)d and ns electrons are close in energy, allowing fine-grained loss.
In non-transition (p-block) elements, oxidation states usually differ by 2 units (e.g. Sn²⁺ and Sn⁴⁺; Pb²⁺ and Pb⁴⁺) because of the inert-pair effect (the ns² pair is reluctant to ionise unless both electrons leave together).
Q 4.14 Describe the preparation of K₂Cr₂O₇ from iron chromite ore. Effect of increasing pH?
Q 4.17 Use the E° values for M²⁺/M and M³⁺/M²⁺ to comment on (i) stability of Fe³⁺ in acid vs Cr³⁺ or Mn³⁺ (ii) ease of oxidation of iron vs Cr or Mn metal.
(i) Stability of M³⁺: E°(Mn³⁺/Mn²⁺) = +1.5 V (very positive — Mn³⁺ is a strong oxidant, hence unstable). E°(Cr³⁺/Cr²⁺) = −0.4 V (Cr³⁺ stable; Cr²⁺ tends to be oxidised). E°(Fe³⁺/Fe²⁺) = +0.8 V (Fe³⁺ moderately stable, intermediate). So order of stability: Cr³⁺ > Fe³⁺ > Mn³⁺.
(ii) Ease of oxidation of metal: E°(M²⁺/M) values are Cr −0.9, Mn −1.2, Fe −0.4 V. The more negative the value, the easier the metal is to oxidise. So: Mn (easiest) > Cr > Fe.
Q 4.18 Predict which of the following will be coloured in aqueous solution: Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺, Co²⁺.
Coloured: Ti³⁺ (d¹), V³⁺ (d²), Mn²⁺ (d⁵), Fe³⁺ (d⁵), Co²⁺ (d⁷) — all have unpaired d-electrons enabling d-d transitions. Colourless: Sc³⁺ (d⁰) and Cu⁺ (d¹⁰) — no d-d transitions possible.
Q 4.19 Compare the stability of the +2 oxidation state for the elements of the first transition series.
+2 stability rises across the series, peaking at Mn²⁺ (d⁵) and remaining important up to Cu²⁺. Sc²⁺ is unknown (Sc prefers +3 = noble gas). Ti²⁺ and V²⁺ are strong reductants. Mn²⁺ is the most stable thanks to the half-filled d⁵. Fe²⁺ is easily oxidised to Fe³⁺ (also d⁵). Co²⁺, Ni²⁺ and Cu²⁺ are all stable. Zn²⁺ is the only state of Zn (d¹⁰). Order of relative stability: roughly Mn²⁺ > Cr²⁺/Fe²⁺ > V²⁺ > Ti²⁺ for the early half; Co²⁺ ≈ Ni²⁺ < Cu²⁺ < Zn²⁺ for the late half.
Q 4.20 Compare the chemistry of actinoids with that of lanthanoids — (i) electronic configuration (ii) atomic and ionic sizes (iii) oxidation state (iv) chemical reactivity.
Configuration: Lanthanoids — [Xe] 4f^x 5d⁰⁻¹ 6s². Actinoids — [Rn] 5f^x 6d⁰⁻¹ 7s². 5f orbitals less buried than 4f.
Sizes: Both show contraction across the series, but actinoid contraction is greater per element (5f shields worse).
Oxidation states: Lanthanoids almost exclusively +3 (rare +2/+4). Actinoids show wide range from +3 up to +7 (especially Np, Pu).
Reactivity: Both metals are reactive, react with H₂O/acids/halogens. Actinoids more reactive when finely divided. Actinoids are radioactive — dictates handling.
Q 4.21 Account for: (i) Cr²⁺ reducing, Mn³⁺ oxidising though both d⁴ (ii) Co²⁺ stable in water but easily oxidised with complexing agents (iii) d¹ very unstable.
