NEET Chemistry: The d- And f-Block Elements – Detailed Notes

Chapter 8: The d- and f-Block Elements

1. Introduction to d-Block Elements (Transition Elements)

• The d-block elements are located between groups 2 and 13 in the modern periodic table.
• They are characterized by the filling of d-orbitals of the penultimate (n-1) shell. Their general electronic configuration is (n−1)d1−10ns1−2.
• Transition elements are defined as those elements that have incompletely filled d-orbitals in their ground state or in any of their common oxidation states.
• Zinc (Zn), Cadmium (Cd), and Mercury (Hg) are d-block elements but are not considered transition elements because their d-orbitals are completely filled (d10 configuration) in their ground state as well as in their common oxidation states (Zn2+, Cd2+, Hg2+). However, they are often discussed with transition metals due to their position in the d-block.

2. General Characteristics of Transition Elements

A. Electronic Configuration:

• General: (n−1)d1−10ns1−2.
• Exceptions: Cr ([Ar]3d54s1), Cu ([Ar]3d104s1) due to extra stability of half-filled and completely filled d-orbitals.
• Filling of (n−1)d orbitals.

B. Metallic Character:

• All transition elements are typical metals.
• They are hard, malleable, ductile, possess high tensile strength, and have high thermal and electrical conductivities.
• High melting and boiling points due to strong metallic bonds (involving both s and d electrons).
• Mercury is an exception, being a liquid at room temperature.

C. Variable Oxidation States:

• Most characteristic property.
• They exhibit multiple oxidation states because the (n−1)d and ns electrons have very similar energies, allowing both to participate in bonding.
• Common oxidation states: +2 and +3 are very common.
• Highest oxidation state is generally observed in the middle of the series (e.g., Mn shows +7).
• Stability of oxidation states:
  • Lower oxidation states are generally ionic.
  • Higher oxidation states are often covalent (e.g., KMnO4​, K2​Cr2​O7​).
  • For the first transition series, +2 and +3 oxidation states are most stable.
  • Stability of +2 oxidation state generally decreases from left to right.

D. Formation of Coloured Ions:

• Most transition metal compounds are coloured in solid or solution states.
• Reason: The presence of incompletely filled (n-1)d orbitals. When visible light falls on the compound, electrons in lower energy d-orbitals absorb light and get excited to higher energy d-orbitals. This process is called d-d transition.
• The colour observed is the complementary colour of the light absorbed.
• Sc3+, Ti4+, Zn2+, Cu+ are colourless as they have d0 or d10 configurations, so no d-d transitions are possible.

E. Formation of Paramagnetic Ions:

• Many transition metal ions are paramagnetic.
• Reason: Presence of unpaired electrons in their d-orbitals.
• Paramagnetism increases with the number of unpaired electrons (maximum for d5 configuration, e.g., Cr3+, Mn2+, Fe3+) and decreases as the number of unpaired electrons decreases.
• d0 and d10 ions (e.g., Sc3+, Ti4+, Zn2+, Cu+) are diamagnetic as they have no unpaired electrons.
• Magnetic moment (μ) is calculated using the spin-only formula: μ=n(n+2)​ BM (Bohr Magneton), where ‘n’ is the number of unpaired electrons.

F. Formation of Complex Compounds (Coordination Compounds):

• Transition metals readily form complex compounds.
• Reasons:
  1. Small size and high effective nuclear charge of the metal ions.
  2. Availability of vacant d-orbitals to accept electron pairs from ligands.
• Examples: [Fe(CN)6​]4−, [Cu(NH3​)4​]2+, [Ni(CO)4​].

G. Catalytic Properties:

• Many transition metals and their compounds act as good catalysts.
• Reasons:
  1. Variable oxidation states: They can change their oxidation states and form unstable intermediates.
  2. Large surface area: Providing a surface for adsorption of reactants.
  3. Ability to form complexes.
• Examples: V2​O5​ in contact process, Finely divided Fe in Haber’s process, Ni in hydrogenation of oils.

H. Formation of Interstitial Compounds:

• Transition metals form interstitial compounds with small atoms like H, C, N, B by trapping them in the interstitial voids of the metal lattice.
• Properties: Non-stoichiometric, hard, high melting points, retain metallic conductivity, chemically inert.
• Examples: TiC, Fe3​H, Mn4​N.

