Exhaustive Guide: Important Compounds & f-Block | Class 12 Chemistry

Compounds & f-Block Elements

Module 3 | CBSE Class 12 Chemistry | Final d- & f-Block Module

Table of Contents

• 1. Potassium Dichromate (K₂Cr₂O₇)
  • 1.1 Preparation
  • 1.2 Properties and Oxidizing Action
• 2. Potassium Permanganate (KMnO₄)
  • 2.1 Preparation
  • 2.2 Properties and Oxidizing Action
• 3. The Lanthanoids (4f Series)
  • 3.1 Lanthanoid Contraction and Consequences
• 4. The Actinoids (5f Series)
• 5. Comparison: Lanthanoids vs Actinoids
• 6. Previous Year Questions (PYQs)

1. Potassium Dichromate (K₂Cr₂O₇)

Potassium dichromate is a very important chemical used extensively as an oxidizing agent in volumetric analysis and in the leather industry.

1.1 Preparation from Chromite Ore

It is generally prepared from chromite ore (FeCr₂O₄) in a three-step process:

1. Fusion: Chromite ore is fused with sodium carbonate (Na₂CO₃) in free excess of air to form yellow Sodium Chromate.
   4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2. Acidification: The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to yield an orange solution of Sodium Dichromate.
   2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
3. Conversion to Potassium Salt: Sodium dichromate is more soluble than potassium dichromate. Hence, it is treated with Potassium Chloride (KCl) to precipitate orange crystals of Potassium Dichromate.
   Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ (orange crystals) + 2NaCl

1.2 Properties and Oxidizing Action

Chromate-Dichromate Equilibrium (pH Dependent):
The chromate (CrO₄^2-, yellow) and dichromate (Cr₂O₇^2-, orange) ions are interconvertible in aqueous solution depending on the pH.
Acidic Medium (Low pH): 2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O (Turns Orange)
Basic Medium (High pH): Cr₂O₇^2- + 2OH^- → 2CrO₄^2- + H₂O (Turns Yellow)

Oxidizing Action in Acidic Medium:
Potassium dichromate acts as a strong oxidizing agent. The dichromate ion is reduced to Cr³⁺ (green).

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O    (E° = +1.33 V)

Important Oxidations:

• It oxidizes iodides to iodine: 6I^- → 3I₂ + 6e^-
• It oxidizes H₂S to sulphur: 3H₂S → 3S + 6H⁺ + 6e^-
• It oxidizes Fe²⁺ to Fe³⁺ (used in titration): 6Fe²⁺ → 6Fe³⁺ + 6e^-

2. Potassium Permanganate (KMnO₄)

A dark purple, almost black, crystalline solid. It is a powerful oxidizing agent used in analytical chemistry (redox titrations) and as a disinfectant.

2.1 Preparation

It is prepared from Pyrolusite ore (MnO₂).

1. Fusion: MnO₂ is fused with an alkali metal hydroxide (KOH) and an oxidizing agent (like KNO₃ or O₂) to give dark green Potassium Manganate (K₂MnO₄).
   2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
2. Electrolytic Oxidation: The green manganate is then oxidized electrolytically in alkaline solution to yield purple permanganate.
   MnO₄^2- (green) → MnO₄^- (purple) + e^-

2.2 Properties and Oxidizing Action

The oxidizing action of KMnO₄ is highly dependent on the pH of the medium.

A. In Acidic Medium:

It acts as a very strong oxidant, reducing from +7 to +2 state (colourless).

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

• Oxalates (C₂O₄^2-) are oxidized to CO₂.
• Fe²⁺ is oxidized to Fe³⁺.
• Iodides (I^-) are oxidized to Iodine (I₂).

B. In Neutral or Faintly Alkaline Medium:

It is reduced from +7 to +4 state (forming brown precipitate of MnO₂).

MnO₄^- + 2H₂O + 3e^- → MnO₂(s) + 4OH^-

Important Exception: In neutral/alkaline medium, KMnO₄ oxidizes Iodide (I^-) directly to Iodate (IO₃^-), not Iodine (I₂).

3. The Lanthanoids (4f Series)

The 14 elements immediately following Lanthanum (La, Z=57), from Cerium (Ce, 58) to Lutetium (Lu, 71), in which the 4f orbitals are progressively filled.

• Electronic Configuration: [Xe] 4f^1-14 5d^0-1 6s².
• Oxidation States: The most common and stable oxidation state is +3. However, some elements show +2 and +4 states if it leads to stable empty (f⁰), half-filled (f⁷), or fully filled (f¹⁴) subshells.
  Examples: Ce⁴⁺ is a strong oxidant (it wants to gain an electron to reach the stable +3 state). Eu²⁺ and Yb²⁺ are strong reducing agents.