Cr²⁺ → Cr³⁺ converts d⁴ → d³ (half-filled t₂g, octahedral CFSE bonus). Mn³⁺ → Mn²⁺ converts d⁴ → d⁵ (half-filled d-shell). Each species is moving towards a more stable d-config — driving Cr²⁺ to be oxidised (reducing) and Mn³⁺ to be reduced (oxidising).
In water, [Co(H₂O)₆]²⁺ is stable. With strong-field ligands (CN⁻, NH₃), the splitting Δ_o increases enough that the oxidation Co²⁺ → Co³⁺ becomes favourable; Co³⁺ low-spin d⁶ has very high CFSE and so is preferred.
d¹ ions (e.g. Ti³⁺) tend to be either oxidised (loss of the d-electron giving the stable d⁰ noble-gas configuration of Ti⁴⁺) or reduced. The single d-electron offers little exchange-energy advantage.
Q 4.22 What is meant by 'disproportionation'? Give two examples in aqueous solution.
Disproportionation is a redox reaction in which the same element in a single oxidation state is simultaneously oxidised to a higher state and reduced to a lower state.
Q 4.23 Which metal in the first transition series exhibits +1 most frequently and why?
Copper. Loss of one 4s electron from Cu (3d¹⁰4s¹) gives Cu⁺ with a stable, fully-filled 3d¹⁰ configuration. Although Cu⁺ disproportionates in water, it is well known in the solid state (Cu₂O, CuCl, CuBr, CuI) and in many complexes.
Q 4.24 Calculate the number of unpaired electrons in: Mn³⁺, Cr³⁺, V³⁺, Ti³⁺. Which is most stable in aqueous solution?
Mn³⁺ = [Ar] 3d⁴ → 4 unpaired
Cr³⁺ = [Ar] 3d³ → 3 unpaired
V³⁺ = [Ar] 3d² → 2 unpaired
Ti³⁺ = [Ar] 3d¹ → 1 unpaired
Most stable in aqueous solution = Cr³⁺ (d³, half-filled t₂g level — extra CFSE in octahedral hydrate).
Q 4.25 Give examples and reasons for: (i) lowest oxide is basic, highest is amphoteric/acidic (ii) highest state in oxides and fluorides (iii) highest state in oxoanions.
Low oxide = high ionic character, low charge → basic; e.g. MnO is basic. High oxide = covalent, multiply-bonded, high oxidation number → acidic; e.g. Mn₂O₇ gives HMnO₄. Cr₂O₃ (intermediate) is amphoteric.
O and F have small size + highest electronegativity. They form strong σ and π bonds with metals, stabilising the highest possible oxidation state. Examples: Mn₂O₇, OsO₄, MnF₄, OsF₆, CrF₆.
The metal centre is surrounded by tetrahedral O atoms with multiple π-bonding from O 2p → metal d, stabilising the high charge. Examples: MnO₄⁻ (Mn⁷⁺), CrO₄²⁻ / Cr₂O₇²⁻ (Cr⁶⁺), VO₄³⁻ (V⁵⁺).
Q 4.26 Indicate the steps in the preparation of: (i) K₂Cr₂O₇ from chromite ore (ii) KMnO₄ from pyrolusite ore.
Q 4.27 What are alloys? Name an alloy containing lanthanoids and its uses.
An alloy is a homogeneous mixture (usually solid solution) of two or more metals — or of metals and non-metals — which often has properties superior to its components.
Example containing lanthanoids: Mischmetall ≈ 95% mixed Ln (mainly Ce, La) + 5% Fe + traces of S, C, Ca, Al. Uses: flints in cigarette lighters and gas-stove ignitors (it sparks on friction); incendiary tracer rounds; alloying additive in Mg-based alloys to improve high-temperature strength.
Q 4.28 What are inner transition elements? Which atomic numbers are inner transition: 29, 59, 74, 95, 102, 104?
Inner transition elements are the f-block elements — lanthanoids (4f, Z 58–71) and actinoids (5f, Z 90–103). Their valence electrons enter inner f-orbitals.