I. Formation of Alloys:

• Transition metals readily form alloys with other metals due to their similar atomic sizes.
• Examples: Brass (Cu, Zn), Bronze (Cu, Sn), Stainless steel (Fe, Cr, Ni, C).

J. Standard Electrode Potentials (E∘):

• Generally high positive E∘ values for M2+/M indicate difficulty in forming M2+ ions.
• ECu2+/Cu∘​ is positive (+0.34 V), meaning Cu does not liberate H2​ from acids. It shows Cu is less reactive than hydrogen.
• EMn2+/Mn∘​, EV2+/V∘​, ECr2+/Cr∘​ are all negative.

3. Trends in First Transition Series (Sc to Zn)

• Atomic and Ionic Radii:
  • Generally decrease across a period due to increasing effective nuclear charge.
  • Towards the end (Fe, Co, Ni), slight increase or almost constant due to increasing electron-electron repulsion.
  • Zn has a slightly larger radius due to d10 configuration leading to increased electron-electron repulsion.
• Ionization Enthalpies:
  • Generally increase across a period (Sc to Zn) due to increasing nuclear charge and decreasing atomic size.
  • Irregularities are observed due to varying stability of d-orbital configurations (d5, d10).
  • Cr and Cu show higher IE1​ due to s1 configuration. Zn shows higher IE1​ and IE2​ due to stable 3d10 configuration.
• Oxidation States: (Discussed above)
• Magnetic Properties: (Discussed above)
• Colour: (Discussed above)

4. Some Important Compounds of Transition Elements

A. Potassium Dichromate (K2​Cr2​O7​)

• Preparation: From chromite ore (FeCr2​O4​).
  1. Oxidation of chromite ore with Na2​CO3​ and air: 4FeCr2​O4​+8Na2​CO3​+7O2​Fusion​8Na2​CrO4​+2Fe2​O3​+8CO2​
  2. Acidification of sodium chromate to sodium dichromate: 2Na2​CrO4​+2H+→Na2​Cr2​O7​+2Na++H2​O
  3. Conversion of sodium dichromate to potassium dichromate (less soluble): Na2​Cr2​O7​+2KCl→K2​Cr2​O7​↓+2NaCl
• Structure: Two tetrahedral CrO4​ units sharing one oxygen atom. Cr-O-Cr bond angle is 126∘.
• Properties:
  • Orange coloured solid.
  • Strong oxidizing agent in acidic medium (Cr2​O72−​→Cr3+). Cr2​O72−​+14H++6e−→2Cr3++7H2​O
  • Interconverts with chromate (CrO42−​) based on pH: Cr2​O72−​OH−​2CrO42−​ (Orange to Yellow in basic medium) 2CrO42−​H+​Cr2​O72−​ (Yellow to Orange in acidic medium)
• Uses: Oxidizing agent in organic reactions, primary standard in volumetric analysis.

B. Potassium Permanganate (KMnO4​)

• Preparation: From pyrolusite ore (MnO2​).
  1. Fusion of MnO2​ with KOH and O2​ (or KNO3​) to form potassium manganate (green): 2MnO2​+4KOH+O2​Heat​2K2​MnO4​+2H2​O
  2. Oxidative disproportionation of manganate to permanganate:
     • Chemical Method (acidification): 3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O
     • Electrolytic Oxidation: Oxidation of manganate (MnO42−​) to permanganate (MnO4−​) at anode in alkaline solution.
• Structure: Tetrahedral MnO4−​ ion. Mn in +7 oxidation state.
• Properties:
  • Dark purple coloured solid.
  • Strong oxidizing agent. Its oxidizing power varies with pH.
    • Acidic Medium: (MnO4−​→Mn2+) MnO4−​+8H++5e−→Mn2++4H2​O
    • Neutral/Weakly Alkaline Medium: (MnO4−​→MnO2​) MnO4−​+2H2​O+3e−→MnO2​+4OH−
    • Strongly Alkaline Medium: (MnO4−​→MnO42−​) MnO4−​+e−→MnO42−​
• Uses: Oxidizing agent in lab and industry, disinfectant, in volumetric analysis to estimate reducing agents.

5. The f-Block Elements (Inner Transition Elements)

• Characterized by the filling of (n−2)f orbitals.
• Comprise two series: Lanthanoids (4f series) and Actinoids (5f series).

