Classical Idea Of Redox Reactions Oxidation And Reduction Reactions

The correct answer is (B) : Cu

( under hot conditions )

Coordination Compound Oxidation States Analysis

Let's analyze both Assertion A and Reason R:

Assertion A: In and , the carbonyl ligand ( ) is neutral. Thus, the oxidation state of Ni and Fe in these complexes is indeed zero. So, Assertion A is correct.

Reason R: Low oxidation states of metals are stabilized by synergistic bonding, also known as backbonding. This involves σ-donation from the ligand to the metal and π-backdonation from the metal to the ligand. Crucially, the ligand must have empty π orbitals to accept electrons from the metal. is a π-acceptor ligand, not a π-donor ligand. So, Reason R is incorrect.

Conclusion:

The correct answer is (3): A is correct, but R is not correct.

The question asks about the element that does not show a variable oxidation state among the given options: Chlorine, Fluorine, Bromine, and Iodine.

To solve this, we must understand the oxidation states of halogens. Halogens can exhibit different oxidation states, but Fluorine is an exception. Let's explore why:

1. Fluorine: Fluorine is the most electronegative element in the periodic table. It has an electronegativity of 3.98 on the Pauling scale. Due to its high electronegativity, fluorine can only exhibit an oxidation state of -1. It cannot show positive oxidation states as it is unable to lose electrons. Therefore, fluorine always has a constant oxidation state of -1.
2. Chlorine: Chlorine can have multiple oxidation states, ranging from -1 to +7, because it can gain, lose, or share electrons under different conditions. It appears in negative oxidation states like -1 and positive states like +1, +3, +5, and +7.
3. Bromine: Similar to chlorine, bromine can also exhibit multiple oxidation states such as -1, +1, +3, +5, and +7.
4. Iodine: Iodine also shows variability in its oxidation states like -1, +1, +3, +5, and +7, similar to other halogens except fluorine.

In conclusion, among the given options, Fluorine is the element that does not show variable oxidation states because it always has an oxidation state of -1. Hence, the correct answer is Fluorine.

To determine which compound cannot act as an oxidising agent, we must understand what an oxidising agent is. An oxidising agent, or oxidant, is a substance that has the ability to oxidize other substances—meaning, it accepts electrons from them. Thus, a good oxidising agent has a tendency to gain electrons and, in the process, is reduced.

Let's analyze each option:

• MnO₄^- (Permanganate Ion): The permanganate ion is a powerful oxidising agent, especially in acidic media. It can be reduced from Mn⁷⁺ to Mn²⁺ (or even MnO₂), which involves a significant gain of electrons. Therefore, MnO₄^- definitely acts as an oxidising agent.
• SO₂^- (Sulfur Dioxide Ion): This is a bit unclear as SO₂ is typically a gas and not commonly known as an ion in this form. However, SO₂ can be both a reducing agent and an oxidising agent under different conditions, but its ionized form, SO₂^2-, acts more as a reducing agent. But for the analysis given in the question, we must consider SO₂ in an oxidising capability context.
• N₃^- (Azide Ion): The azide ion is typically not an oxidising agent. It is more often involved in nucleophilic reaction processes or acts in explosive compounds due to its instability. Thus, it does not generally function as an oxidising agent.
• BrO₃^- (Bromate Ion): Like MnO₄^-, BrO₃^- is a known strong oxidising agent. It can be reduced from Br⁵⁺ to Br^-, accepting electrons in the process.

From this analysis, it's evident that the azide ion (N₃^-) does not typically act as an oxidising agent.

Conclusion: The substance that cannot act as an oxidising agent is N₃^-.

Step 1: The reaction involves the oxidation of toluene with and . is a strong oxidizing agent, typically oxidizing the methyl group ( ) in toluene to a carboxylic acid ( ) group. This results in a formyl group ( ) being left behind in residue (A), as toluene is partially oxidized.

Step 2: The treatment with water and NaHSO ensures further oxidation of the formyl group ( ) into a carboxyl group ( ). Diluting with HCl gives a sodium salt of the carboxyl group ( ) in compound (B).

Thus, the structures of residue (A) and compound (B) are as follows: Residue (A) has a formyl group ( ) and compound (B) has a carboxylate group ( ).