Class 9 Chemistry Chapter 4: The d- and f-Block Elements — Important Questions with Complete Answers

The d- and f-block elements chapter represents a critical shift in Class 9 chemistry—from main group (s- and p-block) patterns to transition metal behaviour. Examiners test this chapter across three key areas: (1) electron configuration and orbital filling rules for transition metals; (2) properties like variable oxidation states, magnetic behaviour, and coloured ion formation; and (3) basic characteristics of lanthanoids and actinoids. In recent CBSE board cycles, questions here account for 8–12 marks in the theory paper and often appear as application-based problems. The 2026–27 pattern emphasises conceptual clarity over rote memorisation. Students who confuse d-block electron configuration with p-block trends, or who cannot explain *why* transition metals form coloured compounds, typically lose 3–4 marks unnecessarily. This guide isolates the exact question types that appear most often and teaches you to recognize the reasoning patterns behind each mark allocation.

**Q1: Which electronic configuration represents a transition metal?** A) [Ar] 3s² 3p⁶ B) [Ar] 3d⁵ 4s² C) [Ar] 3d¹⁰ 4s² D) [Ar] 3d¹⁰ 4s² 4p⁶ **Answer: B** **Explanation:** Transition metals have partially filled d-orbitals (3d¹ to 3d⁹). Options C and D show filled d-orbitals (post-transition or main group metals). Only B shows a half-filled d-subshell, characteristic of transition metals like Mn. **Q2: Lanthanoids are characterized by:** A) Filling of 4f orbitals B) Filling of 5d orbitals C) Filling of 6s orbitals D) Filling of 4d orbitals **Answer: A** **Explanation:** The lanthanoid series (elements 57–71) involve progressive filling of 4f electrons while maintaining [Xe] 4f ⁿ 5d⁰⁻¹ 6s² configuration. **Q3: Which of the following shows maximum magnetic moment?** A) Zn²⁺ B) Cu²⁺ C) Mn²⁺ D) Fe²⁺ **Answer: C** **Explanation:** Mn²⁺ has configuration [Ar] 3d⁵ with five unpaired electrons. Zn²⁺ is d¹⁰ (diamagnetic); Cu²⁺ is d⁹ (1 unpaired); Fe²⁺ is d⁶ (4 unpaired). Maximum unpaired electrons = maximum paramagnetism. **Q4: Why do transition metals form coloured compounds?** A) Due to high nuclear charge B) Due to d–d electron transitions absorbing visible light C) Due to metallic bonding D) Due to high ionization energy **Answer: B** **Explanation:** Transition metal ions like [Cu(H₂O)₄]²⁺ absorb visible light when electrons jump between partially filled d-orbitals (3d ← 3d transitions). This selective absorption causes colour. **Q5: Actinoids differ from lanthanoids primarily in:** A) Number of electrons B) Filling of 5f orbitals instead of 4f C) Chemical inertness D) Inability to form complexes **Answer: B** **Explanation:** Actinoids (elements 89–103) fill 5f orbitals, whereas lanthanoids fill 4f. Both show similar coordination chemistry and variable oxidation states, but radioactivity is more prominent in actinoids.