3.1 Lanthanoid Contraction and Consequences

Lanthanoid Contraction: The steady and regular decrease in atomic and ionic radii of lanthanoids from Cerium to Lutetium with increasing atomic number.

Cause: As we move along the period, the nuclear charge increases. The new electrons enter the inner 4f subshell. The 4f electrons provide very poor shielding (screening) effect. Hence, the effective nuclear charge pulling the outermost electrons increases, causing the size to contract.

Consequences:

1. Similarity in 2nd and 3rd transition series: Elements of the 4d and 5d series have almost identical sizes (e.g., Zr and Hf), making them very difficult to separate.
2. Decrease in Basic Strength: As the size of Ln³⁺ ions decreases from La to Lu, the covalent character of the Ln-OH bond increases. Thus, La(OH)₃ is the most basic while Lu(OH)₃ is the least basic.

4. The Actinoids (5f Series)

The 14 elements following Actinium (Ac, 89), from Thorium (Th, 90) to Lawrencium (Lr, 103), involving the filling of 5f orbitals. All actinoids are radioactive.

• Electronic Configuration: [Rn] 5f^1-14 6d^0-1 7s².
• Oxidation States: Unlike lanthanoids, actinoids show a much wider range of oxidation states (e.g., +3, +4, +5, +6, +7). The +3 state is most common, but higher states are frequent in the earlier elements (e.g., Uranium shows +4, Neptunium shows +5, Plutonium and Americium show +6).

5. Comparison: Lanthanoids vs Actinoids

Property	Lanthanoids (4f)	Actinoids (5f)
Oxidation States	Mainly +3. Limited higher states (+2, +4).	Exhibit a wide range of oxidation states (+3 to +7).
Reason for Oxidation States	Large energy gap between 4f and 5d orbitals.	Comparable energies of 5f, 6d, and 7s orbitals allow all of them to participate in bonding.
Contraction	Lanthanoid Contraction is less pronounced.	Actinoid Contraction is greater from element to element due to poorer shielding by 5f electrons.
Complex Formation	Less tendency to form complexes.	Higher tendency to form complexes.
Radioactivity	Only Promethium (Pm) is radioactive.	All actinoids are radioactive.

6. Previous Year Questions (PYQs)

Part A: Conceptual & Give Reasons (1-2 Marks)

[CBSE 2017, 2021]

Q1. Why does the colour of K₂Cr₂O₇ change when treated with a base?

Answer: The orange dichromate ion (Cr₂O₇^2-) is in pH-dependent equilibrium with the yellow chromate ion (CrO₄^2-). When a base is added (pH increases), the OH^- ions shift the equilibrium to the right, converting the orange dichromate into yellow chromate.

[CBSE 2018, 2020]

Q2. Why are actinoid contractions greater from element to element than lanthanoid contractions?

Answer: This is because the 5f electrons (in actinoids) provide poorer shielding from the nuclear charge than the 4f electrons (in lanthanoids). As a result, the outer electrons in actinoids are pulled more strongly toward the nucleus, leading to a more pronounced contraction in size.

[CBSE 2016, 2019]

Q3. Actinoids exhibit a much larger number of oxidation states than lanthanoids. Explain why.

Answer: In actinoids, the energy levels of the 5f, 6d, and 7s orbitals are very close to one another (comparable energies). Therefore, electrons from all these subshells can readily participate in chemical bonding, allowing them to exhibit a wide range of oxidation states (up to +7).

Part B: Chemical Reactions (3 Marks)

[CBSE 2015, 2022]

Q4. Complete and balance the following chemical equations:
(i) MnO₄^- + C₂O₄^2- + H⁺ →
(ii) Cr₂O₇^2- + Fe²⁺ + H⁺ →
(iii) MnO₄^- + I^- + H₂O → (Neutral medium)

Answer:
(i) 2MnO₄^- + 5C₂O₄^2- + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
(ii) Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(iii) 2MnO₄^- + I^- + H₂O → 2MnO₂ + IO₃^- + 2OH^- (Note: Iodide is oxidized to Iodate in neutral/alkaline medium).

[CBSE Sample Paper 2024]

Q5. Write the chemistry of preparation of potassium dichromate from chromite ore.

Answer:
Step 1: Fusion of chromite ore with sodium carbonate in air.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ (yellow) + 2Fe₂O₃ + 8CO₂
Step 2: Acidification of sodium chromate to sodium dichromate.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ (orange) + 2Na⁺ + H₂O
Step 3: Conversion to Potassium Dichromate by reacting with KCl.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ (orange crystals) + 2NaCl

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