Redox Reactions

\) or \) \)

The question involves determining the change in the oxidation state of manganese in the reaction of potassium permanganate (KMnO₄) with thiosulfate in a neutral or faintly alkaline medium. This can be solved by understanding the oxidation-reduction behavior of manganese in this chemical reaction.

In a neutral or faintly alkaline medium, potassium permanganate (KMnO₄) acts as a strong oxidizing agent. The oxidation state of manganese in KMnO₄ is +7. During the reaction with thiosulfate, manganese is reduced. We need to determine the oxidation state to which it is reduced.

The overall balanced reaction in a neutral or faintly alkaline medium is:

In the above reaction, manganese goes from an oxidation state of +7 in KMnO₄ to an oxidation state of +4 in MnO₂. Thus, the change in oxidation state of manganese is: .

The overall change in the oxidation state of manganese in this reaction is 3.

Hence, the correct answer is 3.

• The formula for chromyl chloride is CrO₂Cl₂.
• The oxidation state of chromium (x) is determined by the equation: x + 2(−2) + 2(−1) = 0.
• Simplifying: x − 4 − 2 = 0 ⇒ x = +6.
• Therefore, the oxidation state of chromium in CrO₂Cl₂ is +6.

Calculation of Oxidation States:

1. Fe(CO)₅: Carbon monoxide (CO) is a neutral ligand, so the oxidation state of Fe in Fe(CO)₅ is :
   Oxidation state of Fe = 0.
2. VO₂⁺: Oxygen has an oxidation state of −2. For VO₂⁺:
   
   The oxidation state of V is +4.
3. WO₃: Each oxygen contributes −2, so for tungsten (W):
   
   The oxidation state of W is +6.

Step 4: Sum of Oxidation States:

The total sum of oxidation states is 10.

Step 1: Reduction Reaction in Acidic Medium The reduction reaction is: In this reaction, 3 electrons are transferred during the reduction of to .

Calculation of Oxidation Number of Mo in

Given Compound:

, which can also be written as .

Step-by-Step Calculation:

1. Let the oxidation number of Mo be .
2. The oxidation numbers of the other components are:
   • : (3 ammonium ions contribute ).
   • : (phosphorus in the phosphate group).
   • : (oxygen).
3. Write the total oxidation number equation for the compound:
4. Simplify the equation:
5. Combine terms:
6. Simplify further:
7. Divide by 12:

Conclusion:

The oxidation number of molybdenum ( ) in is +6.

• Eu²⁺ (Europium in +2 Oxidation State):
  • Eu²⁺ has a tendency to lose an electron and get oxidised to Eu³⁺, which is a more stable oxidation state for europium due to the half-filled f-orbital configuration (4f⁷).
  • This tendency to donate electrons makes Eu²⁺ a reducing agent.
• Ce⁴⁺ (Cerium in +4 Oxidation State):
  • Ce⁴⁺ has a tendency to gain an electron and get reduced to Ce³⁺, which is a more stable oxidation state for cerium due to its 4f¹ configuration.
  • This ability to accept electrons makes Ce⁴⁺ a strong oxidising agent.

Conclusion:

Eu²⁺ acts as a reducing agent, while Ce⁴⁺ acts as an oxidising agent.

The correct answer is 1.

MnO^Θ₄ gains one electron to form MnO₄^-2 in a strongly alkaline medium.

6

Reduction Half Reaction:

Oxidation Half Reaction:

Adding the oxidation half and reduction half, we get the net reaction as:

Thus, .

To find the values of , , , and in the balanced equation of chlorine disproportionation in an alkaline medium, let's follow these steps:

1. Disproportionation involves the simultaneous oxidation and reduction of the same element. In this case, chlorine ( ) is both reduced to chloride ions ( ) and oxidized to hypochlorite ions ( ).
2. First, write the half-reactions for oxidation and reduction:

• Reduction:
• Oxidation:

1. Combine the half-reactions. Make sure the number of electrons lost is equal to the number of electrons gained to maintain electron balance:

• Overall equation after canceling electrons:

1. Balancing the atoms and charges, we have:

From the combined reaction:

• The coefficient of ( ) is 1.
• The coefficient of ( ) is 2.
• The coefficient of ( ) is 1.
• The coefficient of ( ) is 1.