From the list: 59 (Pr — lanthanoid), 95 (Am — actinoid), 102 (No — actinoid) are inner transition elements. Z 29 (Cu) and Z 74 (W) are d-block; Z 104 (Rf) is a transactinide d-block element.
Q 4.29 The chemistry of actinoids is not as smooth as that of lanthanoids. Justify with examples of oxidation states.
Lanthanoids have a single dominant +3 state (only Eu, Yb show +2; Ce, Tb show +4). The chemistry is therefore highly regular across the series.
In contrast, actinoid oxidation states are highly variable and irregular: Th shows +3, +4; Pa: +3, +4, +5; U: +3, +4, +5, +6; Np: +3 to +7; Pu: +3 to +7; Am: +3 to +6; Cm onwards: mostly +3 only. The maximum state climbs to +7 at Np then falls. This irregular pattern arises because the 5f, 6d and 7s sub-shells lie at comparable energies, allowing many different electron-loss patterns. Couple this with radioactivity (only nanogram quantities of late actinoids), and their chemistry becomes far less smooth.
Q 4.30 Last element in the actinoid series? Write its electronic configuration and possible oxidation state.
The last actinoid is lawrencium (Lr, Z = 103).
Configuration: [Rn] 5f¹⁴ 6d¹ 7s².
Possible oxidation state: +3 (loss of the 6d¹ + 7s² electrons gives Lr³⁺ = [Rn] 5f¹⁴, a fully-filled f-shell — extra-stable). Lr does not show higher states because of the very large IE₄ for breaking the f¹⁴ shell.
Q 4.31 Use Hund's rule to derive the electronic configuration of Ce³⁺ and calculate its spin-only magnetic moment.
Ce (Z = 58) atom = [Xe] 4f¹ 5d¹ 6s². Removing 3 electrons (6s², then 5d¹): Ce³⁺ = [Xe] 4f¹. One unpaired electron → n = 1.
μ = √[n(n+2)] = √[1(3)] = √3 = 1.73 BM
Q 4.32 Members of the lanthanoid series with +4 and +2 oxidation states; correlate with electronic configuration.
+4 states: Ce⁴⁺ (4f⁰), Pr⁴⁺ (4f¹), Tb⁴⁺ (4f⁷), Dy⁴⁺ (4f⁸ — close to 4f⁷), Nd⁴⁺ (4f²) — usually only in oxides MO₂. Driven by stability of empty (4f⁰) or half-filled (4f⁷) configurations. +2 states: Eu²⁺ (4f⁷), Yb²⁺ (4f¹⁴), Sm²⁺ (4f⁶ — near half-filled). Driven by stability of half-filled (4f⁷) or fully filled (4f¹⁴) configurations.
Q 4.33 Compare actinoids and lanthanoids: (i) configuration (ii) oxidation states (iii) chemical reactivity.
(See answer to Q 4.20 for full comparison.) Briefly: Lanthanoids — [Xe] 4f^x 6s², common +3 only, less reactive. Actinoids — [Rn] 5f^x 7s², variable +3 to +7, highly reactive (especially when finely divided), all radioactive.
Q 4.34 Write the electronic configurations of elements with atomic numbers 61, 91, 101 and 109.
Z = 61 → Pm (Promethium): [Xe] 4f⁵ 6s²
Z = 91 → Pa (Protactinium): [Rn] 5f² 6d¹ 7s²
Z = 101 → Md (Mendelevium): [Rn] 5f¹³ 7s²
Z = 109 → Mt (Meitnerium): [Rn] 5f¹⁴ 6d⁷ 7s²
Q 4.35 Compare 1st-series transition metals with 2nd and 3rd in the same group: configuration, oxidation states, IE, atomic sizes.