**Q1: Define transition metals. Name any two transition metals in the first transition series and write their electronic configurations.** **Answer:** Transition metals are elements that have incompletely filled d-orbitals in their ground state or in any of their stable oxidation states. Two examples from the first transition series (3d series): 1) **Iron (Fe):** [Ar] 3d⁶ 4s² 2) **Copper (Cu):** [Ar] 3d¹⁰ 4s¹ Note: Cu is sometimes listed because Cu⁺ has [Ar] 3d¹⁰ (completely filled d-orbitals), but elemental Cu is classified as transition metal. **Q2: Write the difference between lanthanoids and actinoids with respect to radioactivity and common oxidation states.** **Answer:** | Property | Lanthanoids | Actinoids | |---|---|---| | **Radioactivity** | Almost all stable (non-radioactive) | All are radioactive | | **Common Oxidation States** | Mainly +3, rarely +2, +4 | +3, +4, +5, +6 (more variable) | Example: Cerium (Ce) is +3 mostly; Uranium (U) exists as +4, +6. **Q3: Why does manganese show multiple oxidation states? Give two examples of its compounds.** **Answer:** Manganese (Z = 25) has configuration [Ar] 3d⁵ 4s². The five 3d electrons are loosely held and can be removed to varying degrees, allowing multiple stable oxidation states: +2, +3, +4, +6, +7. Examples: 1) **MnO** (Mn²⁺) — black oxide 2) **KMnO₄** (Mn⁷⁺) — purple permanganate, strong oxidizing agent (Also acceptable: MnO₂, K₂MnO₄) **Q4: State the position of transition metals in the periodic table and explain why they have higher melting points than main group metals.** **Answer:** Transition metals occupy Groups 3–12 (or 3–11 in IUPAC numbering) in Periods 4, 5, and 6. They have higher melting points because: - Unpaired d-electrons contribute to metallic bonding strength - More free electrons available for bonding → stronger lattice energy Example: Iron (Fe) melts at 1538°C; sodium (Na) melts at 98°C. Both metals, but Fe's d-electrons create a much stronger metallic bond network. **Q5: Explain why Zn²⁺ is colourless while Cu²⁺ is blue in aqueous solution.** **Answer:** **Zn²⁺:** Configuration [Ar] 3d¹⁰ (completely filled d-orbitals). No d–d transitions possible → no visible light absorbed → **colourless**. **Cu²⁺:** Configuration [Ar] 3d⁹ (one unpaired electron in d-orbitals). Electrons can transition from lower to higher d-orbitals (3d ← 3d) and absorb light in the orange–yellow region → complementary **blue colour** observed. General principle: Coloured transition metal ions have partially filled d-orbitals (d¹ to d⁹); fully filled or empty d-orbitals give colourless ions.

**Q1: Transition metals exhibit variable oxidation states. Explain this property using the concept of d-electron removal and provide three examples from the first transition series with different oxidation states.** **Solution:** *Step 1: State the reason for variable oxidation states* Transition metals have d-electrons that are only slightly higher in energy than s-electrons (in the same or adjacent shell). Both d and s electrons can be removed with relatively small energy differences, allowing multiple stable oxidation states. *Step 2: Identify the electron removal pattern* For example, iron [Ar] 3d⁶ 4s² can lose 2, 3, or even 6 electrons depending on conditions: - Fe²⁺: Loses 4s² → [Ar] 3d⁶ - Fe³⁺: Loses 4s² and one 3d¹ → [Ar] 3d⁵ - Fe⁶⁺: (rare) Loses more 3d electrons *Step 3: Provide three examples* 1) **Manganese (Mn):** +2 (MnO), +7 (KMnO₄) 2) **Chromium (Cr):** +2 (CrO), +3 (Cr₂O₃), +6 (K₂CrO₄) 3) **Iron (Fe):** +2 (FeSO₄), +3 (Fe₂(SO₄)₃), +6 (rare) The ease of d-electron removal in response to chemical conditions is unique to transition metals. **Q2: Explain why copper (Cu) is often classified as a transition metal despite having a filled d-orbital ([Ar] 3d¹⁰ 4s¹) in its ground state. How does its +1 oxidation state support this classification?