1. Thus, the balanced redox reaction coefficients are , , , and .

Therefore, the correct set of coefficients is 1, 2, 1, and 1, which matches with the given option: 1, 2, 1, and 1.

To understand this question, we need to look at the redox reaction between permanganate ion and iodide ion in an alkaline medium.

The permanganate ion acts as a strong oxidizing agent. In an alkaline medium, it can oxidize iodide ions to iodate ions according to the balanced chemical equation:

This reaction shows that in an alkaline medium, is oxidized to (iodate ion) when reacted with permanganate ion. The presence of hydroxide ions from the alkaline medium helps stabilize the formation of manganese dioxide ( ).

Let's analyze the options:

1. - Periodate ion, not typically formed in alkaline conditions with this oxidation.
2. - Hypoiodite ion, not the product in this specific reaction.
3. - Molecular iodine, not a product of this reaction in alkaline medium.
4. - Iodate ion, correct product of this reaction.

Thus, the correct answer is that in an alkaline medium, oxidizes to .

The problem asks to balance the given half-reaction for the reduction of dichromate ion ( ) in an acidic medium and identify the unknown coefficients and the product species .

Concept Used:

The balancing of a redox half-reaction in an acidic medium follows a set of sequential rules:

1. Balance all atoms other than oxygen (O) and hydrogen (H).
2. Balance the oxygen atoms by adding water ( ) molecules to the side deficient in oxygen.
3. Balance the hydrogen atoms by adding hydrogen ions ( ) to the side deficient in hydrogen.
4. Balance the net charge on both sides by adding electrons ( ) to the side with the greater positive charge.

We also need to know the product of the reduction of dichromate. In an acidic medium, the dichromate ion ( ), where chromium is in the +6 oxidation state, is reduced to the chromium(III) ion ( ).

Step-by-Step Solution:

Step 1: Identify the species A and write the skeleton equation.

In an acidic medium, the dichromate ion ( ) is a strong oxidizing agent and gets reduced to the chromium(III) ion ( ). Therefore, the species is . The skeleton half-reaction is:

The chromium atoms are already balanced, with 2 on each side.

Step 2: Balance the oxygen atoms.

There are 7 oxygen atoms on the left-hand side (LHS) and none on the right-hand side (RHS). To balance the oxygen atoms, we add 7 water molecules ( ) to the RHS.

Comparing this with the given format , we can identify .

Step 3: Balance the hydrogen atoms.

Now, there are hydrogen atoms on the RHS and none on the LHS. To balance the hydrogen atoms in an acidic medium, we add 14 hydrogen ions ( ) to the LHS.

Comparing this with the given format , we can identify .

Step 4: Balance the charge.

Calculate the total charge on both sides of the equation.

• Total charge on LHS =
• Total charge on RHS =

The charge is not balanced. To balance the charge, we add electrons ( ) to the side with the higher (more positive) charge. We need to add electrons to the LHS.

Comparing this with the given format , we can identify .

Final Result:

The completely balanced half-reaction is:

By comparing this with the given equation , we find:

Therefore, , , , and are respectively: 14, 6, 7, and .

To solve this problem, we need to determine the number of moles of ions required for the reaction where oxidizes oxalate ions to carbon dioxide . The balanced redox reaction in an acidic medium is:

Let's break down the stoichiometric coefficients:

1. 1 mole of reacts with 5/2 = 2.5 moles of .
2. For every 2 moles of , 16 moles of are needed. Therefore, for 1 mole of , 16/2 = 8 moles of are required.

Hence, the number of moles of ions required by 1 mole of is 8.

This result is verified as falling within the provided range of 8 to 8.

The given reactions show the behavior of thiosulphate with iodine and bromine:

• With iodine:
• With bromine:

Let's analyze the reactions:

Thiosulphate with Iodine: In the first reaction, iodine is reduced to iodide , while thiosulphate is oxidized to tetrathionate . This indicates that iodine acts as an oxidizing agent.