Configuration: 1st row uses 3d/4s; 2nd row 4d/5s (more exceptions, e.g. Mo 4d⁵5s¹); 3rd row 5d/6s (also irregular, Pt 5d⁹6s¹).
Oxidation states: Heavier members favour higher oxidation states — Cr(VI) is strong oxidant, but Mo(VI) and W(VI) are stable. Lower oxidation states common only for 1st-row metals; rare in 2nd/3rd row.
Ionisation enthalpies: Generally higher for 2nd/3rd-row elements (relativistic effects increase IE). The gap between IE₁ and IE₂ is smaller for heavier members.
Atomic sizes: 4d > 3d (one extra shell), but 5d ≈ 4d (lanthanoid contraction). Examples: Zr 160 pm ≈ Hf 159 pm; Nb 146 pm ≈ Ta 146 pm.
Q 4.36 Number of 3d electrons in: Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺, Cu²⁺. Show the octahedral hydrate distribution.
Ion
3d electrons
Octahedral high-spin distribution (t₂g, eg)
Ti²⁺
2
(↑)(↑)( ) — 2 unpaired in t₂g
V²⁺
3
(↑)(↑)(↑) — 3 unpaired in t₂g
Cr³⁺
3
(↑)(↑)(↑) — 3 unpaired
Mn²⁺
5
(↑)(↑)(↑) (↑)(↑) — 5 unpaired
Fe²⁺
6
(↑↓)(↑)(↑) (↑)(↑) — 4 unpaired (high-spin)
Fe³⁺
5
(↑)(↑)(↑) (↑)(↑) — 5 unpaired
Co²⁺
7
(↑↓)(↑↓)(↑) (↑)(↑) — 3 unpaired
Ni²⁺
8
(↑↓)(↑↓)(↑↓) (↑)(↑) — 2 unpaired
Cu²⁺
9
(↑↓)(↑↓)(↑↓) (↑↓)(↑) — 1 unpaired
Q 4.37 Comment: elements of the first transition series possess many properties different from those of heavier transition elements.
3d-row metals are smaller than 4d/5d members; bond lengths and unit cells are shorter.
3d-row metals have lower enthalpies of atomisation than 4d/5d (less extensive metal–metal bonding) — hence lower melting points generally.
3d ions tend to be high-spin in weak fields (water); heavier d-block ions are usually low-spin even in weak fields, due to larger Δ.
Lower oxidation states (+2, +3) dominate in 3d row; higher states (+4, +6) dominate in 4d/5d row.
3d ions are more often coloured and paramagnetic; many 4d/5d complexes are diamagnetic and weakly coloured.
3d row has more well-developed aqua-cation chemistry; 4d/5d more often appear as oxocations or in covalent compounds.
Q 4.38 Magnetic moment values: K₄[Mn(CN)₆] = 2.2 BM; [Fe(H₂O)₆]²⁺ = 5.3 BM; K₂[MnCl₄] = 5.9 BM. Comment.
K₄[Mn(CN)₆] contains [Mn(CN)₆]⁴⁻, i.e. Mn²⁺ (d⁵) with strong-field CN⁻ ligands — pairing occurs, leaving only 1 unpaired electron; predicted μ = √3 = 1.73 BM, observed 2.2 BM. Consistent with low-spin d⁵.
[Fe(H₂O)₆]²⁺ contains Fe²⁺ (d⁶) with weak-field H₂O — high-spin with 4 unpaired electrons; predicted μ = √24 = 4.90 BM, observed 5.3 BM (slight orbital contribution). Consistent with high-spin d⁶.
Conclusion: ligand strength controls whether a d⁵/d⁶ centre is high- or low-spin, with profound impact on the magnetic moment.
Quick Concept Recall L1 RememberL2 Understand
Pick a topic and the simulator returns the one-line answer commonly required in objective questions.
Pick a concept to begin.
Activity 8.5 — Build Your Chapter Concept MapL5 EvaluateL6 Create
Aim: Synthesise the entire Chapter 8 into a single connected concept map.
