Problem 50 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A white substance (A) reacts with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous \(\mathrm{NH}_{3}\) or \(\mathrm{NaOH}\) to \((\mathrm{C})\) produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A), (B), (C), (D) and (E). Write the equations of the reactions involved.

Step-by-Step Solution

Answer

\( \mathrm{A} = \mathrm{CaCO}_3 \), \( \mathrm{B} = \mathrm{CO}_2 \), \( \mathrm{C} = \mathrm{CaSO}_4 \), \( \mathrm{D} = \mathrm{C} \) (carbon), \( \mathrm{E} = \mathrm{CO}_2 \).

Step 1: Identify Substance A

The fact that (A) reacts with dilute \( \mathrm{H}_{2}\mathrm{SO}_{4} \) to release a colourless gas and form a solution indicates that (A) might be a carbonate. Let's consider \( \mathrm{CaCO}_3 \) (calcium carbonate) as a possible option.

Step 2: Identify Gas B and solution C

When a carbonate reacts with an acid, it produces carbon dioxide and the corresponding salt in solution. In this case, the colourless gas \((B)\) is likely \( \mathrm{CO}_2 \), and the solution \((C)\) is calcium sulfate \((\mathrm{CaSO}_4)\). \[ \mathrm{CaCO}_3 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{CO}_2 (\text{\(B\)}) + \mathrm{CaSO}_4 (\text{\(C\)}) + \mathrm{H}_2\mathrm{O} \]

Step 3: Determine the Precipitate D

The gas \((B)\), \( \mathrm{CO}_2 \), reacts with acidified \( \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 \) to form a green solution of \( \mathrm{Cr}^{3+} \) ions and a slightly coloured precipitate \((D)\). The likely candidate for \((D)\) is \( \mathrm{C} \), such as \( \mathrm{C}O \) (carbon). This produces a green solution of chromium(III) sulfate. \[ \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 3 \mathrm{SO}_2 + 2 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3 (\text{Green solution}) + K_2SO_4 + 4 \mathrm{H}_2O \]

Step 4: Identify Gas E

When \((D)\), carbon, burns in air, it forms carbon dioxide \((E)\), which we have already identified as \((B)\). Thus \((E)\) is again \( \mathrm{CO}_2 \). \[ \mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \]

Step 5: Identify the Colourless Liquid

The colourless liquid produced from the reaction of \((E)\) with \((B)\) that turns anhydrous copper sulfate blue, indicates that it is water \((\mathrm{H}_2\mathrm{O})\). Anhydrous copper sulfate is a test for the presence of water as it turns from white to blue when hydrated. \[ \mathrm{CuSO}_4(s) + 5 \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}(s) \]

Step 6: Determine Reactions with NH3 and NaOH

The addition of aqueous \( \mathrm{NH}_3 \) or \( \mathrm{NaOH} \) forming a precipitate, which dissolves again in excess, indicates \( \mathrm{C} \) is a transition metal like \( \mathrm{Cu}^{2+} \) or \( \mathrm{Zn}^{2+} \). \( \mathrm{CaSO}_4 \) forms \( \mathrm{Ca(OH)}_2 \) which reacts further to dissolve, however, the common identity as \( \mathrm{Zn}(\mathrm{OH})_2 \) fits other properties better such as amphoteric behavior. Finally, the reaction dissolves in excess due to the formation of complex ions. \[ \mathrm{Ca}\mathrm{SO}_4 + 2 \mathrm{NH}_3 + 2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2(s) + 2 \mathrm{NH}_4\mathrm{SO}_4 \]

Step 7: Summary and Conclusion

Through the reactions described, we can identify the substances: \( \mathrm{A} = \mathrm{CaCO}_3 \), \( \mathrm{B} = \mathrm{CO}_2 \), \( \mathrm{C} = \mathrm{CaSO}_4 \), \( \mathrm{D} = \mathrm{C} \) (carbon), and \( \mathrm{E} = \mathrm{CO}_2 \) (carbon dioxide again). These correspond with the series of reactions described in the prompt.

Key Concepts

carbonate_reaction

When a carbonate compound, such as calcium carbonate (\(\text{CaCO}_3\)), reacts with an acid like sulfuric acid (\(\text{H}_2\text{SO}_4\)), it undergoes a chemical reaction known as a carbonate reaction. This type of reaction is part of a larger category known as acid-base reactions. During this reaction, the carbonate ion \(\text{(CO}_3^{2-})\) interacts with the hydrogen ions \((\text{H}^+\)) from the acid. This produces carbon dioxide gas (\(\text{CO}_2\)), water (\(\text{H}_2\text{O}\)), and a dissolved salt. In our example, the salt formed is calcium sulfate \((\text{CaSO}_4)\).

This is a common laboratory reaction and is often used to demonstrate the release of gas and formation of precipitates. The balanced equation for this specific reaction is:
\[\mathrm{CaCO}_3 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{CO}_2 + \mathrm{CaSO}_4 + \mathrm{H}_2\mathrm{O}\]
Each component of this reaction plays a specific role; carbonates act as the base, acids provide hydrogen ions, and carbon dioxide provides visual confirmation of the reaction's occurrence.

copper_sulfate_test

Anhydrous copper sulfate is a special, white powdery substance that serves as an excellent test for the presence of water. When water molecules bind to anhydrous copper sulfate, it undergoes a transformation by turning blue. This occurs because water molecules hydrate the copper ions within the copper sulfate crystal structure. This test is widely used to indicate moisture or confirm the presence of water in various chemical processes.

Additionally, this is why the compound is often used in laboratories where mechanism detection is critical. If you add a colorless liquid to anhydrous copper sulfate and the substance turns blue, it confirms the presence of water. The transformation is due to the formation of copper sulfate pentahydrate \((\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O})\), which is blue, in contrast to the anhydrous form which is white. The reaction is represented as:
\[\mathrm{CuSO}_4(s) + 5 \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}(s)\]
Through this reaction, the presence of water is visually detectable, due to the stark color change.

precipitation_reactions

Precipitation reactions involve the formation of an insoluble solid from a solution of two soluble salts. This type of reaction occurs when ions in aqueous solutions combine to form an insoluble compound or precipitate. Precipitation reactions are central in many chemical analyses and synthetic procedures. When a solution containing calcium sulfate \((\mathrm{CaSO}_4)\) is treated with a reagent such as ammonia (\(\mathrm{NH}_3\)) or sodium hydroxide \((\mathrm{NaOH})\), a precipitation reaction occurs forming calcium hydroxide \((\mathrm{Ca(OH)}_2\)).

This initial precipitate may then dissolve upon the addition of excess reagent, demonstrating the solution's ability to form a new complex ion in situ. This process is an example of converting a metal ion into a soluble complex by reacting it with an excess of a complexing agent. The equations for these reactions illustrate this dynamic:
\[\mathrm{CaSO}_4 + 2 \mathrm{NH}_3 + 2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ca(OH)}_2(s) + 2 \mathrm{NH}_4\mathrm{SO}_4\]
Sometimes, it's crucial in experiments to finally dissolve the precipitate in an excess reagent, which leads to further insights into the compound's chemical behavior.