Thiosulphate with Bromine: In the second reaction, bromine is reduced to bromide . Here, the thiosulphate is completely oxidized to sulfate . This shows that bromine acts as a stronger oxidizing agent compared to iodine.

Conclusion:

The difference in behavior is due to the relative strengths of iodine and bromine as oxidizing agents. Bromine is a stronger oxidant than iodine, which is why it can oxidize thiosulphate to a greater extent, all the way to sulfate.

Based on this analysis, the correct statement that justifies the dual behavior of thiosulphate is:

Bromine is a stronger oxidant than iodine.

Let's analyze the assertion and reason given in the question:

1. Assertion (A): In aqueous solutions, is reducing while is oxidising in nature.
2. Reason (R): Extra stability to half-filled electronic configuration is observed than incompletely filled electronic configuration.
3. Both and are transition metal ions. Their behavior in redox reactions is influenced by their electronic configurations.
4. has the electronic configuration of . It tends to lose one electron to form the more stable configuration, where all five 3d subshells are half-filled. Therefore, acts as a reducing agent.
5. has the electronic configuration of . By gaining an electron, it attains the stable half-filled configuration . Hence, acts as an oxidizing agent.
6. The reason provided is that the stability of the half-filled 3d subshell is significant in determining the redox behavior of these ions.

From the analysis above, we can conclude:

• Both the assertion and reason are true.
• The reason correctly explains why the assertion is true (i.e., the redox behavior of these ions is indeed due to the stability of their electronic configurations).

Therefore, the most appropriate answer is: Both (A) and (R) are true and (R) is the correct explanation of (A).

To solve this problem, we need to determine the number of moles of O₃ and O₂ involved when 1 mole of PbS is oxidized. The chemical reaction for oxidizing PbS using O₃ is:

PbS + 4O₃ → PbSO₄ + 4O₂

Step-by-step explanation:

From the balanced reaction, 1 mole of PbS requires 4 moles of O₃ to be oxidized to PbSO₄.

During this process, it produces 4 moles of O₂.

Given these stoichiometric relationships, the number of moles of O₃ (“X” moles) used is 4, and the number of moles of O₂ (“Y” moles) produced is also 4.

The problem asks for the sum X + Y, which is:

X + Y = 4 + 4 = 8

Verification:

The computed value of X + Y = 8 falls within the expected range of [8,8].

To evaluate the given statements and determine which is correct, let's analyze each statement based on known chemical reactions and principles.

1. Statement I: solid undergoes disproportionation reaction under alkaline conditions to form and .
   • Disproportionation is a specific type of redox (reduction-oxidation) reaction in which a single substance is simultaneously reduced and oxidized to form two different products.
   • In alkaline conditions, elemental sulfur can actually disproportionate. The reaction is:

• This confirms that under alkaline conditions, sulfur can indeed form (sulfide) and (thiosulfate).

1. Statement II: can undergo disproportionation reaction under acidic conditions.
   • Perchlorate ion is known for being very stable, especially under acidic conditions. It is already in its highest oxidation state, where chlorine is +7.
   • Generally, in a disproportionation reaction, a substance must be capable of existing in a higher as well as a lower oxidation state. However, chlorine in cannot oxidize further (higher than +7 is unfeasible under normal conditions).
   • Therefore, no disproportionation reaction of occurs under acidic conditions, confirming that Statement II is incorrect.

Based on the analysis above:

• Statement I is correct as sulfur in can undergo disproportionation to form and under alkaline conditions.
• Statement II is incorrect as does not undergo disproportionation under acidic conditions.

Therefore, the correct answer is: Statement I is correct but statement II is incorrect.

The heat of formation of is the heat change when 1 mole of is formed from its elements in their standard states.
By using the given reactions, the heat of formation is found to be .
Final Answer: .

Step 1: The reaction is a combination reaction (A-II).

Step 2: The reaction is a decomposition reaction (B-III).

Step 3: The reaction is a redox displacement reaction (C-I).

Step 4: The reaction is a displacement reaction (D-IV).