** **Solution:** *Step 1: Address the apparent contradiction* Copper in its neutral state has [Ar] 3d¹⁰ 4s¹—technically a filled d-subshell. However, the IUPAC definition of transition metals includes elements with d-orbitals in any **stable oxidation state**, not just ground state. *Step 2: Examine the +1 oxidation state* Cu⁺ has configuration [Ar] 3d¹⁰ (loses the 4s¹ electron). Although d¹⁰ is filled, Cu⁺ is extremely stable and common in compounds like CuCl, Cu₂O, and [Cu(NH₃)₂]⁺. *Step 3: Justify the transition metal classification* Because copper readily forms Cu⁺ and Cu²⁺ ions (Cu²⁺ = [Ar] 3d⁹), and these oxidation states involve d-orbital chemistry (d-electron removal and d–d transitions), copper is classified as a transition metal. Its +2 oxidation state (d⁹) directly shows incomplete d-orbital character. *Conclusion:* Classification is based on d-orbital involvement in chemistry, not just electron count. **Q3: Lanthanoids show a progressive decrease in atomic and ionic radii across the series (known as lanthanide contraction). Explain the cause and one important consequence of this phenomenon.** **Solution:** *Step 1: Define lanthanide contraction* As atomic number increases from La (Z=57) to Lu (Z=71), atomic and ionic radii decrease more sharply than expected based on main group trends. This occurs despite the addition of electrons. *Step 2: Explain the cause* As 4f electrons are filled (La to Lu), the 4f orbitals have poor shielding power for outer electrons. Each additional 4f electron is attracted strongly by the increasing nuclear charge, pulling the electron cloud inward. Simultaneously, earlier-filled 4f electrons don't shield the nucleus effectively from outer electrons, resulting in net contraction. Mathematically: Nuclear charge ↑ faster than shielding ↑ → radii ↓ *Step 3: State one important consequence* **Chemical consequence:** The lanthanides become progressively smaller from La to Lu. This means: - Ionic radii decrease: La³⁺ (1.06 Å) > Lu³⁺ (0.86 Å) - Charge density increases, strengthening electrostatic interactions - Basic strength of oxides and hydroxides decreases (La₂O₃ is more basic than Lu₂O₃) - Coordination chemistry and complex formation tendencies change systematically This contraction also affects the chemistry of 5d transition metals, which come immediately after lanthanoids and have smaller-than-expected radii. **Q4: Compare the chemistry of actinoids with lanthanoids. Why are actinoids considered more complex in their chemistry?** **Solution:** *Step 1: List structural similarities* Both lanthanoids (4f ⁿ 5d⁰⁻¹ 6s²) and actinoids (5f ⁿ 6d⁰⁻¹ 7s²) are inner transition elements; both exhibit progressive filling of f-orbitals; both show lanthanide (or actinide) contraction. *Step 2: Highlight key chemical differences* | Feature | Lanthanoids | Actinoids | |---|---|---| | **Oxidation States** | Mainly +3 | +3 to +6 (variable) | | **Radioactivity** | Mostly stable | All radioactive | | **Ionic Character** | Predominantly ionic compounds | More covalent character | | **Magnetic Properties** | Paramagnetic (unpaired 4f) | Highly paramagnetic + ferromagnetic | *Step 3: Explain complexity in actinoids* Actinoids are more complex because: 1) **Multiple oxidation states:** 5f electrons are closer in energy to 6d and 7s; hence, multiple electrons can be removed depending on oxidation–reduction conditions. Example: U exhibits +3, +4, +5, +6. 2) **Radioactivity:** All actinoids are radioactive, decaying via α, β, or γ emission. This adds nuclear chemistry dimensions absent in lanthanoids. 3) **Covalent character:** 5f orbitals are larger and more diffuse than 4f; actinoid compounds often show greater covalent character and form more complex organometallic compounds. 4) **Coordination:** Actinoids form stable complexes with chelating ligands (e.g., [UO₂(NO₃)₄]²⁻), whereas lanthanoid complex chemistry is more straightforward. *Conclusion:* The 5f orbitals' larger size and closer energy proximity to valence s/d orbitals make actinoid chemistry significantly richer and more unpredictable.

**Q1: Define transition metals. Explain the electronic configuration of the first transition series (from Sc to Zn). How do their properties differ from main group metals? Provide examples to support your answer.** **Full Solution:** *Part A: Definition and scope* Transition metals are elements characterized by the presence of incompletely filled d-orbitals in their ground state or in any stable oxidation state. They occupy d-block of the periodic table (Groups 3–12, Periods 4, 5, 6). *Part B: Electronic configurations of the first transition series (3d series)* The first transition series spans Sc (Z=21) to Zn (Z=30). All have [Ar] 4s² configuration, but vary in 3d filling: - **Sc:** [Ar] 3d¹ 4s² - **Ti:** [Ar] 3d² 4s² - **V:** [Ar] 3d³ 4s² - **Cr:** [Ar] 3d⁵ 4s¹ (exception: half-filled d is more stable) - **Mn:** [Ar] 3d⁵ 4s² - **Fe:** [Ar] 3d⁶ 4s² - **Co:** [Ar] 3d⁷ 4s² - **Ni:** [Ar] 3d⁸ 4s² - **Cu:** [Ar] 3d¹⁰ 4s¹ (exception: filled d is more stable) - **Zn:** [Ar] 3d¹⁰ 4s² *Part C: Properties compared to main group metals* 1) **Variable Oxidation States:** Main group metals (e.g., Na, Mg, Al) show fixed oxidation states (+1, +2, +3). Transition metals show multiple: Fe (+2, +3), Mn (+2 to +7). *Reason:* d- and s-electrons have similar ionization energies; both can be removed. 