CBSE Board Class 12 Chemistry Important Reactions List for Boards

Understanding inorganic chemistry is a cornerstone of the CBSE Board Class 12 Chemistry Important Reactions List for Boards. Mastering the fundamental principles of oxidation states, identifying common oxidizing agents, and recognizing specific reaction patterns is important for achieving high scores. By focusing on these core concepts, students can move beyond rote memorization to a logical understanding essential for success in both board and competitive examinations.

Fundamentals of Oxidation and Reduction

The study of inorganic chemistry in the CBSE Board Class 12 Chemistry Important Reactions List for Boards relies heavily on a clear understanding of electron transfer. By mastering how elements shift between oxidation states, you can predict the products of complex redox reactions without relying solely on rote memorization. The following principles outline the behavior of key reagents and the rules governing their transformations.

1. Oxidizing Agents (Oxidants)

An oxidizing agent is a substance that causes the oxidation of another species in a reaction, meaning it increases the oxidation number of the element it reacts with. For instance, an oxidant can convert Cr³⁺ to Cr⁶⁺.

Key characteristics and examples:

• Common Examples: O₂, KNO₃, K₂Cr₂O₇, KMnO₄, and S₂O₈²⁻ (Peroxodisulfate).

• Defining Property: Oxidizing agents typically contain an element in its maximum possible oxidation state. An element at its highest oxidation state can only be reduced (its oxidation number decreases), and a substance that gets reduced is an oxidizing agent.

• KNO₃: Nitrogen is in the +5 state, its maximum.

• K₂Cr₂O₇: Chromium is in the +6 state, its maximum.

• KMnO₄: Manganese is in the +7 state, its maximum.

• S₂O₈²⁻: This is an oxidant due to a peroxide linkage (O-O) where oxygen has a -1 oxidation state. It is unstable and readily breaks to form SO₄²⁻, with oxygen going to its more stable -2 state, thereby reducing the agent itself.

2. Rules for Oxidation States and Reaction Types

• Maximum Oxidation Number: For many elements, it equals the Group Number. For p-block elements, it is often Group Number - 10 (Memory Tip: For Group 15, the max oxidation state is 15 - 10 = +5).

• Identifying Agent Type:

• If an element is in its maximum oxidation state, the compound acts as an oxidizing agent.

• If an element is in its minimum oxidation state, the compound acts as a reducing agent.

• Disproportionation Reaction (DPP): A reaction where a single species undergoes both oxidation and reduction simultaneously. This occurs when the element is in an intermediate oxidation state.

• Example 1 (NCERT): Cu⁺ disproportionates in aqueous solution. Cu⁺ → Cu²⁺ (Oxidation: +1 to +2) and Cu⁺ → Cu⁰ (Reduction: +1 to 0).

• Example 2 (NCERT): Manganate ion (Mn⁶⁺) disproportionates in acidic medium. Mn⁶⁺ → Mn⁷⁺ (in MnO₄⁻) (Oxidation) and Mn⁶⁺ → Mn⁴⁺ (in MnO₂) (Reduction).

Reactions of Copper (Cu)

The chemistry of copper is unique due to the stability differences between its +1 (cuprous) and +2 (cupric) oxidation states. Understanding how copper interacts with different anions allows students to predict whether a simple salt will form or if a complex redox transformation will occur.

1. Reaction of Cu²⁺ with Halides: A Comparative Analysis

The reaction of Cu²⁺ with halide ions shows distinct behavior.

• Cases with F⁻, Cl⁻, and Br⁻: These proceed via simple combination.

• Cu²⁺ + 2F⁻ → CuF₂

• Cu²⁺ + 2Cl⁻ → CuCl₂

• Cu²⁺ + 2Br⁻ → CuBr₂

• Case with I⁻ (Iodide): A simple combination to form CuI₂ does not occur. Instead, a redox reaction takes place.

• Cu²⁺ acts as an oxidizing agent, oxidizing iodide (I⁻) to iodine (I₂). (2I⁻ → I₂ + 2e⁻)

• Cu²⁺ itself is reduced to Cu⁺. (Cu²⁺ + e⁻ → Cu⁺)

• The final product is CuI, not CuI₂.

• Reason: The oxidizing power order is F₂ > Cl₂ > Br₂ > I₂. Iodine (I₂) is the weakest oxidizing agent. Cu²⁺ is strong enough to oxidize I⁻ to I₂ but not strong enough to oxidize F⁻, Cl⁻, or Br⁻.

