Problem 116 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the products \(\mathrm{A}\) and \(\mathrm{B}\) here. (1) \(2 \mathrm{Ag}^{+}\)(excess) \(+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) \(\stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{A}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (2) \(2 \mathrm{NO}_{2}^{-}+2 \mathrm{I}^{-} \stackrel{\text { acid medium }}{\longrightarrow} \mathrm{B}+\mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (a) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=\mathrm{N}_{2}\) (b) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=\mathrm{N}_{2} \mathrm{O}\) (c) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=2 \mathrm{NO}_{2}\) (d) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=2 \mathrm{NO}\)

Step-by-Step Solution

Answer

There seems to be a discrepancy; \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{~S}\) and \(\mathrm{B} = \mathrm{N}_{2}\), which matches no given option exactly.

Step 1: Interpret Reaction (1)

The first reaction involves thiosulfate (\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)) reacting with an excess of silver ions (\(\mathrm{Ag}^+\)) to form \(\mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), which in water produces \(\mathrm{A}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Here, \(\mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is a complex that can decompose in water to give silver sulfide (\(\mathrm{Ag}_{2} \mathrm{~S}\)) and sulfuric acid. Therefore, \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{~S}\).

Step 2: Analyze Reaction (2)

In the second reaction, nitrite ions (\(\mathrm{NO}_{2}^{-}\)) and iodide ions (\(\mathrm{I}^-\)) react in an acid medium to produce \(\mathrm{B}\), iodine (\(\mathrm{I}_{2}\)), and water. Typically, nitrite in an acidic medium undergoes reduction to form nitrogen-containing gases. Here, under acidic conditions, \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{I}^{-}\) will produce nitrogen gas (\(\mathrm{N}_{2}\)), iodine, and water. Thus, \(\mathrm{B} = \mathrm{N}_{2}\).

Step 3: Match to Given Options

Based on the analysis, \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{~S}\) and \(\mathrm{B} = \mathrm{N}_{2}\). Comparing with the provided choices, this matches option (b): \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{~S}\), \(\mathrm{B} = \mathrm{N}_{2} \mathrm{O}\). However, \(\mathrm{B}\) should be \(\mathrm{N}_{2}\), which is not accurately listed.

Key Concepts

Silver Thiosulfate Reaction

The silver thiosulfate reaction is a fascinating example of how silver ions chemically interact with thiosulfate ions. When you mix silver ions \(\mathrm{Ag}^{+}\) in excess with thiosulfate ions \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), they initially form a compound known as silver thiosulfate \(\mathrm{Ag}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\). This is not the final product in the given environment, so it undergoes further transformation.
In the presence of water, this compound decomposes. The breakdown results in forming silver sulfide \(\mathrm{Ag}_{2} \mathrm{S}\) and sulfuric acid \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Silver sulfide is a black precipitate, typically insoluble in water, marking a clear endpoint to this reaction.Key points include:

• Excess silver ions reaction with thiosulfate
• Formation and decomposition of silver thiosulfate
• Final products: silver sulfide and sulfuric acid

Nitrite and Iodide Reaction

The interaction between nitrite ions \(\mathrm{NO}_{2}^{-}\) and iodide ions \(\mathrm{I}^{-}\) in an acid medium underscores essential principles of inorganic chemistry. Under acidic conditions, these two ions come together to enact a series of transformations resulting in the formation of iodine \(\mathrm{I}_{2}\), water, and another product \(\mathrm{B}\).
This interaction holds particular curiosity: nitrite ions are known to undergo reduction reactions to form nitrogen-containing compounds. In acidic solutions, this often formalizes as nitrogen gas \(\mathrm{N}_{2}\).
When this reaction occurs, you'll notice iodine as a distinctive purple or black solid, depending on concentration, and the aqueous reaction leads to the generation of water as a by-product.Highlights:

• Acidic medium as a key condition
• Transformation to form iodine, water, and nitrogen gas
• Nitrite's characteristic conversion under reduction conditions

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, are a fundamental element of chemistry that involve the transfer of electrons between two species. These reactions are characterized by the change in oxidation states of the reactants, indicating which species are oxidized and reduced.
In the reactions we've examined, oxidation-reduction principles govern the transformations. In reaction \(\#2\), nitrite undergoes a reduction process, leading to the formation of nitrogen gas. Conversely, iodide ions are oxidized to produce iodine. This dual process of electron exchange is key in balancing the overall chemical equation.
Understanding redox reactions involves recognizing:

• Oxidizing agent: the species that gets reduced
• Reducing agent: the species that gets oxidized
• Electron transfer signifying redox dynamics

Products of Chemical Reactions

Chemical reactions result in the formation of products, which are substances created by the rearrangement of atoms and bonds from the reactants.
In the studied reactions, the products tell us much about the conditions of the chemical processes. The first reaction leads to the formation of silver sulfide and sulfuric acid, straightforward products of the breakdown of \(\mathrm{Ag}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\).
The second reaction is slightly more complex, delivering iodine, water, and nitrogen gas from initially different reactants. Identifying the correct products is crucial for understanding the reaction mechanisms at play.Key Takeaways:

• Recognition of product formation in chemical reactions
• Importance of reaction conditions in determining products
• Connections between reactant transformations and final products