A. Lanthanoids (4f-series)

• Elements from Cerium (Ce, Z=58) to Lutetium (Lu, Z=71).
• General electronic configuration: [Xe]4f1−145d0−16s2.
• Lanthanoid Contraction:
  • Gradual decrease in atomic and ionic radii (especially Ln3+ ions) from Ce to Lu.
  • Reason: Poor shielding effect of 4f electrons. As atomic number increases, nuclear charge increases, but the 4f electrons provide very poor shielding for the valence electrons. This leads to a stronger pull of the nucleus on the outermost electrons, causing a contraction in size.
  • Consequences:
    1. Elements of the second (4d) and third (5d) transition series in the same group have very similar atomic radii (e.g., Zr and Hf have almost identical radii, leading to similar chemical properties and difficulty in separation).
    2. Basicity of hydroxides decreases from Ce(OH)3​ to Lu(OH)3​ (as ionic size decreases, covalent character increases, basicity decreases).
• Oxidation States: Most common oxidation state is +3. Some elements also show +2 and +4 oxidation states (if they lead to stable f0, f7, or f14 configurations).
  • Ce4+ (f0) is a strong oxidizing agent.
  • Eu2+ (f7) and Yb2+ (f14) are strong reducing agents.
• Colour: Many lanthanoid ions are coloured due to f-f transitions.
• Chemical Reactivity: All are highly electropositive and reactive metals. React with water, acids, etc.
• Uses: Alloys (Mischmetal), catalysts.

B. Actinoids (5f-series)

• Elements from Thorium (Th, Z=90) to Lawrencium (Lr, Z=103).
• General electronic configuration: [Rn]5f1−146d0−17s2.
• Actinoid Contraction: Similar to lanthanoid contraction, a gradual decrease in atomic and ionic radii occurs from Th to Lr due to poor shielding by 5f electrons. This contraction is more pronounced than lanthanoid contraction due to even poorer shielding by 5f electrons.
• Oxidation States: Exhibit a larger number of oxidation states than lanthanoids (e.g., Th (+4), Pa (+5), U (+3, +4, +5, +6), Np (+3, +4, +5, +6, +7), Pu (+3, +4, +5, +6, +7)). This is because 5f, 6d, and 7s orbitals are very close in energy, allowing participation of more electrons in bonding. The most common state is +3.
• Colour: Most actinoid ions are coloured due to f-f transitions.
• Radioactivity: All actinoids are radioactive. Uranium and Thorium are naturally occurring, while others are synthetic.
• Chemical Reactivity: Highly reactive metals.
• Uses: Nuclear energy (Uranium, Plutonium), smoke detectors (Americium).

6. Differentiating Lanthanoids and Actinoids

Feature	Lanthanoids (4f-series)	Actinoids (5f-series)
Electronic Config.	[Xe]4f1−145d0−16s2	[Rn]5f1−146d0−17s2
Oxidation States	Predominantly +3; sometimes +2,+4	Predominantly +3; wider range (+4,+5,+6,+7)
Radioactivity	Only Promethium (Pm) is radioactive	All are radioactive
Magnetic Prop.	Less complex paramagnetism	More complex paramagnetism
Colour	Less intense colours	More intense colours
Complex Formation	Less tendency to form complexes	Greater tendency to form complexes
Basicity of Hyd.	More basic (e.g., La(OH)3​ strong base)	Less basic (e.g., U(OH)3​ weak base)
Contraction	Lanthanoid contraction (less pronounced)	Actinoid contraction (more pronounced)

NEET Chemistry: The d- and f-Block Elements – Practice Questions

I. Multiple Choice Questions (MCQs)

1. Question: Which of the following elements is not a transition element? a) Iron (Fe) b) Copper (Cu) c) Zinc (Zn) d) Chromium (Cr)

2. Question: The general electronic configuration of d-block elements is: a) (n−1)d1−10ns2 b) (n−1)d1−10ns1−2 c) (n−2)f1−14(n−1)d0−1ns2 d) ns2np1−6

3. Question: Which of the following ions is diamagnetic? a) Ti3+ (Atomic No. of Ti = 22) b) Mn2+ (Atomic No. of Mn = 25) c) Cu2+ (Atomic No. of Cu = 29) d) Sc3+ (Atomic No. of Sc = 21)

4. Question: The reason for the colour of transition metal compounds is: a) Formation of covalent bonds b) Presence of completely filled d-orbitals c) d-d transitions d) Large atomic size

5. Question: Which of the following oxidation states is most common for lanthanoids? a) +2 b) +3 c) +4 d) +5

6. Question: Lanthanoid contraction is due to: a) Poor shielding effect of 5f electrons b) Poor shielding effect of 4f electrons c) Strong shielding effect of 4f electrons d) Strong shielding effect of 5f electrons

7. Question: Which of the following is a consequence of lanthanoid contraction? a) Zr and Hf have significantly different atomic radii. b) The basicity of lanthanoid hydroxides increases from Ce to Lu. c) Elements of 4d and 5d series in the same group have almost similar radii. d) Stability of +2 oxidation state increases across the series.

