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

Question

Identify the products A and 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

(b) \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{S}, \mathrm{B} = \mathrm{N}_{2} \mathrm{O}\).

Step 1: Reaction Analysis of First Equation

Examine the given reactions. In the first reaction: \(2 \mathrm{Ag}^{+}\) ions react with \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to form \(\mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\). This compound then reacts with \(\mathrm{H}_{2} \mathrm{O}\) to release \(\mathrm{A}\) along with \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Since thiosulfate (\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)) decomposes in acidic conditions, it's logical to get silver sulfide (\(\mathrm{Ag}_{2} \mathrm{S}\)) because sulfide ions are often formed from the decomposition of thiosulfate. Hence, \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{S}\).

Step 2: Reaction Analysis of Second Equation

For the second reaction, \(2 \mathrm{NO}_{2}^{-}\) reacts with \(2 \mathrm{I}^{-}\) in an acidic medium. This can yield nitrogen oxides and iodine. In an acidic medium, the reduction of \(\mathrm{NO}_{2}^{-}\) commonly results in dinitrogen monoxide (\(\mathrm{N}_{2}\, \mathrm{O}\)), not a decrease beyond the \(\mathrm{NO}\) level due to the structure and available options. Thus, \(\mathrm{B} = \mathrm{N}_{2}\mathrm{O}\).

Step 3: Match the Products with Options

Compare the identified products \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{S}\) and \(\mathrm{B} = \mathrm{N}_{2}\, \mathrm{O}\) with options provided. Option (b) matches this pattern: \(\mathrm{A} = \mathrm{Ag}_{2} \mathrm{S}, \mathrm{~B} = \mathrm{N}_{2} \mathrm{O}\).

Key Concepts

Silver Chemistry

Silver, represented as Ag on the periodic table, is a transition metal known for its high electrical conductivity and luster. In chemical reactions, silver often forms compounds by donating its 1+ or 2+ oxidation states, typically resulting in cationic silver ions like Ag\(^+\). These ions readily form compounds with anions such as sulfides, chlorides, and more.

In our exercise, silver ions (Ag\(^+\)) react with thiosulfate ions (S\(_2\)O\(_3\)\(^{2-}\)) to create silver thiosulfate (Ag\(_2\)S\(_2\)O\(_3\)). This compound can further decompose in water under acidic conditions to form silver sulfide (Ag\(_2\)S). Silver sulfide is a black compound, frequently encountered as tarnish on silver surfaces.

Understanding silver chemistry involves recognizing the tendencies of silver ions to form various inorganic compounds by acting as a reducing agent or forming precipitates. In analytical chemistry, this behavior is used to detect and quantify silver and its compounds through precipitation reactions.

Thiosulfate Decomposition

Thiosulfate, with its formula S\(_2\)O\(_3\)\(^{2-}\), is a fascinating ion used for its reducing properties and ability to form complexes. It is commonly seen in the context of photography and water treatment processes.

When thiosulfate decomposes, especially in acidic environments, it is prone to breaking down into sulfur and sulfide ions. This is because the thiosulfate ion is an integral sulfur-oxygen compound that is unstable under acidic conditions.

In the given reaction, thiosulfate reacts with silver ions to form silver thiosulfate, which further decomposes to produce silver sulfide (Ag\(_2\)S) along with a by-product like sulfuric acid (H\(_2\)SO\(_4\)). This kind of decomposition highlights the complex interplay between thiosulfate and metal ions in solution chemistry. Thus, recognizing the conditions that promote thiosulfate decomposition is key in understanding its role in various chemical and industrial processes.

Nitrogen Oxides Formation

Nitrogen oxides are a group of gases composed of nitrogen and oxygen in different ratios. They play essential roles in the nitrogen cycle and are important in environmental chemistry.

In the exercise, the nitrite ions (NO\(_2\)\(^{-}\)) react in an acidic medium with iodide ions (I\(^{-}\)), leading to the formation of nitrogen oxides and iodine (I\(_2\)). Under these conditions, nitrite often converts to dinitrogen monoxide (N\(_2\)O), a gas commonly known as nitrous oxide or "laughing gas."

This process illustrates the importance of conditions such as pH in driving specific chemical reactions. Nitrogen oxides form due to the reduction of nitrite ions, showcasing the delicate balance of oxidation and reduction processes in chemistry. These reactions have significant implications for the atmospheric chemistry and are often considered when discussing air pollution control and industrial emissions.