Problem 49 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

When a white crystalline compound \(\mathbf{X}\) is heated with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) and concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\), a reddish brown gas \(\mathbf{A}\) is evolved. On passing A into caustic soda solution, a yellow coloured solution of \(\mathbf{B}\) is obtained. Neutralizing the solution \(\mathbf{B}\) with acetic acid and on subsequent addition of lead acetate, a yellow precipitate \(\mathbf{C}\) is obtained. When \(\mathbf{X}\) is heated with NaOH solution, a colourless gas is evolved and on passing this gas into \(\mathrm{K}_{2} \mathrm{HgI}_{4}\) solution, a reddish brown precipitate \(\mathbf{D}\) is formed. Identify \(\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}\) and \(\mathbf{X}\). Write the equations of reactions involved.

Step-by-Step Solution

Answer

A is Cl₂, B is NaOCl, C is Pb(OCl)₂, D is (NH₂Hg)HgI₂, X is NH₄Cl.

Step 1: Identify Red Gas (A)

When compound X is heated with \( \text{K}_2\text{Cr}_2\text{O}_7 \) and concentrated \( \text{H}_2\text{SO}_4 \), a reddish-brown gas \( A \) is evolved. The combination of potassium dichromate and sulfuric acid is a common method to produce chlorine gas \( \text{Cl}_2 \), which is reddish-brown. Thus, \( A \) is \( \text{Cl}_2 \).

Step 2: Reaction of Gas A with Caustic Soda

When \( \text{Cl}_2 \) (gas \( A \)) is passed into caustic soda (NaOH) solution, a yellow solution of \( B \) is obtained. This happens due to the formation of a mixture of sodium chloride (\( \text{NaCl} \)) and sodium hypochlorite (\( \text{NaOCl} \)) in water, which gives a pale yellow color. Thus, \( B \) can be considered as \( \text{NaOCl} \).

Step 3: Neutralizing Solution B

The yellow solution \( B \) is neutralized with acetic acid which will not largely affect \( \text{NaOCl} \). Adding lead acetate to the neutralized solution precipitates \( \text{Pb(OCl)}_2 \), which is yellow. Hence, \( C \) is \( \text{Pb(OCl)}_2 \).

Step 4: Heating X with NaOH

When \( X \) is heated with \( \text{NaOH} \), a colorless gas is released. This gas must be \( \text{NH}_3 \) because it reacts with \( \text{K}_2\text{HgI}_4 \) to form a reddish-brown precipitate. Hence, \( X \) likely contains \( \text{NH}_4^+ \) ions, meaning \( X \) can be \( \text{NH}_4\text{Cl} \).

Step 5: Reaction of Gas from Step 4 with Solution

The colorless gas \( \text{NH}_3 \), when passed into \( \text{K}_2\text{HgI}_4 \) (Nessler's reagent), results in a reddish-brown precipitate \( D \). This precipitate is famously known as \( \text{HgI}_2 \) or Millon's base from Nessler's reagent test for \( \text{NH}_4^+ \). Thus, \( D \) is \( \text{(NH}_2\text{Hg} )\) in reddish-brown color.

Step 6: Conclusion

Thus, \( A \) is \( \text{Cl}_2 \), \( B \) is \( \text{NaOCl} \), \( C \) is \( \text{Pb(OCl)}_2 \), \( D \) is \( \text{(NH}_2\text{Hg} ) \), and \( X \) is \( \text{NH}_4\text{Cl} \). The reactions are:1. \( \text{NH}_4\text{Cl} + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{Cl}_2 \) (A)2. \( \text{Cl}_2 + \text{NaOH} \rightarrow \text{NaCl} + \text{NaOCl} \) (B)3. \( \text{NaOCl} + \text{Pb(CH}_3\text{COO)}_2 \rightarrow \text{Pb(OCl)}_2 \) (C)4. \( \text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NH}_3 \)5. \( \text{NH}_3 + \text{K}_2\text{HgI}_4 \rightarrow \text{(NH}_2\text{Hg} ) \text{HgI}_2 \) (D)The compound \( X \) is \( \text{NH}_4\text{Cl} \).

Key Concepts

Reaction Mechanism

Understanding a reaction mechanism is crucial when identifying chemical processes, as it involves analyzing each step of a chemical transformation. In our given exercise, the reaction mechanism begins by heating a white crystalline compound, X, with \(\text{K}_2\text{Cr}_2\text{O}_7\) and concentrated \(\text{H}_2\text{SO}_4\). This combination results in the release of chlorine gas \(\text{Cl}_2\), which is a reddish-brown gas.ewlineChemical reactions often proceed through a series of subtle molecular changes involving bond-breaking and bond-making events. The mechanism involving chlorine gas production illustrates a typical oxidation-reduction process, where \(\text{K}_2\text{Cr}_2\text{O}_7\) acts as an oxidizing agent, causing the release of chlorine from \(\text{NH}_4\text{Cl}\).ewlineFurther, when this chlorine gas is introduced to a caustic soda solution, a new product is formed through a substitution reaction. Understanding this process helps students visualize how initial reactants convert to specific products through defined pathways. This process is meticulously observed by scientists for predicting and controlling chemical reactions.

Chlorine Gas Production

Chlorine gas production, as highlighted in this exercise, is an important concept in inorganic chemistry. Chlorine gas \(\text{Cl}_2\) is generated from white crystalline compound \(X\) when heated with a potent oxidizing mixture of \(\text{K}_2\text{Cr}_2\text{O}_7\) and concentrated sulfuric acid \(\text{H}_2\text{SO}_4\). This reaction is not only a demonstration of a common laboratory practice but also highlights the dynamic nature of chemical transformations.ewlineThis specific reaction is a popular method to produce chlorine gas, and its recognition supports understanding of oxidation reactions where chlorine is liberated from compounds like \(\text{NH}_4\text{Cl}\). Chlorine is initially present as chloride ions in the compound, and upon heating, these ions are oxidized to form diatomic chlorine gas.ewlineThe reddish-brown appearance of chlorine gas is a useful identifying feature. Beyond its visual presence, chlorine gas plays an essential role in various chemical production processes, such as the synthesis of bleach when reacted with \(\text{NaOH}\). Understanding these reactions provides foundational knowledge for innovations in industrial and environmental chemistry.

Lead Precipitate Reactions

Lead precipitate reactions serve as an excellent example of precipitation and neutralization reactions. When chlorine-derived yellow solution \(B\) (which is primarily \(\text{NaOCl}\)) is neutralized with acetic acid and treated with lead acetate, a curious and significant chemical change occurs, resulting in the formation of a yellow precipitate \(C\), which is \(\text{Pb(OCl)}_2\).ewlineThis yellow precipitate formation is a result of the reaction between sodium hypochlorite and lead ions. Precipitation reactions are a subcategory of double replacement reactions where two ionic compounds in solution react to form an insoluble product. As the lead ions interact with hypochlorite ions, an insoluble lead compound forms, which precipitates out of the solution as a solid.ewlineSuch reactions provide insight into ionic interactions, where the solubility rules come into play, allowing chemists to predict product formation. They also emphasize the practicalities of manipulating chemical reactions to isolate specific compounds, which is a vital skill in experimental chemistry.