8. Question: Potassium permanganate (KMnO4​) acts as an oxidizing agent in acidic medium. The oxidation state of Mn changes from: a) +7 to +4 b) +7 to +2 c) +7 to +6 d) +6 to +2

9. Question: The conversion of chromate ion (CrO42−​) to dichromate ion (Cr2​O72−​) occurs in: a) Strongly basic medium b) Strongly acidic medium c) Neutral medium d) Both acidic and basic medium

10. Question: Which of the following elements is radioactive among the lanthanoids? a) Cerium (Ce) b) Gadolinium (Gd) c) Promethium (Pm) d) Europium (Eu)

11. Question: Which of the following pairs of elements have nearly identical atomic radii due to lanthanoid contraction? a) Ti, Zr b) Zr, Hf c) V, Nb d) Cr, Mo

12. Question: Which of the following statements is incorrect regarding actinoids? a) All actinoids are radioactive. b) They show a wider range of oxidation states compared to lanthanoids. c) They have a greater tendency to form complexes than lanthanoids. d) They show less pronounced actinoid contraction compared to lanthanoid contraction.

13. Question: The spin-only magnetic moment for Mn2+ ion (Atomic No. of Mn = 25) is: a) 1.73 BM b) 3.87 BM c) 4.90 BM d) 5.92 BM

14. Question: Which transition metal compound is used as a catalyst in the contact process for the manufacture of sulphuric acid? a) Fe2​O3​ b) MnO2​ c) V2​O5​ d) Ni

15. Question: Interstitial compounds are formed when small atoms are trapped in the crystal lattice of transition metals. Which of the following is NOT a characteristic of interstitial compounds? a) High melting points b) High reactivity c) Hardness d) Retain metallic conductivity

II. Assertion-Reason Type Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is NOT the correct explanation of A. c) A is true but R is false. d) A is false but R is true.

16. Assertion (A): Zinc, cadmium, and mercury are not considered transition elements. Reason (R): They have completely filled d-orbitals in their ground state and in their common oxidation states.

17. Assertion (A): Transition metals exhibit variable oxidation states. Reason (R): The electrons in (n−1)d and ns orbitals have very similar energies and both can participate in bonding.

18. Assertion (A): Cr2+ is a stronger reducing agent than Fe2+. Reason (R): Cr2+ has d4 configuration while Fe2+ has d6 configuration. Cr2+ oxidises to Cr3+ (d3) which is stable.

19. Assertion (A): Actinoids show a wider range of oxidation states than lanthanoids. Reason (R): The 5f, 6d, and 7s orbitals have comparable energies and are more exposed for bonding.

20. Assertion (A): Most transition metal ions are paramagnetic. Reason (R): They have unpaired electrons in their d-orbitals.

III. Short Answer / Conceptual Questions

21. Question: Why do transition elements show variable oxidation states?

22. Question: How does the colour of K2​Cr2​O7​ solution change when it is made alkaline? Write the balanced chemical equation for the reaction.

23. Question: Explain the term ‘lanthanoid contraction’. What are its two important consequences?

24. Question: Calculate the spin-only magnetic moment of Fe3+ ion (Atomic No. of Fe = 26).

25. Question: What are interstitial compounds? Why are they formed by transition metals?

26. Question: Give two reasons why transition metals form a large number of complex compounds.

27. Question: Complete the following reactions: a) MnO42−​+H+→⋯+… (unbalanced) b) Cr2​O72−​+Fe2++H+→⋯+… (unbalanced)

28. Question: Differentiate between lanthanoids and actinoids based on two points.

29. Question: Why is Cu+ diamagnetic while Cu2+ is paramagnetic? (Atomic No. of Cu = 29)

30. Question: Describe the preparation of potassium permanganate from pyrolusite ore.

Answers and Explanations

I. Multiple Choice Questions (MCQs) – Answers

1. Answer: c) Zinc (Zn) Explanation: Zinc (Zn) has a completely filled d-orbital (3d10) in its ground state ([Ar]3d104s2) and in its common oxidation state (Zn2+: [Ar]3d10). Therefore, it does not meet the definition of a transition element (having incompletely filled d-orbitals).

2. Answer: b) (n−1)d1−10ns1−2 Explanation: The general electronic configuration for d-block elements involves the filling of (n−1)d orbitals, and the ns orbital can have one or two electrons (e.g., Cr has 4s1, Cu has 4s1).

3. Answer: d) Sc3+ (Atomic No. of Sc = 21) Explanation: A diamagnetic ion has no unpaired electrons. a) Ti3+: Z=22, [Ar]3d14s2→Ti3+ is [Ar]3d1 (1 unpaired electron) – Paramagnetic. b) Mn2+: Z=25, [Ar]3d54s2→Mn2+ is [Ar]3d5 (5 unpaired electrons) – Paramagnetic. c) Cu2+: Z=29, [Ar]3d104s1→Cu2+ is [Ar]3d9 (1 unpaired electron) – Paramagnetic. d) Sc3+: Z=21, [Ar]3d14s2→Sc3+ is [Ar]3d0 (0 unpaired electrons) – Diamagnetic.

4. Answer: c) d-d transitions Explanation: The colour of transition metal compounds is primarily due to d-d transitions. In the presence of ligands or a crystal field, the d-orbitals split into different energy levels. When white light falls on the compound, electrons absorb energy from the visible region and jump from a lower energy d-orbital to a higher energy d-orbital. The colour observed is the complementary colour of the light absorbed.

5. Answer: b) +3 Explanation: The most common and stable oxidation state for all lanthanoids is +3. Other oxidation states (+2,+4) are observed for a few elements to achieve stable f0,f7, or f14 configurations, but +3 is universal.

6. Answer: b) Poor shielding effect of 4f electrons Explanation: Lanthanoid contraction is the gradual decrease in atomic and ionic radii across the lanthanoid series. This is caused by the poor shielding (screening) effect of the 4f electrons. As the nuclear charge increases with atomic number, the 4f electrons are not effective in shielding the outer electrons, leading to a stronger effective nuclear charge and thus a contraction in size.

7. Answer: c) Elements of 4d and 5d series in the same group have almost similar radii. Explanation: The lanthanoid contraction causes the atomic radii of elements in the 5d transition series to be almost identical to those of the corresponding elements in the 4d series. For example, Zr (4d) and Hf (5d) have very similar atomic radii and chemical properties, making their separation difficult.

8. Answer: b) +7 to +2 Explanation: In acidic medium, potassium permanganate (KMnO4​) acts as a strong oxidizing agent. The manganese in MnO4−​ is in the +7 oxidation state. It gets reduced to Mn2+ (the +2 oxidation state) in acidic solution.

9. Answer: b) Strongly acidic medium Explanation: Chromate ions (CrO42−​, yellow) exist predominantly in basic medium, while dichromate ions (Cr2​O72−​, orange) exist predominantly in acidic medium. The interconversion is pH-dependent: 2CrO42−​+2H+⇌Cr2​O72−​+H2​O. Thus, adding acid converts chromate to dichromate.

10. Answer: c) Promethium (Pm) Explanation: Among the lanthanoids, only Promethium (Pm) is a radioactive element. All other lanthanoids are stable.

11. Answer: b) Zr, Hf Explanation: Due to lanthanoid contraction, the elements of the 4d and 5d transition series that are in the same group have very similar atomic and ionic radii. Zirconium (Zr) and Hafnium (Hf) are in Group 4 and have almost identical atomic radii, leading to very similar chemical properties.

12. Answer: d) They show less pronounced actinoid contraction compared to lanthanoid contraction. Explanation: Actinoid contraction is more pronounced than lanthanoid contraction. This is because 5f electrons provide even poorer shielding than 4f electrons, leading to a greater cumulative effect of nuclear charge. All other statements are correct regarding actinoids.

13. Answer: d) 5.92 BM Explanation: For Mn2+ (Atomic No. 25), the electronic configuration is [Ar]3d5. There are 5 unpaired electrons (n=5). Using the spin-only formula: μ=n(n+2)​=5(5+2)​=5×7​=35​≈5.92 BM.

14. Answer: c) V2​O5​ Explanation: Vanadium pentoxide (V2​O5​) is used as a catalyst in the contact process for the industrial manufacture of sulphuric acid. It catalyzes the oxidation of sulfur dioxide to sulfur trioxide (2SO2​+O2​V2​O5​​2SO3​).

15. Answer: b) High reactivity Explanation: Interstitial compounds are generally quite chemically inert and stable. They are known for their extreme hardness, high melting points, and retention of metallic conductivity, but not high reactivity.

II. Assertion-Reason Type Questions – Answers

16. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: The definition of a transition element requires incompletely filled d-orbitals in their ground state or in any common oxidation state. Zinc, cadmium, and mercury have full d-orbitals (d10) in their common oxidation states (+2), which means they do not fit this definition. Thus, R correctly explains A.

17. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Transition metals exhibit variable oxidation states because their (n−1)d and ns orbitals are very close in energy. This allows electrons from both sets of orbitals to participate in chemical bonding, leading to multiple stable oxidation states.

18. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Cr2+ has a d4 configuration. It acts as a stronger reducing agent because it can readily lose one electron to achieve the more stable d3 configuration of Cr3+, which benefits from a half-filled t2g​ sub-level (in octahedral complexes). Fe2+ (d6) is less prone to this type of stable configuration upon oxidation.

19. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Actinoids indeed show a wider range of oxidation states compared to lanthanoids. This is due to the fact that 5f, 6d, and 7s orbitals in actinoids are of comparable energies and are more exposed for bonding compared to 4f orbitals in lanthanoids, allowing more electrons to participate in bonding.

20. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Paramagnetism arises from the presence of unpaired electrons in a substance. Most transition metal ions have incompletely filled d-orbitals, which means they have one or more unpaired electrons. These unpaired electrons give rise to paramagnetic behaviour.

III. Short Answer / Conceptual Questions – Answers

21. Answer: Transition elements exhibit variable oxidation states primarily due to two reasons:

1. Similar Energies of (n−1)d and ns Orbitals: The energies of the (n−1)d and ns orbitals are very close. This allows electrons from both these subshells to participate in bonding.
2. Formation of Stable Configurations: They tend to achieve stable d0, d5 (half-filled), or d10 (fully-filled) configurations by losing or gaining electrons, leading to different oxidation states.

22. Answer: When potassium dichromate (K2​Cr2​O7​) solution (orange) is made alkaline, it converts to potassium chromate (K2​CrO4​) solution (yellow). Balanced chemical equation: Cr2​O72−​ (Orange)+2OH−→2CrO42−​ (Yellow)+H2​O

23. Answer:

• Lanthanoid Contraction: It refers to the steady and gradual decrease in the atomic and ionic radii of the lanthanoid elements (from Cerium, Z=58 to Lutetium, Z=71) as the atomic number increases.
• Reason: This contraction is caused by the poor shielding effect (or screening effect) of the 4f electrons. As we move across the series, the nuclear charge increases steadily, but the 4f electrons are ineffective in shielding the outer valence electrons from the increasing nuclear pull. This results in a stronger attraction of the nucleus towards the outermost electrons, leading to a contraction in size.
• Two Important Consequences:
  1. Similar Atomic Radii of 4d and 5d Transition Series Elements: Due to lanthanoid contraction, the elements of the 5d transition series have atomic radii very similar to their corresponding elements in the 4d series (e.g., Zr and Hf have almost identical sizes). This similarity in size leads to very similar chemical properties, making their separation difficult.
  2. Decreased Basicity of Lanthanoid Hydroxides: As the ionic radius of Ln3+ ions decreases across the series (Ce3+ to Lu3+), the covalent character of the Ln−OH bond in Ln(OH)3​ increases, and consequently, their basic strength decreases. Thus, Ce(OH)3​ is more basic than Lu(OH)3​.