Procedure:
Place "d- and f-Block" at the centre of a page.
Branch outwards into d-block and f-block.
From d-block, branch to: position, configuration, properties (radii, IE, oxidation states, magnetism, colour, complexes), important compounds (K₂Cr₂O₇, KMnO₄), and applications.
From f-block, branch to: lanthanoids and actinoids; under each, write configuration, oxidation states, contraction, key uses.
Draw cross-links: e.g. lanthanoid contraction → 4d ≈ 5d size in d-block; variable oxidation states → catalysis.
Reflect: Which two concepts appear in the most number of links? Why are they central?
The two most-linked concepts are usually "partly-filled d-orbitals" and "variable oxidation states". They sit at the heart of catalysis, magnetism, colour, complex formation, and the very definition of "transition element". Almost every phenomenon in the chapter can be traced back to one of these two ideas — making them the conceptual anchors of d-block chemistry.
Final Mixed Practice — Competency-Based Questions L3L4
Stimulus: A school chemistry club prepares a 0.02 M KMnO₄ solution and uses it to titrate FeSO₄ in dilute H₂SO₄. They also test K₂Cr₂O₇ as an alternative oxidant for the same Fe²⁺ titration.
Q1. (MCQ) Which of these is the highest oxidation state of Mn known in a chemical compound?
(a) +4 in MnO₂
(b) +6 in K₂MnO₄
(c) +7 in KMnO₄
(d) +5 in Mn(III) compounds
(c) +7 in KMnO₄ — equal to its group number (7).
Q2. (MCQ) The colour of [Cu(H₂O)₆]²⁺ is blue because:
(a) Cu²⁺ is paramagnetic
(b) Cu²⁺ has 1 unpaired d-electron undergoing d-d transition
(c) H₂O is blue
(d) Cu²⁺ has fully filled d¹⁰
(b) Cu²⁺ is d⁹ → one unpaired d-electron undergoes a d-d transition absorbing in the orange-yellow region; the complementary blue is transmitted.
Q3. (SA) One mole of KMnO₄ in acidic medium can oxidise how many moles of (i) Fe²⁺ and (ii) C₂O₄²⁻?
Q4. (LA) Why is the third ionisation enthalpy of Mn unusually high but that of Fe relatively low?
Mn²⁺ has the configuration [Ar] 3d⁵ — half-filled, exceptionally stable. Removing a third electron destroys this stability, costing extra energy → high IE₃. Fe²⁺ has [Ar] 3d⁶; removing the sixth d-electron creates the stable d⁵ configuration of Fe³⁺, so IE₃ is comparatively low.
Q5. (HOT) A student finds that adding NaOH to an orange solution of K₂Cr₂O₇ turns it yellow. Subsequent addition of dilute HCl restores the orange colour. Explain.
The system involves the equilibrium: Cr₂O₇²⁻ + 2 OH⁻ ⇌ 2 CrO₄²⁻ + H₂O. Adding NaOH shifts the equilibrium to the right (yellow CrO₄²⁻). Adding HCl removes OH⁻ (forms water), shifting back to orange Cr₂O₇²⁻. The Cr oxidation state stays +6 throughout — only the degree of condensation changes with pH.
Final Mixed Practice — Assertion–Reason L4L5
Choose: A) Both A and R true and R explains A · B) Both true but R does not explain A · C) A true, R false · D) A false, R true.
Assertion (A): KMnO₄ is a stronger oxidising agent in acidic medium than in alkaline medium.
Reason (R): In acid medium, MnO₄⁻ accepts 5 electrons (Mn⁷⁺ → Mn²⁺) and has E° = +1.52 V, much greater than the +0.56 V for the alkaline reduction (Mn⁷⁺ → Mn⁶⁺).
A. Both true, R explains A.
B. Both true, R does not explain A.
C. A true, R false.
D. A false, R true.
Answer: A. The higher reduction potential in acid medium reflects greater oxidising power.
