Problem 105 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A white amorphous powder (A) on heating yields a colourless, noncombustible gas (B) and a solid (C). The latter compound assumes a yellow colour on heating and changes to white on cooling. ' \(C\) ' dissolves in dilute acid and the resulting solution gives a white precipitate on adding \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) solution. A' dissolves in dilute HCl with the evolution of gas, which is identical in all respects with ' \(\mathrm{B}\) '. The gas 'B' turns lime water milky, but the milkiness disappears with the continuous passage of gas. The solution of ' \(A\) ', as obtained above, gives a white precipitate (D) on the addition of excess of \(\mathrm{NH}_{4} \mathrm{OH}\) and passing \(\mathrm{H}_{2} \mathrm{~S}\). Another portion of the solution gives initially a white precipitate (E) on the addition of sodium hydroxide solution, which dissolves on futher addition of the base. Identify the compounds \(\mathrm{A}, \mathrm{B}, \mathrm{D}\), and \(\mathrm{E}\).

Step-by-Step Solution

Answer

A is ZnCO3, B is CO2, D is ZnS, E is Zn(OH)2.

Step 1: Identifying gas B

On heating compound \( A \), it releases a colorless, noncombustible gas \( B \). