d- and f-block elements for CBSE Class 12 Board Exams:

 D- and F-block elements for CBSE Class 12 Board Exams:

1. Why do transition elements show variable oxidation states?

Answer:

  • Transition elements have partially filled d-orbitals, allowing them to lose different numbers of electrons.
  • The energy difference between (n-1)d and ns orbitals is small, so both contribute to bonding.
  • Example:
    • Iron (Fe): +2 (Fe²⁺) and +3 (Fe³⁺)
    • Manganese (Mn): +2, +3, +4, +6, +7 (in KMnO₄)

2. Why are Zn, Cd, and Hg not considered transition elements?

Answer:

  • Transition elements should have an incomplete d-subshell in their ground state or oxidation state.
  • Zn, Cd, and Hg have a completely filled dⁱ⁰ configuration in all oxidation states.
  • Hence, they do not show typical properties like variable oxidation states, colored compounds, and catalytic behavior.

3. Why do transition metals form colored compounds?

Answer:

  • Due to the d-d electronic transitions within partially filled d-orbitals.
  • In presence of a ligand, the d-orbitals split into different energy levels.
  • When light falls, electrons jump from a lower energy level to a higher one, absorbing a part of visible light.
  • The remaining light is reflected, which gives color to the compound.
  • Example: Cu²⁺ (blue), Fe³⁺ (yellow-brown), MnO₄⁻ (purple).

4. Why do transition metals and their compounds act as good catalysts?

Answer:

  • Variable oxidation states help in redox reactions.
  • Large surface area provides active sites for reactants.
  • Ability to form complexes with reactants lowers the activation energy.
  • Example: Fe in Haber’s process, V₂O₅ in Contact process.

5. Why do transition metals form complexes easily?

Answer:

  • Due to small size and high charge density.
  • Presence of vacant d-orbitals allows them to accept lone pairs from ligands.
  • Example: [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺.

6. Why is the enthalpy of atomization high for transition metals?

Answer:

  • Due to strong metallic bonding (involvement of d-electrons).
  • More unpaired electrons = Stronger bonding = Higher enthalpy of atomization.
  • Example: Tungsten (W) has a very high melting point due to strong bonding.

7. What is lanthanoid contraction? What are its consequences?

Answer:

  • Lanthanoid contraction: The gradual decrease in atomic and ionic radii of lanthanides across the period due to poor shielding by 4f-electrons.
  • Consequences:
    1. Similar atomic size of Zr (160 pm) and Hf (159 pm).
    2. Increase in density of later lanthanoids.
    3. Greater stability of +3 oxidation state.

8. Why do actinoids show a greater range of oxidation states than lanthanoids?

Answer:

  • Due to poor shielding of 5f-electrons, making them more available for bonding.
  • Actinoids exhibit oxidation states from +3 to +6, while lanthanoids show only +3.
  • Example: Uranium (U) shows +3, +4, +5, +6 oxidation states.

9. Why does KMnO₄ act as a strong oxidizing agent in acidic, basic, and neutral media?

Answer:

  • Mn in KMnO₄ is in +7 oxidation state, which is highly unstable.

  • In acidic medium:

    MnO4−+8H++5e−→Mn2++4H2OMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

    (Purple → Colorless)

  • In neutral medium:

    MnO4−+4H++3e−→MnO2+2H2OMnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O

    (Purple → Brown)

  • In basic medium:

    MnO4−+e−→MnO42−MnO₄⁻ + e⁻ → MnO₄^{2-}

    (Purple → Green)

10. Explain the oxidation behavior of K₂Cr₂O₇ in acidic medium.

Answer:

  • Chromate ions (Cr2O72−)(Cr₂O_7^{2-}) act as oxidizing agents.

  • In acidic medium:

    Cr2O72−+14H++6e−→2Cr3++7H2OCr₂O_7^{2-} + 14H⁺ + 6e⁻ → 2Cr^{3+} + 7H₂O

    (Orange → Green)

  • Used in oxidation of Fe²⁺ to Fe³⁺, I⁻ to I₂, SO₃²⁻ to SO₄²⁻.

11. Why do transition metals exhibit high tensile strength and hardness?

Answer:

  • Due to strong metallic bonding arising from delocalized d-electrons.
  • More unpaired electrons = Stronger bonding.
  • Example: Tungsten (W) is used in filaments due to high melting point and strength.

12. Why do actinoids have poor shielding effect compared to lanthanoids?

Answer:

  • 5f-electrons in actinoids have diffused orbitals, leading to less effective shielding.
  • Higher effective nuclear charge pulls electrons closer, reducing atomic size.
  • This results in higher reactivity and greater oxidation states.

13. Applications of lanthanides and actinides in industries and nuclear reactors.

Answer:

  • Lanthanides:

    1. CeO₂ – Used for glass polishing.
    2. Nd-Fe-B magnets – Used in hard drives and wind turbines.
    3. La in camera lenses – Increases clarity.

  • Actinides:

    1. Uranium (U) and Plutonium (Pu) – Used in nuclear reactors.
    2. Thorium (Th) – Used in nuclear power generation.
    3. Americium (Am) – Used in smoke detectors.

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