24. Answer: To calculate the spin-only magnetic moment for Fe3+ ion (Atomic No. of Fe = 26):

1. Electronic Configuration of Fe: [Ar]3d64s2
2. Electronic Configuration of Fe3+ ion: To form Fe3+, two electrons are removed from the 4s orbital and one electron from the 3d orbital. Fe3+ is [Ar]3d5.
3. Number of Unpaired Electrons (n): In the 3d5 configuration, all five d-orbitals contain one electron each (according to Hund’s rule). So, n=5.
4. Spin-only Magnetic Moment (μ): Using the formula μ=n(n+2)​ BM. μ=5(5+2)​=5×7​=35​ μ≈5.92 BM (Bohr Magnetons)

25. Answer:

• Interstitial Compounds: These are compounds formed when small atoms (like Hydrogen (H), Carbon (C), Nitrogen (N), or Boron (B)) are trapped in the interstitial voids (or spaces) within the crystal lattice of transition metals.
• Reason for Formation by Transition Metals: Transition metals have crystal lattices with sufficiently large interstitial sites that can accommodate these small atoms. Also, their variable valency and metallic bonding characteristics allow for the stabilization of these structures.

26. Answer: Transition metals form a large number of complex compounds (coordination compounds) due to the following reasons:

1. Small Size and High Effective Nuclear Charge: Transition metal ions generally have a small size and a high positive charge density (high effective nuclear charge). This enables them to attract and hold electron pairs donated by ligands effectively.
2. Availability of Vacant d-Orbitals: They possess vacant d-orbitals of appropriate energy to accept lone pairs of electrons from ligands (electron-pair donors) to form coordinate bonds. This allows for the formation of stable complexes.

27. Answer: a) Conversion of Manganate to Permanganate in Acidic Medium (Oxidative disproportionation): 3MnO42−​ (Green)+4H+→2MnO4−​ (Purple)+MnO2​+2H2​O b) Oxidation of Fe2+ by Dichromate in Acidic Medium: Cr2​O72−​+6Fe2++14H+→2Cr3++6Fe3++7H2​O

28. Answer: | Feature | Lanthanoids (4f-series) | Actinoids (5f-series) | | :——————- | :————————————– | :—————————————— | | Oxidation States | Predominantly +3; sometimes +2,+4 | Predominantly +3; wider range (+4,+5,+6,+7) | | Radioactivity | Only Promethium (Pm) is radioactive | All are radioactive | | Complex Formation | Less tendency to form complexes | Greater tendency to form complexes | | Contraction | Lanthanoid contraction (less pronounced) | Actinoid contraction (more pronounced) |

29. Answer:

• Copper (Cu): Atomic Number = 29.
  • Electronic Configuration of Cu: [Ar]3d104s1.
• Cu+ ion: Formed by losing one 4s electron.
  • Electronic Configuration of Cu+: [Ar]3d10.
  • Since all electrons in the 3d orbital are paired (a completely filled d-orbital), Cu+ has no unpaired electrons. Therefore, Cu+ is diamagnetic.
• Cu2+ ion: Formed by losing one 4s electron and one 3d electron.
  • Electronic Configuration of Cu2+: [Ar]3d9.
  • In the 3d9 configuration, there is one unpaired electron in the 3d orbital. Therefore, Cu2+ is paramagnetic.

30. Answer: Potassium permanganate (KMnO4​) is prepared from pyrolusite ore (MnO2​) in two main steps:

1. Conversion of MnO2​ to Potassium Manganate (K2​MnO4​): Finely powdered pyrolusite ore (MnO2​) is fused with solid KOH (potassium hydroxide) in the presence of an oxidizing agent like atmospheric oxygen (O2​) or potassium nitrate (KNO3​). This reaction produces potassium manganate, which is green in colour. 2MnO2​+4KOH+O2​Heat/Fusion​2K2​MnO4​ (Green)+2H2​O
2. Oxidation of Potassium Manganate to Potassium Permanganate (KMnO4​): Potassium manganate can be converted to potassium permanganate by either chemical oxidation (acidification) or electrolytic oxidation.
   • Chemical Oxidation (Disproportionation): The green solution of potassium manganate is acidified using a dilute acid (like H2​SO4​). Manganate ion undergoes disproportionation, where a part of it oxidizes to permanganate and another part reduces to manganese dioxide. 3K2​MnO4​+4HCl→2KMnO4​ (Purple)+MnO2​+2KCl+2H2​O (Ionic reaction: 3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O)
   • Electrolytic Oxidation: In this method, a solution of potassium manganate is electrolyzed. Manganate ions are oxidized at the anode to permanganate ions, while water is reduced at the cathode to produce hydrogen gas. This is a common industrial method. At Anode: MnO42−​→MnO4−​+e− At Cathode: 2H2​O+2e−→H2​+2OH−