2) **Higher Melting/Boiling Points:** Transition metals: Fe (1538°C), Cu (1085°C) Main group metals: Na (98°C), Mg (650°C) *Reason:* Unpaired d-electrons participate in metallic bonding, creating a stronger lattice. 3) **Coloured Ions and Compounds:** [Cu(H₂O)₄]²⁺ is blue; [Fe(H₂O)₆]²⁺ is pale green; [Fe(H₂O)₆]³⁺ is yellow-brown. Main group ions (Na⁺, Mg²⁺, Al³⁺) are colourless. *Reason:* d–d electron transitions in partially filled d-orbitals absorb visible light. 4) **Paramagnetism & Ferromagnetism:** Fe, Co, Ni are ferromagnetic (permanent magnetic moment). Most transition metals show paramagnetism (unpaired d-electrons). Main group metals are mostly diamagnetic. 5) **Complex Ion Formation:** Transition metals form stable coordination complexes: [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺. Main group metals rarely form such complexes. *Reason:* Empty d-orbitals can accept electron pairs from ligands. 6) **Catalytic Activity:** Transition metals (Fe in Haber process, Pt in catalytic converters) are effective catalysts. Main group metals are not. *Reason:* Variable oxidation states and complex formation enable catalytic cycles. *Part D: Summary table* | Property | Transition Metals | Main Group Metals | |---|---|---| | d-orbital occupancy | Partially filled | Filled or empty | | Oxidation states | Multiple | Usually one or two | | Melting point | Generally high | Varies | | Colour | Often coloured ions | Colourless ions | | Magnetism | Paramagnetic/ferromagnetic | Diamagnetic | | Catalytic activity | High | Low | **Q2: Lanthanoids and actinoids are inner transition elements. Compare their chemistry, explaining why actinoids are more complex. Discuss their applications and the lanthanide contraction phenomenon.** **Full Solution:** *Part A: Definition and position* Lanthanoids (elements 57–71: La to Lu) and actinoids (elements 89–103: Ac to Lr) are inner transition elements whose valence electrons enter f-orbitals. They occupy positions in the extended periodic table, typically shown separately below the main table. *Part B: Electronic configurations* **Lanthanoids:** [Xe] 4f ⁿ 5d⁰⁻¹ 6s² Example: **Ce** [Xe] 4f¹ 5d¹ 6s² (Z=58) or [Xe] 4f² 6s² (depending on state) **Actinoids:** [Rn] 5f ⁿ 6d⁰⁻¹ 7s² Example: **U** [Rn] 5f³ 6d¹ 7s² (Z=92) *Part C: Comparative chemistry* 1) **Oxidation States:** - **Lanthanoids:** Primarily +3 (most common); rarely +2 (Eu, Yb) or +4 (Ce, Tb) - **Actinoids:** +3, +4, +5, +6 regularly; example U: +3 (UCl₃), +4 (UO₂), +6 (UO₃, UF₆) *Reason:* 5f orbitals are larger and closer in energy to valence 6d/7s; more electrons can participate in bonding. 2) **Radioactivity:** - **Lanthanoids:** Mostly stable (non-radioactive); Pm is slightly radioactive. - **Actinoids:** All radioactive; half-lives vary from U-238 (4.5 billion years) to Lr-262 (3.6 hours). 3) **Magnetic Properties:** - **Lanthanoids:** Paramagnetic due to unpaired 4f electrons; magnetic moments follow Curie law. - **Actinoids:** Highly paramagnetic; some show ferromagnetic ordering (Np, Pu compounds). 4) **Covalent Character:** - **Lanthanoids:** Primarily ionic compounds; limited covalency. - **Actinoids:** Greater covalent character in compounds due to larger f-orbitals; form stable organometallic complexes like (C₅H₅)₃U. 5) **Complex Formation:** - **Lanthanoids:** Coordination number ≈ 8–9; simple outer-sphere complexes. - **Actinoids:** Variable coordination; stable inner-sphere complexes with chelating ligands (e.g., [UO₂(NO₃)₄]²⁻). *Part D: Why actinoids are more complex* 1. **Overlap of 5f, 6d, and 7s energies:** All three subshells are close in energy, making multiple electrons available for bonding. In lanthanoids, 4f electrons are deeply buried, and only +3 is typical. 