Problem 62 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The gas liberated on heating a mixture of two salts with \(\mathrm{NaOH}\), gives a reddish brown precipitate with an alkaline solution of \(\mathrm{K}_{2}\left[\mathrm{HgI}_{4}\right]\). The aqueous solution of the mixture on treatment with \(\mathrm{BaCl}_{2}\) gives a white precipitate which is sparingly soluble in conc. HCl. On heating the mixture with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) and conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\), red vapours of \(A\) are produced. The aqueous solution of the mixture gives a deep blue colouration of \(B\) with potassium ferricyanide solution. Identify the radicals in the given mixture and write the balanced equations for the formation of \(A\) and \(B\).

Step-by-Step Solution

Answer

The radicals are \( \text{NH}_4^+ \), \( \text{SO}_4^{2-} \), \( \text{Cl}^- \), and \( \text{Fe}^{2+} \). Equations: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Cl}^- \rightarrow 2\text{Cr}^{3+} + 3\text{Cl}_2(g) + 7\text{H}_2\text{O} \) and \( 3\text{Fe}^{2+} + 2\text{[Fe(CN)_6]^{3-}} \rightarrow \text{Fe}_3[\text{Fe(CN)}_6]_2 \).

Step 1: Analyze the reaction with K2[HgI4]

The question states that the gas liberated when the salts are heated with \( \text{NaOH} \) gives a reddish-brown precipitate with an alkaline solution of \( \text{K}_{2}[\text{HgI}_{4}] \). This suggests the presence of ammonia gas (\( \text{NH}_3 \)), indicating a nitrate (\( \text{NO}_3^- \)) or nitrite (\( \text{NO}_2^- \)) ion in the mixture. \( \text{NH}_3 \) reacts with \( \text{K}_{2}[\text{HgI}_{4}] \) to form \( \text{HgO} \, \cdot \text{NH}_2\text{I} \), which is a reddish-brown precipitate.

Step 2: Analyze the reaction with BaCl2

The formation of a white precipitate that is sparingly soluble in conc. \( \text{HCl} \) upon treatment with \( \text{BaCl}_{2} \) suggests the presence of sulfate ions (\( \text{SO}_4^{2-} \)). Barium sulfate (\( \text{BaSO}_4 \)) is a white precipitate that is known for being insoluble in dilute acids and sparingly soluble or insoluble in concentrated hydrochloric acid.

Step 3: Analyze the reaction with K2Cr2O7 and H2SO4

The production of red vapors when the mixture is heated with \( \text{K}_{2}\text{Cr}_{2}\text{O}_{7} \) and concentrated \( \text{H}_2\text{SO}_4 \) indicates the presence of a chloride ion (\( \text{Cl}^- \)). \( \text{Cl}^- \) can be oxidized by \( \text{K}_{2}\text{Cr}_{2}\text{O}_{7} \) in acidic conditions to form \( \text{Cl}_2 \) gas, which appears as red fumes.

Step 4: Analyze the reaction with Potassium Ferricyanide

The deep blue color obtained with potassium ferricyanide suggests the presence of ferrous ions (\( \text{Fe}^{2+} \)). Ferrous ions react with potassium ferricyanide to form a deep blue precipitate known as Turnbull's Blue, which confirms the presence of iron as a radical in the mixture.

Step 5: Write balanced equations for the formation of A and B

**For the formation of \( A \):**The red vapors (\( A \)) are from the reaction where chloride ions form chlorine gas:\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Cl}^- \rightarrow 2\text{Cr}^{3+} + 3\text{Cl}_2(g) + 7\text{H}_2\text{O} \]**For the formation of \( B \):**The deep blue color, \( B \), is from the formation of Turnbull's Blue:\[ 3\text{Fe}^{2+} + 2\text{[Fe(CN)_6]^{3-}} \rightarrow \text{Fe}_3[\text{Fe(CN)}_6]_2 \]

Key Concepts

Radical identification

Identifying radicals in a mixture involves recognizing specific ions responsible for particular reactions. In the given exercise, the liberation of gas upon heating and reactions with specific reagents serves to identify these radicals.