2. Reaction of Copper with Dilute HCl

Copper metal does not react with dilute HCl.

• Cu + dil. HCl → No Reaction

• Reason: The standard electrode potential (E°) for the Cu²⁺/Cu couple is positive. This favors reduction (Cu²⁺ → Cu). The oxidation of Cu (Cu → Cu²⁺) by H⁺ is thermodynamically unfavorable.

General Reaction Heuristics (BKTs)

Mastering inorganic chemistry often requires recognizing patterns rather than memorizing individual equations. These "Best Known Tricks" (BKTs) serve as reliable mental models that help you predict the behavior of various compounds under specific conditions, such as heating or dehydration.

"पानी निकालो" (Remove Water) Rule

When a compound containing hydrogen and oxygen is heated, a common reaction pathway is the elimination of water (H₂O) (Memory Tip: Think "पानी निकालो" to remember to remove water when heating such compounds).

• H₂CO₃ (aq) → H₂O + CO₂

• H₂SO₄ → H₂O + SO₃

• H₂SO₃ → H₂O + SO₂

• (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O (8H atoms lead to 4H₂O)

• 2HCrO₄⁻ → Cr₂O₇²⁻ + H₂O (Key for dichromate formation)

Preparation and Reactions of K₂Cr₂O₇ (Potassium Dichromate)

Potassium dichromate is one of the most important reagents in inorganic chemistry, serving as a primary standard in volumetric analysis and a versatile oxidizing agent in organic synthesis. Understanding its synthesis from ore and its pH-dependent behavior is crucial for mastering d-block chemistry.

1. Chromate vs. Dichromate

• Chromate: CrO₄²⁻ (contains one Cr atom, yellow).

• Dichromate: Cr₂O₇²⁻ (contains two Cr atoms, orange).

• In both ions, the oxidation state of Chromium is +6.

2. Preparation of Potassium Dichromate

The process starts from chromite ore (FeCr₂O₄ or FeO·Cr₂O₃) where Iron is +2 and Chromium is +3.

1. Ore Concentration and Oxidation: Oxidize Cr³⁺ to Cr⁶⁺ by reacting chromite ore with an oxidizing agent (O₂) and a base (Na₂CO₃). This forms yellow sodium chromate (Na₂CrO₄). Iron also oxidizes to Fe₂O₃.
   4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

2. Conversion of Chromate to Dichromate: Acidify the sodium chromate solution with H⁺. This converts the yellow chromate to orange dichromate (Cr₂O₇²⁻).
   2CrO₄²⁻ (yellow) + 2H⁺ → Cr₂O₇²⁻ (orange) + H₂O

3. Formation of Potassium Dichromate: Treat the sodium dichromate solution with potassium chloride (KCl). Potassium dichromate is less soluble than sodium dichromate and crystallizes out.
   Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

3. K₂Cr₂O₇ as an Oxidizing Agent

In an acidic medium, K₂Cr₂O₇ is a powerful oxidizing agent. The Cr⁶⁺ in dichromate readily reduces to the more stable Cr³⁺ state.

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

4. BKTs: Predicting Products of K₂Cr₂O₇ Reactions (in Acidic Medium)

These rules simplify product prediction (Memory Tip: Use these rules when K₂Cr₂O₇ acts as an oxidizing agent in acidic solutions).

• Rule 1: Oxidation of Negative Ions: Negative oxidation state + Oxidizing Agent → Zero oxidation state product.

• I⁻ → I₂ (iodine, 0 state)

• Cl⁻ → Cl₂ (chlorine, 0 state)

• Br⁻ → Br₂ (bromine, 0 state)

• S²⁻ (from H₂S) → S (sulfur, 0 state)

• Rule 2: Oxidation of p-block and d-block Cations:

• d-block elements: Oxidation state increases by +1. (e.g., Fe²⁺ → Fe³⁺)

• p-block elements: Oxidation state increases by +2. (e.g., Sn²⁺ → Sn⁴⁺, S⁺⁴ → S⁺⁶, N⁺³ → N⁺⁵)

• Rule 3: Oxidation of Organic Compounds:

• Alcohols are oxidized to Carboxylic Acids. (e.g., C₂H₅OH → CH₃COOH)

• Note: In all these reactions, dichromate is reduced to Cr³⁺.