Final Conclusion: The correct matching is Option (3), A-II, B-III, C-I, D-IV.

Steam volatile compounds are typically those that are capable of being vaporized or evaporated easily at relatively low temperatures. Generally, compounds with hydrogen bonding or low molecular weight are steam volatile.
Among the options:
- (A) : This is a nitrophenol compound, which is steam volatile due to the phenolic group that can form hydrogen bonds and is low molecular weight.
- (B) : This is a nitroaniline compound, which can also be steam volatile due to the amine group that can form hydrogen bonds.
- (C) : This is an amphenol compound, and although it contains an amine and phenolic group, it has a higher molecular weight and is less likely to be steam volatile.
- (D) : This is phenol, which is steam volatile but not included in the right matching with other steam volatile candidates. Thus, the correct answer is (1).

The question asks about the relative reactivity of different carbonyl compounds in a nucleophilic addition reaction. Understanding the electronic effects and steric factors that influence the reactivity is essential in predicting the outcome.

The general order of reactivity of carbonyl compounds towards nucleophilic addition is influenced by two main factors:

• Electronic effects: Electron-withdrawing groups enhance the electrophilicity of the carbonyl carbon, thereby increasing its reactivity.
• Steric effects: Smaller groups on the carbonyl carbon make it more accessible to nucleophiles, enhancing reactivity.

Let's analyze each compound:

1. Acetophenone: Contains a phenyl group which is electron-donating via resonance, reducing the electrophilicity of the carbonyl carbon. Moreover, the bulkiness of the phenyl group can cause steric hindrance, further reducing reactivity.
2. p-Tolualdehyde: Has a methyl group which is slightly electron-donating, but since the carbonyl group is still more accessible than in acetophenone, it is more reactive.
3. Benzaldehyde: Lacks additional bulky substitutions and electron-donating groups compared to acetophenone, thus increasing its reactivity compared to both acetophenone and p-tolualdehyde.
4. p-Nitrobenzaldehyde: Contains a nitro group, which is a strong electron-withdrawing group. This significantly enhances the electrophilicity of the carbonyl carbon, making it the most reactive among the given compounds.

Therefore, considering these effects, the correct order of reactivity for nucleophilic addition reactions is:

This matches with the given correct option:

acetophenone p-tolualdehyde benzaldehyde p-nitrobenzaldehyde

To determine the product (A) formed in the given reaction sequence, let's analyze each step of the transformation:

1. Initial Reaction: The starting compound is propyne (CH₃-C≡CH).
2. Step 1: The reaction with , involves oxymercuration. In the presence of these reagents, propyne undergoes hydration, which typically leads to the formation of a ketone. In this case, it converts to acetone (CH₃-C(=O)-CH₃).
3. Step 2: The subsequent reaction with HCN introduces a cyanohydrin formation. The compound adds across the carbonyl group of acetone forming 2-hydroxy-2-methylpropanenitrile.
4. Step 3: The final step with H₂/Ni involves hydrogenation. The nitrile group (-CN) is reduced to an amine group (-CH₂NH₂).

So, the final product (A) formed is 2-amino-2-methylpropanol, depicted as:

This corresponds to the correct answer option.

Let's analyze each oxidation reaction:

A. :
Both and can oxidize iodide to iodine in acidic medium.
This reaction is possible with both oxidizing agents.

B. :
Both and can oxidize sulfide to sulfur in acidic medium.
This reaction is possible with both oxidizing agents.

C. :
Both and can oxidize to in acidic medium.
This reaction is possible with both oxidizing agents.

D. :
Both and can oxidize iodide to iodate in acidic medium, but it requires a stronger oxidizing power.
This reaction is possible under strong conditions, but not always consistent in standard acidic conditions.

E. :
Thiosulfate oxidation to sulfate is more complex and specific to certain oxidizing power and conditions. typically does not oxidize thiosulfate to sulfate quantitatively in acidic medium, while may do so under vigorous conditions.
This reaction is not consistently carried out by both oxidizing agents.

Therefore, reactions A, B, and C are the only ones that are consistently carried out by both and under standard acidic conditions.

Final Answer:
The final answer is .