Assertion (A): The maximum oxidation state shown by a 3d-series element equals its group number up to manganese.
Reason (R): Up to Mn the maximum state involves losing all the available 4s and 3d electrons.
A. Both true, R explains A.
B. Both true, R does not explain A.
C. A true, R false.
D. A false, R true.
Answer: A. e.g. Mn (group 7) reaches +7 in MnO₄⁻ by losing 2 (4s) + 5 (3d) electrons. Beyond Mn, IE values become too high to lose all d-electrons.
Assertion (A): All actinoids are radioactive while only one lanthanoid is.
Reason (R): Most actinoid nuclei have neutron-to-proton ratios outside the band of stability.
A. Both true, R explains A.
B. Both true, R does not explain A.
C. A true, R false.
D. A false, R true.
Answer: A. Promethium (Z = 61) is the only naturally radioactive lanthanoid (no stable isotope). All actinoids (Z = 89 to 103) lie in the heavy-nuclei region where stable isotopes do not exist.
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Frequently Asked Questions - NCERT Exercises and Solutions: The d- and f-Block Elements
What are the key NCERT exercise types in Chapter 4 The d- and f-Block Elements?
NCERT Class 12 Chemistry Chapter 4 The d- and f-Block Elements exercises cover definitions, structure-property relationships, reaction mechanisms, numerical problems, and predict-the-product questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should focus on reaction mechanisms, IUPAC nomenclature, and stoichiometric reasoning to score full marks.
How should students approach reaction mechanism questions in The d- and f-Block Elements?
For reaction mechanism questions in NCERT Class 12 Chemistry Chapter 4 The d- and f-Block Elements: (1) identify the substrate, reagent, solvent, and conditions, (2) draw curly-arrow electron movement at each step, (3) label intermediates (carbocation, carbanion, free radical, transition state), (4) state stereochemistry where relevant. The MyAiSchool solutions show full curly-arrow mechanisms for every multi-step reaction.
What are the most-asked CBSE board questions from Chapter 4?
From NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements), the most-asked CBSE board questions test conceptual understanding, structure-property logic, distinguishing tests, IUPAC naming, and short numerical problems. 5-mark questions usually combine structural reasoning + mechanism + application. The MyAiSchool exercise set tags each question by mark weight and Bloom level for prioritized prep.
How do I balance chemical equations in NCERT exercises?
For balancing equations in NCERT Class 12 Chemistry Chapter 4: (1) write skeletal equation with correct formulas, (2) balance atoms other than H, O first, (3) balance O, then H (using H2O for organic reactions; or in acidic/basic medium for redox), (4) balance charge using e- in redox, (5) cross-check atom count and charge on both sides. The MyAiSchool solutions include balanced redox half-reactions where applicable.
What are common mistakes students make in Chapter 4 exercises?
Common mistakes in NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements) include: (1) incorrect IUPAC names (wrong locants or suffix), (2) skipping reaction conditions (catalyst, temperature, solvent), (3) wrong mechanism arrows, (4) sign errors in numerical (Ecell, Kc, ΔG), (5) forgetting stereochemistry (retention/inversion/racemisation). The MyAiSchool solutions flag these traps for each question.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Chemistry Chapter 4 The d- and f-Block Elements solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with chemical reasoning, curly-arrow mechanisms, IUPAC naming verifications, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Verified against NCERT and CBSE marking schemes.
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Part A - Physics (Section A) · Part A - Physics (Section B) · Part B - Chemistry (Section A) · Part B - Chemistry (Section B) · Part C - Mathematics (Section A) · Part C - Mathematics (Section B)
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Part A - Physics (Section A) · Part A - Physics (Section B) · Part B - Chemistry (Section A) · Part B - Chemistry (Section B) · Part C - Mathematics (Section A) · Part C - Mathematics (Section B)