Class 12 : Chemistry (English) – Chapter 4: The d- and f-Block Elements

EXPLANATION & SUMMARY

✨ Introductio
🔵 The d-block elements lie in the middle of the periodic table (Groups 3–12) and are also called transition elements because they show transition in properties from highly reactive s-block to less reactive p-block.
🟢 The f-block elements are placed separately at the bottom as lanthanoids (Z = 58–71) and actinoids (Z = 90–103), forming the inner transition elements.
🟠 These two blocks are important because they contribute to metallurgy, catalysis, alloys, electronic devices, and nuclear chemistry.

🌿 The d-Block Elements
🔹 Position and Electronic Configuration
💡 General electronic configuration: (n−1)d¹–¹⁰ ns⁰–²
✏ Example: Sc (Z=21) → [Ar] 3d¹ 4s²
➡ Key points:
In d-block, last electron enters (n−1)d orbital.
Variable oxidation states arise because both (n−1)d and ns orbitals have comparable energies.
Transition elements: those d-block elements that have partially filled d orbitals in their ground state or in common oxidation states.

🔹 General Characteristics
🧪 1. Variable Oxidation States
Due to similar energy of (n−1)d and ns orbitals.
Mn shows maximum (+2 to +7).
Stability depends on lattice enthalpy, hydration enthalpy, and electronic factors.
⚡ 2. Formation of Coloured Ions
Colour due to d–d transitions (electronic transitions between split d-orbitals in presence of ligands).
Colour intensity depends on ligands (Crystal Field Theory).
🔴 3. Catalytic Properties
d-block metals act as catalysts because they:
Can change oxidation states.
Provide surface for adsorption.
Examples: V₂O₅ in Contact process, Fe in Haber process.
🟡 4. Magnetic Properties
Paramagnetism due to unpaired d electrons.
Magnitude of magnetic moment (μ) = √(n(n+2)) BM (n = no. of unpaired e⁻).
Some compounds (like Mn²⁺, Fe³⁺) are strongly paramagnetic.
🔵 5. Alloy Formation
Close atomic sizes allow atoms to substitute each other.
Examples: Steel (Fe + C + others), brass (Cu + Zn), bronze (Cu + Sn).
🟢 6. Formation of Complex Compounds
Strong tendency to form complexes due to:
Small cationic size,
High nuclear charge,
Availability of vacant d orbitals.
Example: [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺.

🔹 Periodic Trends
💡 Across period (3d → 4d → 5d → 6d series):
Atomic radii: Decrease → slight increase due to electron–electron repulsion and lanthanoid contraction.
Ionisation enthalpy: Generally increases left to right.
Standard electrode potential: Depends on stability of oxidation states.

🔹 Group-wise Highlights
🟦 Group 3 (Sc, Y, La, Ac)
Show +3 oxidation state, form colourless ions.
🟩 Group 4 (Ti, Zr, Hf)
Stable +4 state.
TiO₂: Photocatalyst, used in paints.
🟨 Group 5 (V, Nb, Ta)
Exhibit +5 stable state.
V₂O₅: Catalyst in Contact process.
🟧 Group 6 (Cr, Mo, W)
Cr: +3 and +6 stable.
K₂Cr₂O₇: Oxidising agent.
🟥 Group 7 (Mn, Tc, Re)
Mn shows oxidation states +2 to +7.
KMnO₄: Powerful oxidising agent.
🟪 Group 8–10 (Fe, Co, Ni, Ru, Rh, Pd, Os, Ir, Pt)
Ferrous metals → important in metallurgy.
Fe: Forms haemoglobin.
Pt, Pd: Catalysts in hydrogenation.
🟫 Group 11 (Cu, Ag, Au)
Known as coinage metals.
Stable in +1 oxidation state.
Cu²⁺, Ag⁺ form coloured complexes.
⬛ Group 12 (Zn, Cd, Hg)
Show +2 state.
Filled d¹⁰ configuration → not true transition metals.
Hg forms amalgams.

🌿 The f-Block Elements
🔹 General Features
💡 f-block includes lanthanoids and actinoids.
General configuration: (n−2)f¹–¹⁴ (n−1)d⁰–¹ ns².
Placed separately to keep periodic table structure intact.

🔹 Lanthanoids (Z=58–71)
🟦 Common oxidation state: +3.
🟩 Gradual decrease in atomic size with increase in atomic number = lanthanoid contraction.
Cause: Poor shielding of f-electrons.
Consequences:
Similar properties among lanthanoids.
Difficulty in separation.
Contraction affects d-block (Zr and Hf size almost same).
💡 Uses of Lanthanoids
CeO₂: Oxidising agent, used in polishing glass.
Mischmetal (La + Ce + others): In lighter flints.
Nd: In lasers, magnets.

🔹 Actinoids (Z=90–103)
🟨 Show wide range of oxidation states (+3 to +6, sometimes +7).
🟧 Strongly paramagnetic due to 5f electrons.
🟥 Show actinoid contraction (similar to lanthanoids).
🟦 Most actinoids are radioactive.
💡 Important examples
Th, U, Pu → nuclear fuels.
Am, Cm → artificial elements used in research.

🔹 Comparison of Lanthanoids and Actinoids
🟢 Lanthanoids: Mostly +3 oxidation state, less variable.
🔵 Actinoids: More variable oxidation states.
🟠 Lanthanoids: Mostly stable, few radioactive.
🔴 Actinoids: Mostly radioactive.
💡 Both: Show contraction due to poor shielding of f-electrons.

📝 Summary
d-Block
d-block lies in groups 3–12, general config (n−1)d¹–¹⁰ns⁰–².
Properties: Variable oxidation states, coloured ions (d–d transitions), catalysis, paramagnetism, alloy formation, complex formation.
Trends: Atomic size decreases across, slight irregularities due to lanthanoid contraction.
Important compounds: KMnO₄, K₂Cr₂O₇, TiO₂, Cu²⁺ complexes, coinage metals.
f-Block
f-block → lanthanoids + actinoids, general config (n−2)f¹–¹⁴(n−1)d⁰–¹ns².
Lanthanoids: Stable +3 oxidation, lanthanoid contraction → similar properties. Uses: CeO₂ (polishing), mischmetal, Nd magnets.
Actinoids: Variable oxidation states, radioactive, nuclear fuels.
Actinoid contraction similar to lanthanoid contraction.
Comparison
Lanthanoids: Limited oxidation variability, mostly non-radioactive.
Actinoids: More oxidation variability, mostly radioactive.
Both show contraction, complex formation, paramagnetism.

🎯 Quick Recap
🟢 d-block: Variable oxidation states, coloured ions, catalysts, alloys.
🔵 Lanthanoids: +3 state, contraction, polishing, magnets.
🟠 Actinoids: Radioactive, nuclear fuels, wide oxidation states.
🔴 Both: Contraction → similar sizes, strong complex formation.

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QUESTIONS FROM TEXTBOOK

Question 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
💡 Rule: Remove ns electrons first, then (n−1)d / (n−2)f.
🔵 Cr³⁺: Cr = [Ar] 3d⁵ 4s¹ → Cr³⁺ = [Ar] 3d³
🟢 Pm³⁺: Pm = [Xe] 4f⁵ 6s² → Pm³⁺ = [Xe] 4f⁴
🟠 Cu⁺: Cu = [Ar] 3d¹⁰ 4s¹ → Cu⁺ = [Ar] 3d¹⁰
🔴 Ce⁴⁺: Ce = [Xe] 4f¹ 5d¹ 6s² → Ce⁴⁺ = [Xe]
🟡 Co²⁺: Co = [Ar] 3d⁷ 4s² → Co²⁺ = [Ar] 3d⁷
🔷 Lu²⁺: Lu = [Xe] 4f¹⁴ 5d¹ 6s² → Lu²⁺ = [Xe] 4f¹⁴ 5d¹
🔶 Mn²⁺: Mn = [Ar] 3d⁵ 4s² → Mn²⁺ = [Ar] 3d⁵
✴ Th⁴⁺: Th = [Rn] 6d² 7s² → Th⁴⁺ = [Rn]

Question 4.2
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?
Answer
🟦 Mn²⁺ = 3d⁵ (half-filled) → extra stability; oxidation to Mn³⁺ (3d⁴) breaks half-filled symmetry → unfavourable.
🟩 Fe²⁺ = 3d⁶; oxidation to Fe³⁺ (3d⁵) gains half-filled stability → favourable.
📌 Hence Mn²⁺ resists oxidation, Fe²⁺ is readily oxidised to Fe³⁺.

Question 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.
Answer
🔵 From Sc → Mn the common +2 state arises by loss of 4s² only, leaving (n−1)dⁿ unchanged.
🟢 With increasing nuclear charge, removal of an extra d-electron (to make +3 or higher) becomes progressively harder.
🟠 Therefore, relative stability shifts towards +2 (Ti²⁺ < V²⁺ < Cr²⁺ < Mn²⁺), with Mn²⁺ (d⁵) being most stable due to half-filled shell.

Question 4.4
To what extent do the electronic configurations decide the stability of oxidation states in the first series of transition elements? Illustrate with examples.
Answer
🟦 Highest OS ≈ group number up to Mn: V(+5), Cr(+6), Mn(+7) because (n−1)d and ns electrons are available for bonding.
🟩 d⁵ / d¹⁰ stability modulates trends:
Mn²⁺ (d⁵), Fe³⁺ (d⁵) are specially stable.
Cu⁺ (d¹⁰) is stabilized; Zn²⁺ (d¹⁰) is dominant.
🟨 Later elements (Fe, Co, Ni, Cu) show lower maximum OS (e.g., Fe up to +6, Co up to +3/+4, Ni up to +4, Cu up to +3) due to rising ionisation enthalpy of removing d-electrons.
📌 Thus configuration (half-/fully-filled d) and ns–(n−1)d energy closeness jointly control stability.

Question 4.5
What may be the stable oxidation state of the transition element with the following ground-state d-electron configurations: 3d³, 3d⁵ and 3d⁷?
Answer
🔵 3d³ (like V): total valence e⁻ = d³ + s² → +5 often most stable (vanadates).
🟢 3d⁵ (like Mn): valence = 5 + 2 → +7 most stable (permanganate).
🟠 3d⁷ (like Co): although group number = 9, very high OS is not accessible; +2 and +3 are most stable (Co²⁺, Co³⁺).
📌 Reason: Maximum OS ≈ group number only up to Mn; thereafter it falls because removing more d-electrons becomes difficult.

Question 4.6
Name the oxometal anions of the first-row transition metals in which the metal exhibits oxidation state equal to its group number.
Answer
🟦 V (+5) → vanadate: VO₃⁻ / VO₄³⁻
🟩 Cr (+6) → chromate/dichromate: CrO₄²⁻, Cr₂O₇²⁻
🟨 Mn (+7) → permanganate: MnO₄⁻
(For Ti (+4), titanates such as TiO₃²⁻ exist but the classic equal-group anions emphasized are V, Cr, Mn.)

Question 4.7
What is lanthanoid contraction? What are its consequences?
Answer
🔵 Definition: Steady decrease in ionic/atomic radii of Ln³⁺ (Ce³⁺ → Lu³⁺) due to poor shielding by 4f electrons.
🔶 Consequences:
(i) Zr–Hf and other 4d/5d pairs have almost identical sizes → very similar chemistry → difficult separation.
(ii) Basicity of Ln(OH)₃ decreases from La³⁺ to Lu³⁺.
(iii) Ionic radii shrink across lanthanoids → affects complex stability and colour.

Question 4.8
What are the characteristics of the transition elements and why are they called transition elements? Which d-block elements may not be regarded as transition elements?
Answer
🟦 Key characteristics: variable OS, formation of coloured ions/complexes, paramagnetism, catalytic activity, alloy formation, high enthalpy of atomisation.
🟩 Called “transition”: they form a transition between s- and p-blocks; more precisely, they have partially filled (n−1)d subshell in ground state or common OS.
🟠 Not regarded: Zn, Cd, Hg (Group 12) — d¹⁰ in ground as well as common +2 state; hence lack many transition-metal features.

Question 4.9
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer
🔵 Transition: general (n−1)d¹–⁹ ns¹–² (partially filled d either in atom or common ion).
🟢 Non-transition: d-subshell either absent (s-/p-block) or completely filled (d¹⁰) in atom and common ions (e.g., Zn²⁺ = d¹⁰).
📌 Result: transition metals show variable OS, coloured ions, paramagnetism—features largely absent in non-transition elements.

Question 4.10
What are the different oxidation states exhibited by the lanthanoids?
Answer
🟦 Predominant: +3 for all lanthanoids.
🟩 Also observed (stabilised by f⁰ / f⁷ / f¹⁴ configurations):
+4: Ce⁴⁺, Pr⁴⁺, Tb⁴⁺ (→ f⁰ / f¹ / f⁷ tendencies).
+2: Sm²⁺, Eu²⁺, Yb²⁺ (→ f⁶, f⁷, f¹⁴).
📌 Others occur rarely and are strongly oxidising/reducing to revert to +3.

Question 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) Transition metals generally form coloured compounds.
(iv) Transition metals and their compounds act as good catalysts.
Answer
🧲 (i) Paramagnetism: presence of unpaired d-electrons ⇒ magnetic moment μ = √(n(n+2)) BM.
🔗 (ii) High atomisation enthalpy: strong metallic bonding due to extensive d–d overlap and participation of multiple valence electrons.
🎨 (iii) Colour: d–d transitions under ligand field; sometimes charge-transfer transitions.
⚗ (iv) Catalysis: variable OS ↔ ease of redox, formation of intermediate complexes, and large surface for adsorption.

Question 4.12
What are interstitial compounds? Why are such compounds well known for transition metals?
Answer
🔵 Definition: Small atoms (H, C, N, B) occupy interstices (voids) in a metal lattice without disturbing its framework → non-stoichiometric phases (e.g., TiC, Fe₃C, VN).
🟢 Why common for TMs:
(i) Close-packed lattices with octa/tetra-voids of suitable size,
(ii) High coordination & metallic bonding tolerate guests,
(iii) Resulting phases are hard, chemically inert, high-melting, yet often conducting.

Question 4.13
How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer
🟦 Transition metals: successive OS often differ by 1 due to similar energies of ns and (n−1)d electrons.
Examples: V (+2 to +5), Cr (+2, +3, +6), Mn (+2 to +7), Fe (+2, +3, +6).
🟩 Non-transition p-block: common OS generally differ by 2 (inert-pair effect).
Examples: Sn (+2, +4), Pb (+2, +4), S (−2, +4, +6).
📌 Transition metals thus show greater multiplicity and continuity of oxidation states.

Question 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?
Answer
🟦 Preparation steps
Fusion: FeCr₂O₄ + Na₂CO₃ + O₂ → Na₂CrO₄ + Fe₂O₃
Leaching: Na₂CrO₄ (yellow solution) separated from Fe₂O₃.
Acidification: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O (orange).
Crystallisation: Addition of KCl → K₂Cr₂O₇ (orange crystals).
🟢 Effect of pH
Acidic medium → dichromate (Cr₂O₇²⁻, orange) stable.
Basic medium → chromate (CrO₄²⁻, yellow) predominates.
📌 Colour change with pH = yellow ⇌ orange.

Question 4.15
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(i) I⁻ (ii) Fe²⁺ (iii) H₂S
Answer
🟦 General reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
(i) With iodide
Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂
(ii) With Fe²⁺
Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
(iii) With H₂S
Cr₂O₇²⁻ + 14H⁺ + 3H₂S → 2Cr³⁺ + 7H₂O + 3S↓
📌 Potassium dichromate acts as a strong oxidising agent in acidic medium.

Question 4.16
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with:
(i) Fe²⁺ (ii) SO₂ (iii) Oxalic acid? Write the ionic equations.
Answer
🟦 Preparation
Pyrolusite (MnO₂) fused with KOH + O₂ → K₂MnO₄ (green).
Disproportionation in acidic medium: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.
Purple KMnO₄ crystals obtained.
🟢 Reactions in acidic medium (MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O)
(i) With Fe²⁺:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
(ii) With SO₂:
2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺
(iii) With oxalic acid (C₂O₄²⁻):
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O
📌 KMnO₄ is a very strong oxidising agent.

Question 4.17
For M³⁺/M and M³⁺/M²⁺ systems, E° values are:
Cr³⁺/Cr: −0.9 V, Cr³⁺/Cr²⁺: −0.4 V, Mn³⁺/Mn: −1.2 V, Mn³⁺/Mn²⁺: +1.5 V, Fe³⁺/Fe: −0.4 V, Fe³⁺/Fe²⁺: +0.8 V.
Use this data to comment upon:
(i) Stability of Fe³⁺ in acid vs Cr³⁺/Mn³⁺
(ii) Ease of oxidation of Fe vs Cr or Mn
Answer
🟦 (i) Fe³⁺/Fe²⁺ = +0.8 V → Fe³⁺ reduced easily → Fe³⁺ is stable in acid.
Cr³⁺/Cr²⁺ = −0.4 V → Cr³⁺ less easily reduced, less stable.
Mn³⁺/Mn²⁺ = +1.5 V → Mn³⁺ strongly oxidising, unstable in solution.
🟢 (ii) Fe³⁺/Fe = −0.4 V → Fe easier to oxidise than Cr (−0.9) or Mn (−1.2).
📌 Thus Fe oxidises more readily; Mn³⁺ is unstable, Fe³⁺ stable in acid.

Question 4.18
Predict which of the following will be coloured in aqueous solution: Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺, Co²⁺.
Answer
💡 Colour arises from d–d transitions (partially filled d-orbitals).
Coloured: Ti³⁺ (d¹), V³⁺ (d²), Mn²⁺ (d⁵), Fe³⁺ (d⁵), Co²⁺ (d⁷)
Colourless: Cu⁺ (d¹⁰), Sc³⁺ (d⁰)
📌 Coloured ions ⇌ presence of partially filled d-orbitals.

Question 4.19
Compare stability of +2 oxidation state for first-series transition elements.
Answer
🟦 Early elements (Sc, Ti, V, Cr): +2 less stable (easily oxidised).
🟩 Middle (Mn, Fe, Co, Ni): +2 quite stable (Mn²⁺, Fe²⁺, Co²⁺, Ni²⁺).
🟨 Cu²⁺ stable but Cu⁺ disproportionates.
📌 Stability rises across period, peaks near Mn–Ni, then declines.

Question 4.20
Compare chemistry of actinoids with lanthanoids.
Answer
🟦 Electronic config: Both involve filling of f-orbitals (4f vs 5f).
🟩 Oxidation states: Lanthanoids mainly +3; actinoids variable (+3 → +6).
🟨 Radioactivity: Lanthanoids mostly stable; actinoids mostly radioactive.
🟧 Reactivity: Actinoids more reactive, especially towards air/water.
🟪 Contraction: Both show contraction (lanthanoid vs actinoid).
📌 Overall, actinoids have richer redox chemistry and radioactivity.

Question 4.21
How would you account for:
(i) Mn³⁺ strongly reducing, Mn²⁺ stable
(ii) Cobalt(III) unstable in aqueous solution
(iii) d⁴ configuration unstable in ions
Answer
(i) Mn²⁺ (d⁵) = half-filled stability ⇒ Mn³⁺ reduces to Mn²⁺.
(ii) Co³⁺ has very high hydration enthalpy but in water reduces to Co²⁺ (more stable).
(iii) d⁴ tends to convert to more stable d³ (Cr³⁺) or d⁵ (Mn²⁺).

Question 4.22
What is meant by disproportionation? Give two examples.
Answer
🔵 Definition: When a species simultaneously undergoes oxidation and reduction.
🧪 Examples:
Cu⁺ → Cu²⁺ + Cu
2MnO₄²⁻ + 2H₂O → MnO₄⁻ + MnO₂ + 4OH⁻
📌 Characteristic of elements showing intermediate oxidation states.

Question 4.23
Which metal in first series of transition metals exhibits +1 OS most frequently and why?
Answer
🟦 Copper → [Ar] 3d¹⁰ 4s¹.
🟩 Loses one 4s electron → Cu⁺ (3d¹⁰) stable due to filled d-subshell.
📌 Hence +1 oxidation state common for Cu.

Question 4.24
Calculate number of unpaired electrons in gaseous Mn³⁺, Cr³⁺, V³⁺, Ti³⁺. Which is most stable in aqueous solution?
Answer
Mn³⁺ (Z=25): [Ar] 3d⁴ → 4 unpaired.
Cr³⁺ (Z=24): [Ar] 3d³ → 3 unpaired.
V³⁺ (Z=23): [Ar] 3d² → 2 unpaired.
Ti³⁺ (Z=22): [Ar] 3d¹ → 1 unpaired.
🟢 Stability: Mn³⁺ unstable (reduces to Mn²⁺, half-filled d⁵ stable).
📌 Most stable: Cr³⁺ (due to t₂g³ half-filled stability in octahedral field).

Question 4.25
Give examples and reasons for:
(i) Lowest oxide of TM basic; highest oxide acidic.
(ii) Highest OS in oxides/fluorides.
(iii) Highest OS in oxoanions.
Answer
(i) MnO (Mn²⁺) → basic; Mn₂O₇ (Mn⁷⁺) → acidic.
(ii) OsO₄, RuO₄, MnF₄, PtF₆ → high OS.
(iii) Mn in MnO₄⁻ (+7), Cr in Cr₂O₇²⁻ (+6).
📌 As OS increases, oxides become more covalent & acidic.

Question 4.26
Indicate steps in preparation of:
(i) K₂Cr₂O₇ from chromite ore
(ii) KMnO₄ from pyrolusite
Answer
(i) Chromite ore → Na₂CrO₄ (fusion) → Cr₂O₇²⁻ (acidification) → K₂Cr₂O₇ (crystallisation).
(ii) Pyrolusite (MnO₂) + KOH + O₂ → K₂MnO₄ (green) → disproportionation → KMnO₄ (purple).


Question 4.27
What are alloys? Name an important alloy containing some lanthanoid metals. Mention its uses.
Answer
🔵 Alloy: A homogeneous mixture of a metal with other metals or non-metals.
🟢 Important lanthanoid alloy: Mischmetal (≈ 95% lanthanoids, mostly Ce + La + Nd, + Fe and traces of other elements).
🟠 Uses:
In lighter flints.
As reducing agent in extraction of metals.
In special steels.

Question 4.28
What are inner transition elements? Decide which atomic numbers among 29, 59, 74, 95, 102, 104 are inner transition elements.
Answer
🔵 Definition: Elements in which last electron enters (n−2)f orbitals → f-block (lanthanoids + actinoids).
🟢 Given atomic numbers:
29 (Cu) → d-block
59 (Pr) → f-block
74 (W) → d-block
95 (Am) → f-block
102 (No) → f-block
104 (Rf) → d-block
📌 Inner transition elements here: 59, 95, 102.

Question 4.29
The chemistry of actinoids is not so smooth as that of lanthanoids. Justify with examples.
Answer
🟦 Lanthanoids: Almost exclusively +3 oxidation state → regular, smooth chemistry.
🟩 Actinoids: Variable oxidation states due to similar energies of 5f, 6d, 7s orbitals.
🟨 Examples:
U: +3, +4, +5, +6
Np: +3 → +7
Pu: +3 → +7
📌 Hence actinoid chemistry is complex and irregular, unlike lanthanoids.

Question 4.30
Which is the last element in the series of the actinoids? Write its electronic configuration. Comment on the possible oxidation state.
Answer
🟦 Last element: Lawrencium (Lr, Z=103)
🟢 Electronic configuration: [Rn] 5f¹⁴ 6d¹ 7s²
🟠 Possible oxidation state: +3 (most stable, like other actinoids).

Question 4.31
Use Hund’s rule to derive the electronic configuration of Ce³⁺. Calculate its magnetic moment using the spin-only formula.
Answer
🔵 Ce (Z=58): [Xe] 4f² 6s²
🟢 Ce³⁺ = [Xe] 4f¹ (removal of 2 × 6s + 1 × 4f).
🟠 One unpaired electron ⇒ n = 1
📌 μ = √(n(n+2)) = √3 = 1.73 BM

Question 4.32
Name lanthanoids that show +4 OS and those that show +2 OS.
Answer
🟦 +4: Ce, Pr, Tb (stabilised by f⁰, f¹, f⁷)
🟩 +2: Sm, Eu, Yb (stabilised by f⁶, f⁷, f¹⁴)

Question 4.33
Compare chemistry of actinoids with lanthanoids on: (i) electronic configuration (ii) oxidation states (iii) chemical reactivity.
Answer
🟦 Electronic config: Lanthanoids → 4f filling; Actinoids → 5f filling.
🟩 Oxidation states: Lanthanoids → almost always +3; Actinoids → variable (+3 to +6/+7).
🟨 Reactivity: Lanthanoids → less reactive; Actinoids → more reactive, radioactive.

Question 4.34
Write electronic configurations of elements with Z = 61, 91, 101, 109.
Answer
🟦 61 (Pm): [Xe] 4f⁵ 6s²
🟩 91 (Pa): [Rn] 5f² 6d¹ 7s²
🟨 101 (Md): [Rn] 5f¹³ 7s²
🟧 109 (Mt): [Rn] 5f¹⁴ 6d⁷ 7s²

Question 4.35
Compare general characteristics of first-series transition metals with second and third series.
Answer
🟦 Electronic config: All → (n−1)d orbitals; heavier series show relativistic effects.
🟩 Oxidation states: Wider range in 4d/5d than 3d.
🟨 Ionisation enthalpies: Increase down series but irregular due to lanthanoid contraction.
🟧 Atomic radii: 4d and 5d elements have almost same radii (lanthanoid contraction).
📌 5d elements are more stable in higher oxidation states.

Question 4.36
Write no. of 3d electrons in hydrated ions: Ti³⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺. Indicate splitting in octahedral field.
Answer
Ti³⁺: d¹
V²⁺: d³
Cr³⁺: d³
Mn²⁺: d⁵
Fe²⁺: d⁶
Co²⁺: d⁷
Ni²⁺: d⁸
Cu²⁺: d⁹
🟢 In octahedral field → d splits into t₂g (lower, 3 orbitals) and e_g (higher, 2 orbitals).
📌 Filling depends on ligand strength (CFSE).

Question 4.37
Comment: Elements of first transition series differ from heavier transition series.
Answer
🟦 3d elements: more variety in oxidation states, less stable in high OS.
🟩 4d/5d: higher enthalpies of atomisation, higher melting points.
🟨 5d metals (due to lanthanoid contraction) are closer in radii to 4d metals, thus chemical similarity (Zr–Hf).
🟧 First series show more pronounced paramagnetism and coloured compounds due to smaller crystal field splitting.

Question 4.38
What can be inferred from magnetic moment values of: [Mn(CN)₆]³⁻ = 2.2 BM, [Fe(H₂O)₆]²⁺ = 5.3 BM, [K₂MnCl₆] = 5.9 BM?
Answer
🔵 [Mn(CN)₆]³⁻: μ = 2.2 BM ≈ 1 unpaired e⁻ → low-spin complex (strong field CN⁻).
🟢 [Fe(H₂O)₆]²⁺: μ = 5.3 BM ≈ 4 unpaired e⁻ → high-spin (weak field H₂O).
🟠 [K₂MnCl₆]: μ = 5.9 BM ≈ 5 unpaired e⁻ → high-spin Mn(IV) with weak field Cl⁻.
📌 Magnetic moment values reveal ligand field strength and spin state.

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OTHER IMPORTANT QUESTIONS FOR EXAMS

(CBSE MODEL QUESTIONS PAPER)

ESPECIALLY MADE FROM THIS LESSON ONLY

✨ Section A (Q1–Q16, 1 mark each = 16)
Options for Assertion–Reason (AR) questions:
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false.
A is false, but R is true.

Question 1. Which of the following is not considered a transition element?
Fe
Zn
Cr
Ni
Answer: 2

Question 2. The stable oxidation state of Mn in KMnO₄ is:
+2
+4
+6
+7
Answer: 4

Question 3. Which ion is colourless in aqueous solution?
Ti³⁺
Cu²⁺
Zn²⁺
Fe³⁺
Answer: 3

Question 4. Assertion (A): Mn²⁺ compounds are more stable than Fe²⁺ compounds towards oxidation.
Reason (R): Mn²⁺ has a d⁵ half-filled configuration.
Answer: 1

Question 5. Which element among the following shows maximum number of oxidation states?
Cr
Mn
Fe
Co
Answer: 2

Question 6. The pink colour of KMnO₄ solution is due to:
d–d transitions
f–f transitions
Charge transfer transitions
Ionisation
Answer: 3

Question 7. Which is the strongest oxidising agent?
Cr³⁺
Fe³⁺
MnO₄⁻
Cu²⁺
Answer: 3

Question 8. Assertion (A): Lanthanoid contraction is due to poor shielding of 4f electrons.
Reason (R): 4f orbitals shield outer electrons less effectively.
Answer: 1

Question 9. Which of the following is not an actinoid?
Th
U
Eu
Pu
Answer: 3

Question 10. Which ion shows maximum paramagnetism?
Fe²⁺ (d⁶)
Mn²⁺ (d⁵)
Co²⁺ (d⁷)
Ni²⁺ (d⁸)
Answer: 2

Question 11. The alloy mischmetal contains mainly:
La and Ce
Ce and Th
U and Pu
Nd and Pm
Answer: 1

Question 12. Which lanthanoid ion is colourless?
La³⁺
Ce³⁺
Nd³⁺
Tb³⁺
Answer: 1

Question 13. Assertion (A): Co³⁺ is unstable in aqueous solution.
Reason (R): It gets easily reduced to Co²⁺.
Answer: 1

Question 14. The last element of the actinoid series is:
No
Lr
Md
Es
Answer: 2

Question 15. Which compound is used in the Contact process as catalyst?
Fe₂O₃
V₂O₅
TiO₂
MnO₂
Answer: 2

Question 16. Which ion is low-spin complex in CN⁻ ligand field?
Mn³⁺
Mn²⁺
Fe³⁺
Co³⁺
Answer: 4

✨ Section B (Q17–Q21, 2 marks each = 10)
Question 17. Write two reasons why transition metals form coloured compounds.
Answer
🟦 Due to d–d electronic transitions under crystal field splitting.
🟩 Due to charge transfer transitions (ligand to metal or metal to ligand).

Question 18. Why is Zn not considered a transition element?
Answer
🟦 Zn has 3d¹⁰ 4s² configuration → completely filled d-orbital.
🟩 In Zn²⁺ also configuration is 3d¹⁰ (no partially filled d-orbital).

Question 19. Write two uses of KMnO₄.
Answer
🟦 Used as a disinfectant in water treatment.
🟩 Strong oxidising agent in organic and inorganic reactions.

Question 20. State two consequences of lanthanoid contraction.
Answer
🟦 Atomic radii of Zr (4d) and Hf (5d) become almost identical.
🟩 Decrease in basicity of hydroxides La(OH)₃ → Lu(OH)₃.

Question 21. Why is Eu²⁺ stable while Ce²⁺ is unstable?
Answer
🟦 Eu²⁺ has 4f⁷ half-filled stability.
🟩 Ce²⁺ tends to oxidise to Ce³⁺ (stable f¹).

✨ Section C (Q22–Q28, 3 marks each = 21)
Question 22. Explain why actinoids show more variable oxidation states than lanthanoids.
Answer
🟦 In actinoids, energies of 5f, 6d, 7s orbitals are comparable → more redox options.
🟩 Lanthanoids mainly show +3 (stable).
🟨 Examples: U (+3, +4, +5, +6), Pu (+3 → +7).

Question 23. Give three differences between d-block and f-block elements.
Answer
🟦 Orbitals filled: d-block → (n−1)d, f-block → (n−2)f.
🟩 Oxidation states: d-block variable (+2 to +7), f-block mainly +3 (actinoids wider).
🟨 Radioactivity: d-block stable; most actinoids radioactive.

Question 24. Write three points on catalytic behaviour of transition metals.
Answer
🟦 Provide variable oxidation states → alternate pathways.
🟩 Large surface area for adsorption of reactants.
🟨 Intermediate complex formation lowers activation energy.

Question 25. Give reasons:
(i) La³⁺ is colourless but Nd³⁺ is coloured.
(ii) Mn shows highest oxidation state in oxoanions.
(iii) Cu shows +1 state more stable than +2.
Answer
🟦 La³⁺ = 4f⁰ (no unpaired e⁻) → colourless. Nd³⁺ = 4f³ (unpaired) → coloured.
🟩 Mn (+7 in MnO₄⁻) stabilised by high electronegativity of oxygen.
🟨 Cu⁺ = 3d¹⁰ stable (filled subshell).

Question 26. Explain with example the disproportionation reaction in transition metals.
Answer
🟦 Definition: One species undergoes oxidation and reduction simultaneously.
🟩 Example: 2Cu⁺ → Cu²⁺ + Cu
🟨 Example: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

Question 27. Give three uses of lanthanoids.
Answer
🟦 Mischmetal (La, Ce, Nd alloys) → lighter flints.
🟩 CeO₂ → polishing glass, catalyst in petroleum cracking.
🟨 Nd → powerful magnets and lasers.

Question 28. Write three reasons why transition metals form alloys.
Answer
🟦 Atomic radii are comparable → atoms replace each other.
🟩 Strong metallic bonding allows mixture.
🟨 Close-packed lattices permit solid-solution alloys (brass, bronze, stainless steel).

✨ Section D (Q29–Q30, 4 marks each = 8)
Question 29.
Read the passage and answer:
The dichromate–chromate equilibrium is pH dependent. Potassium dichromate (K₂Cr₂O₇) is a strong oxidising agent in acidic medium and is used in volumetric analysis.
(i) Write equilibrium between chromate and dichromate ions.
(ii) What happens when NaOH is added to acidic K₂Cr₂O₇ solution?
(iii) Write ionic equation for oxidation of I⁻ by Cr₂O₇²⁻ in acidic medium.
Answer
🟦 (i) 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O
🟩 (ii) Adding NaOH increases pH → orange dichromate converts to yellow chromate.
🟨 (iii) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂

Question 30.
Study the data and answer:
Oxidation states of Mn are +2, +4, +6, +7. Compounds like MnO, MnO₂, K₂MnO₄, KMnO₄ are known.
(i) Which is the most stable state in aqueous solution? Why?
(ii) Write equation for reduction of MnO₄⁻ to Mn²⁺ in acid.
(iii) Write one use of KMnO₄ based on its oxidising property.
Answer
🟦 (i) +2 state most stable due to half-filled d⁵ stability.
🟩 (ii) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
🟨 (iii) KMnO₄ used as oxidant in removal of Fe²⁺ from water.

✨ Section E (Q31–Q33, 5 marks each = 15)
Question 31.
Explain preparation and properties of KMnO₄.
OR
Explain preparation and properties of K₂Cr₂O₇.
Answer
🟦 Preparation of KMnO₄
Fusion: MnO₂ + KOH + O₂ → K₂MnO₄ (green).
Disproportionation: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.
Crystallisation: Purple KMnO₄ obtained.
🟩 Properties
Strong oxidising agent in acidic/neutral/alkaline media.
Example: In acidic medium:
2MnO₄⁻ + 10Fe²⁺ + 16H⁺ → 2Mn²⁺ + 5Fe³⁺ + 8H₂O
Used in bleaching, disinfection, volumetric analysis.
✅ OR K₂Cr₂O₇: prepared from chromite ore; strong oxidising agent; oxidises I⁻, Fe²⁺, H₂S.

Question 32.
Discuss variable oxidation states in d-block and f-block elements.
OR
Explain magnetic and catalytic properties of transition metals.
Answer
🟦 Variable Oxidation States
d-block: Due to closeness of (n−1)d and ns energies.
Range: V (+2 to +5), Cr (+2, +3, +6), Mn (+2 to +7).
Stabilised by half-/fully-filled configurations (Mn²⁺ d⁵, Zn²⁺ d¹⁰).
🟩 f-block
Lanthanoids: mainly +3, occasional +2/+4.
Actinoids: wider range (+3 to +6/+7).
✅ OR Magnetic + Catalysis
Magnetic: Paramagnetism from unpaired d electrons.
Catalytic: Multiple oxidation states, large surface area, adsorption, complex formation.

Question 33.
Explain lanthanoid contraction and its consequences.
OR
Compare chemistry of lanthanoids and actinoids.
Answer
🟦 Lanthanoid contraction
Steady decrease in ionic radii La³⁺ → Lu³⁺.
Cause: Poor shielding of 4f electrons.
🟩 Consequences
Similar sizes of Zr–Hf, Nb–Ta, → chemical similarity.
Separation of lanthanoids is difficult.
Decrease in basicity of Ln(OH)₃.
Greater stability of later lanthanoid complexes.
✅ OR Lanthanoids vs Actinoids: Both show contraction, actinoids more variable oxidation states, more radioactive, greater covalency.

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