Preparation and Reactions of KMnO₄ (Potassium Permanganate)

Potassium permanganate is perhaps the most famous oxidizing agent used in chemical titrations and organic transformations. Known for its intense purple color and versatile reactivity across different pH levels, its preparation and properties are essential topics for both theory and practical examinations.

1. Manganate vs. Permanganate

• Potassium Permanganate (KMnO₄): Contains MnO₄⁻ ion, Mn in +7 oxidation state.

• Potassium Manganate (K₂MnO₄): Contains MnO₄²⁻ ion, Mn in +6 oxidation state.

2. Preparation of Potassium Permanganate

1. From Pyrolusite Ore (MnO₂): Fuse MnO₂ (Mn in +4 state) with KOH and an oxidizing agent (O₂ or KNO₃). This oxidizes Mn⁴⁺ to Mn⁶⁺, forming dark green potassium manganate (K₂MnO₄).
   2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

2. Conversion of Manganate to Permanganate: Convert green K₂MnO₄ (Mn⁶⁺) to KMnO₄ (Mn⁷⁺).

• a) Disproportionation: In neutral or weakly acidic medium, Mn⁶⁺ disproportionates.
  3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

• b) Commercial Method (Oxidation): Oxidize manganate with a stronger oxidizer like chlorine (Cl₂) or ozone (O₃). This avoids MnO₂ byproduct.

• c) Lab Method: Use a powerful oxidizer like peroxodisulfate (S₂O₈²⁻) to oxidize Mn²⁺ salt to permanganate.

3. KMnO₄ as an Oxidizing Agent

KMnO₄ is a very strong oxidizing agent in acidic, basic, and neutral media due to Mn being in its highest oxidation state (+7). The reduction product depends on the medium:

• Acidic Medium: Mn⁷⁺ is reduced to Mn²⁺.

• Basic or Neutral Medium: Mn⁷⁺ is reduced to Mn⁴⁺ (precipitated as MnO₂).

4. Reactions of KMnO₄

• Heating Effect: KMnO₄ decomposes on heating.
  2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

• Reactions in Acidic Medium: Reaction patterns are identical to K₂Cr₂O₇. The same BKTs apply.

• Fe²⁺ → Fe³⁺

• Sn²⁺ → Sn⁴⁺

• SO₂ / SO₃²⁻ (S⁺⁴) → SO₄²⁻ (S⁺⁶)

• NO₂⁻ (N⁺³) → NO₃⁻ (N⁺⁵)

• Negative ions → Zero state (I⁻ → I₂, Br⁻ → Br₂, Cl⁻ → Cl₂, S²⁻ → S)

• Oxalate (C₂O₄²⁻) → CO₂

• Alcohols → Carboxylic Acids

• Note: In these reactions, MnO₄⁻ is reduced to Mn²⁺.

• Reactions in Basic or Neutral Medium: Products differ from acidic medium.

• I⁻ (Iodide) → IO₃⁻ (Iodate) (Important Exception, unlike I₂ in acidic medium).

• S₂O₃²⁻ (Thiosulfate) → SO₄²⁻ (Sulfate)

• Mn²⁺ → MnO₂ (Mn⁺⁴)

• Note: In these reactions, MnO₄⁻ is reduced to MnO₂.

Summary of Key Redox Reactions

This table summarizes product formation when reacting with K₂Cr₂O₇ (in acid) or KMnO₄ (in acid/base).

Reactions of F-Block Elements (Lanthanides)

Lanthanides (Ln) have a common and most stable oxidation state of +3. Their reactions are generally predictable based on this state.

• With Oxygen: 2Ln + O₂ → Ln₂O₃ (forms lanthanide(III) oxide)

• With Sulfur: 2Ln + 3S → Ln₂S₃ (forms lanthanide(III) sulfide)

• With Nitrogen: 2Ln + N₂ → 2LnN (forms lanthanide(III) nitride)

• With Halogens (X₂): 2Ln + 3X₂ → 2LnX₃ (forms lanthanide(III) halide)

• With Acid (H⁺): 2Ln + 6H⁺ → 2Ln³⁺ + 3H₂ (liberates hydrogen gas)

• With Water: 2Ln + 6H₂O → 2Ln(OH)₃ + 3H₂ (forms lanthanide(III) hydroxide and H₂ gas)