• The gas, when reacting with an alkaline solution of \( \text{K}_2[\text{HgI}_4] \), yielding a reddish-brown precipitate indicates ammonia (\( \text{NH}_3 \)). This points towards the potential presence of nitrate (\( \text{NO}_3^- \)) or nitrite (\( \text{NO}_2^- \)) ions.
• Reaction with \( \text{BaCl}_2 \), producing a white precipitate that dissolves sparingly in concentrated \( \text{HCl} \), suggests sulfate ions (\( \text{SO}_4^{2-} \)).
• The formation of red vapors on heating with \( \text{K}_2\text{Cr}_2\text{O}_7 \) and \( \text{H}_2\text{SO}_4 \) points towards chloride ions (\( \text{Cl}^- \)).
• The deep blue color with potassium ferricyanide indicates ferrous ions (\( \text{Fe}^{2+} \)). This reaction forms Turnbull's Blue.

These reactions help identify the radicals within a complex mixture, allowing chemists to determine the specific ions present.

Reddish-brown precipitate formation

When you think of a reddish-brown precipitate, it often refers to the presence of mercury compounds. Here, ammonia gas released upon heating with \( \text{NaOH} \) reacts with an alkaline solution of \( \text{K}_2[\text{HgI}_4] \). This reaction forms the compounds \( \text{HgO} \cdot \text{NH}_2\text{I} \), responsible for this distinct reddish-brown color. The reaction can be summarized:- Ammonia, in the presence of \( \text{K}_2[\text{HgI}_4] \), forms an aggregated complex of mercury ammonia iodide.- This specific reaction often signifies ammonium or certain nitrogen-containing radicals present in the mixture.- The color and formation of this precipitate is a classical qualitative test in inorganic chemistry for identifying ammonium ions.

Barium sulfate solubility

Barium sulfate (\( \text{BaSO}_4 \)) is a white precipitate commonly known for its low solubility. This property makes it ideal for detecting sulfate ions in a solution. When an aqueous solution containing the mixture is treated with \( \text{BaCl}_2 \), a white precipitate forms:- \( \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \)- This precipitate is sparingly soluble in concentrated acids like \( \text{HCl} \), maintaining its integrity even under such conditions.Knowing the properties of barium sulfate aids in recognizing sulfate ions. Detecting such characteristic precipitation helps in identifying sulfate presence in diverse chemical settings. Additionally, the unique insolubility of \( \text{BaSO}_4 \) in concentrated \( \text{HCl} \) confirms the existence of \( \text{SO}_4^{2-} \) ions.

Red vapors formation

Red vapors in chemistry experiments frequently indicate chlorine gas evolution. This is particularly evident when chlorides are oxidized. In the given problem:- The mixture is heated with \( \text{K}_2\text{Cr}_2\text{O}_7 \) and concentrated \( \text{H}_2\text{SO}_4 \).- Chloride ions (\( \text{Cl}^- \)) oxidize to form \( \text{Cl}_2 \) gas, which dissipates as red vapors.The reaction is as follows:\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Cl}^- \rightarrow 2\text{Cr}^{3+} + 3\text{Cl}_2(g) + 7\text{H}_2\text{O} \]This process is critical for confirming chloride presence, as the red vapors offer a visual cue to the chloride ions being oxidized. The occurrence of such vapors further assists chemists in distinguishing chloride ions within mixed ionic compounds.

Deep blue coloration

The deep blue coloration encountered in the reaction between an aqueous solution of the mixture and potassium ferricyanide is a hallmark of ferrous ion presence. This unique interaction results in a complex known as Turnbull's Blue. - In this process, ferrous ions (\( \text{Fe}^{2+} \)) combine with ferricyanide ions creating a deep blue precipitate.- The chemical equation for this reaction is: \[ 3\text{Fe}^{2+} + 2[\text{Fe(CN)}_6]^{3-} \rightarrow \text{Fe}_3[\text{Fe(CN)}_6]_2 \]This striking blue color is not only visually identifiable but serves as a practical test for identifying iron(II) ions in a solution. Recognizing such complementary reactions and their consequent color changes is essential in experimental chemistry, helping to neatly pinpoint the presence of specific ions.