2. **Radioactive decay chemistry:** Actinoids undergo α, β⁻, β⁺, electron capture. This nuclear instability couples to chemistry—decay products create new isotopes, altering behaviour. 3. **Actinyl formation:** Actinoids readily form AnO₂ⁿ⁺ units (e.g., UO₂²⁺, PuO₂²⁺). These actinyl ions show unique spectroscopy and reactivity absent in lanthanoids. 4. **f-d hybridization:** 5f–6d hybridization in actinoid bonding increases orbital participation and chemical versatility. *Part E: Lanthanide contraction* **Definition:** As atomic number increases from La to Lu, both atomic and ionic radii decrease more steeply than expected from periodic trends. **Cause:** As 4f electrons are progressively filled, increasing nuclear charge is insufficiently shielded by these f-electrons (poor shielding due to diffuse nature). Outer electrons experience stronger effective nuclear charge, contracting the atom. **Magnitude:** La³⁺ (1.06 Å) > Lu³⁺ (0.86 Å) — a decrease of ~0.20 Å across 14 elements. **Consequences:** - **Density:** Lanthanides increase in density from La to Lu (despite shrinking size, mass increases). - **Basicity:** La₂O₃ (most basic) > Lu₂O₃ (least basic); basicity decreases due to increased charge density of smaller ions. - **Crystal structures:** Later lanthanides favour lower coordination numbers. - **Effect on 5d metals:** Lanthanide contraction carries over into Period 6 transition metals (W, Re, Os, Ir, Pt). Hf has unexpectedly small radius (similar to Zr), affecting 5d metal chemistry. *Part F: Applications* **Lanthanoid applications:** - Nd-Fe-B magnets (powerful permanent magnets in motors, headphones) - Rare earth phosphors (CRT displays, fluorescent lamps) - Catalysts (petroleum cracking, automotive catalytic converters) - Optical materials (Nd-doped lasers for surgical/industrial cutting) **Actinoid applications:** - Nuclear fuel: U-235 (fission in reactors); Pu-239 (weapons-grade fissile material) - Medical: Am-241 (smoke detectors); radiotherapy (U, Pu isotopes) - Research: Transuranic element synthesis and nuclear physics. **Q3: Explain why transition metals form coloured compounds while many main group metal ions are colourless. Use crystal field theory concepts and provide three detailed examples.** **Full Solution:** *Part A: Fundamental principle* Colour in transition metal compounds arises from **electronic transitions between d-orbital energy levels**. When visible light interacts with a complex ion, an electron absorbs a photon and jumps from a lower-energy d-orbital to a higher-energy d-orbital (d → d transition). The wavelength of absorbed light determines the complementary colour observed. *Part B: Why main group metals don't show colour* Main group metal ions (e.g., Na⁺, Mg²⁺, Al³⁺, Ca²⁺) have: - Completely filled valence shells (e.g., Na⁺ = [Ne], Mg²⁺ = [Ne]) - No unpaired electrons - No partially filled d- or p-orbitals available for electronic transitions within the visible region Energy gaps for transitions in these ions lie in the UV region (λ < 400 nm), outside human vision. Hence, they appear **colourless**. *Part C: Crystal field splitting in transition metal complexes* When transition metal ions are placed in a ligand field (octahedral, tetrahedral, or square planar), the d-orbitals split into groups of different energies (e.g., in octahedral geometry: t₂g and eg levels). **Example: Octahedral [Fe(H₂O)₆]³⁺** - Fe³⁺ = [Ar] 3d⁵ with five unpaired electrons - Ligand field splits 3d into: lower t₂g (3 orbitals) and upper eg (2 orbitals) - Δ (crystal field splitting energy) = hν where ν corresponds to a visible wavelength - An electron in a lower d-orbital absorbs light and transitions to a higher d-orbital - The unabsorbed wavelengths are transmitted, giving the observed colour *Part D: Three detailed examples* **Example 1: [Cu(H₂O)₄]²⁺ (blue)** - Cu²⁺ = [Ar] 3d⁹; one unpaired electron in d-orbitals - In octahedral field, the 3d⁹ electron configuration places one electron in the eg level and eight in t₂g - d–d transition absorbs light in the orange–yellow region (~600 nm) - Transmitted light is **blue** (complementary to orange) - This is the signature colour of copper(II) solutions **Example 2: [Fe(H₂O)₆]²⁺ (pale green)** - Fe²⁺ = [Ar] 3d⁶; four unpaired electrons - d–d transitions have Δ ≈ 10,000–15,000 cm⁻¹, absorbing red light - Transmitted light is **pale green** (cyan hue) **Example 3: [Fe(H₂O)₆]³⁺ (yellow-brown)** - Fe³⁺ = [Ar] 3d⁵; five unpaired electrons (high-spin, one electron per d-orbital) - Δ ≈ 20,000–25,000 cm⁻¹; absorbs blue light - Transmitted light is **yellow-brown** (orange-brown) *Part E: Factors affecting colour intensity and shade* 1. **Number of unpaired electrons:** More unpaired d-electrons typically allow more d–d transitions → deeper colour. 2. **Nature of ligand:** Ligand field strength determines Δ. Stronger field ligands (CN⁻, CO) produce larger Δ, shifting absorption to shorter wavelengths. - [Fe(CN)₆]⁴⁻ is deep red (strong field) - [Fe(H₂O)₆]²⁺ is pale green (weak field) 3. **Oxidation state:** Different oxidation states have different d-electron counts, hence different absorption profiles. 4. **Geometry:** Tetrahedral vs. octahedral vs. square planar geometries split d-orbitals differently, altering transition energies. *Part F: Quantitative relationship* Δ = hc/λ where Δ is crystal field splitting, h is Planck's constant, c is light speed, λ is the absorbed wavelength. For visible light (400–700 nm), Δ ≈ 10,000–25,000 cm⁻¹, which matches typical transition metal d–d transition energies. This is why colour is observed. Main group elements have much larger or smaller energy gaps, placing transitions outside the visible region.

**Case Study: The Extraction and Applications of Lanthanoids** Rare earth elements (lanthanoids and scandium/yttrium) are critical for modern technology. A mining company in India extracts lanthanoid-rich monazite ore containing ~65% rare earth phosphates. After initial processing, a mixture of lanthanoid oxides is obtained. The separation is challenging because lanthanoids have nearly identical chemical properties due to lanthanide contraction. **Background Data:** - Lanthanoid ionic radii (Å): La³⁺ = 1.06, Nd³⁺ = 0.99, Gd³⁺ = 0.94, Lu³⁺ = 0.86 - Standard reduction potentials (Ln³⁺ + 3e⁻ → Ln) are nearly constant: E° ≈ –2.2 to –2.4 V - Solubility products (Ksp) of lanthanoid hydroxides decrease from La(OH)₃ to Lu(OH)₃ **Questions:** **(i) Why is it difficult to separate lanthanoids by redox reactions?** (2 marks) *Solution:* Since all Ln³⁺/Ln reduction potentials are nearly identical (~–2.2 to –2.4 V), redox chemistry cannot differentiate between lanthanoids. A reducing agent strong enough to reduce one Ln³⁺ will reduce all of them simultaneously. Classical electrochemistry fails. Only small differences in ionic size (lanthanide contraction) allow separation via other methods. **(ii) Fractional crystallization of lanthanoid double salts (e.g., Ln(NO₃)₃·2NH₄NO₃) exploits lanthanide contraction. Explain why smaller lanthanoids crystallize preferentially.** (3 marks) *Solution:* As atomic number increases (La → Lu), ionic size decreases due to lanthanide contraction. Smaller La³⁺ ions (1.06 Å) form larger double salt crystals with weaker lattice energy than smaller Lu³⁺ (0.86 Å). During cooling, the larger double salt (La compound) reaches saturation first and precipitates preferentially. By repeated crystallization, heavier (smaller) lanthanoids gradually concentrate in solution while lighter ones are removed as crystals. This fractional crystallization, though labour-intensive, was historically the primary separation method. **(iii) A lanthanoid sample contains La, Nd, Gd, and Lu in equal molar ratios. When treated with NaOH solution, a gelatinous precipitate forms initially, then partially dissolves. Explain this observation.** (2 marks) *Solution:* Lanthanoid hydroxides [Ln(OH)₃] are amphoteric: - Initially, Ln³⁺(aq) + 3OH⁻(aq) → Ln(OH)₃(s) ↓ (all lanthanoids form white/colourless gelatinous precipitate) - With excess NaOH, smaller lanthanoids (Lu, Gd) form soluble [Ln(OH)₄]⁻ complexes due to higher charge density of smaller ions, which stabilize complex formation: Lu(OH)₃(s) + OH⁻(aq) → [Lu(OH)₄]⁻(aq) - Larger lanthanoids (La, Nd) remain as insoluble hydroxides Partial dissolution = selective precipitation, exploiting size differences. **(iv) Neodymium (Nd) is used in Nd-Fe-B permanent magnets (~1 T field strength). Explain why Nd³⁺ contributes to the magnetic moment and why these magnets are superior to ferrite magnets (Fe₃O₄).** (3 marks) *Solution:* **Nd³⁺ magnetic contribution:** Nd has electronic configuration [Xe] 4f³. The three unpaired 4f electrons have high orbital angular momentum and spin angular momentum. In the Nd-Fe-B crystal lattice, the 4f electrons couple with Fe d-electrons via exchange interaction, enhancing the total magnetic moment. Nd³⁺ acts as a source of unpaired spins. **Why Nd-Fe-B > Fe₃O₄:** 1. **Higher magnetic moment:** Fe₃O₄ relies only on Fe d-electrons (~4.1 μB per formula unit). Nd-Fe-B combines Fe (d⁶, ~2.6 μB) and Nd³⁺ (4f³, ~3.6 μB), yielding total ~20 μB for Nd₂Fe₁₄B, much higher field strength. 2. **Higher Curie temperature:** Nd-Fe-B has Tc ≈ 312°C (vs. Fe₃O₄ Tc ≈ 586°C, but operating range is broader). 3. **Higher remanence:** Once magnetized, Nd-Fe-B retains magnetization far better, requiring weaker external field for devices. 4. **Energy product:** Nd-Fe-B has 40–50 times higher maximum energy product than ferrite, enabling smaller, lighter magnets in motors, hard disk drives, and renewable energy generators. **Conclusion:** Lanthanoid chemistry directly enables modern technology through magnetic applications unavailable from single d-metal systems.

At CBSETUTOR.ai, we understand that mastering Chapter 4 requires more than memorizing facts—it demands pattern recognition, conceptual clarity, and the ability to apply knowledge to unseen scenarios. Our AI-powered tutoring system is trained on years of CBSE board papers, mock tests, and educator feedback to deliver precisely the practice you need. **Adaptive Learning Engine:** Our AI analyzes your responses to 1-mark MCQs in real time. When you choose the wrong reason for lanthanide contraction, the tutor doesn't simply mark it wrong—it maps the gap in your conceptual understanding, flags related misconceptions (e.g., confusing nuclear charge effects with electron shielding), and serves you targeted 2-mark and 3-mark questions to reinforce the exact concept you struggled with. This personalized sequencing means you spend zero time on what you already know. **Board-Pattern Question Bank:** We've indexed every transition metal, lanthanoid, and actinoid question type from CBSE papers (2015–2024), including deleted-curriculum topics. When you practice, you're solving variants of questions that have actually appeared—not generic textbook problems. Our system flags high-likelihood topics (e.g., "3-mark questions on why Cu is classified as transition metal" appear once every two board cycles) and ensures you see them early. **Step-by-Step Solution Walkthroughs:** For 5-mark questions, our AI doesn't hand you a pre-written answer. Instead, it guides you through solution scaffolding: *What does the question demand? Why is that property unique to transition metals? What electron configuration supports your claim? What's the examiner looking for in your conclusion?* This Socratic dialogue builds genuine problem-solving skill, not rote recall. **Misconception Detection:** Common Class 9 errors (e.g., "Zn²⁺ is colourless because Zn is not a transition metal," which conflates ground state and oxidation state definitions) are systematically addressed. When you enter an answer, our algorithm checks not just correctness but reasoning. If you've matched the right answer for the wrong reason, we surface that disconnect and drill the underlying concept. **Daily Practice Recommendations:** Based on your chapter progress, the AI prescribes a 25–minute daily drill: typically 2 MCQs (checking speed and conceptual clarity), 1 short-answer (2 marks), and 1 partial 3-mark question. This spacing maximizes retention and builds confidence without burnout. You see each question type at the right cognitive load level, at the right time. **Examiner-Aligned Vocabulary:** We flag the exact phrasing examiners reward. For instance, in explaining variable oxidation states, saying "both d and s electrons can be removed" is cleaner than "electrons are loosely held." Our AI highlights these nuances in feedback, helping you write answers that match official marking schemes. **Start a 3-day free trial at cbsetutor.ai** to see how our AI adapts to your Chapter 4 gaps. No payment required—just your learning style preferences and a brief diagnostic test. By Day 3, you'll have completed 12–15 questions, received personalized feedback on each, and understood exactly which concepts need more time before your next class test. Join thousands of CBSE Class 9 students who've raised their Chemistry score by 12–18% with targeted, AI-driven practice.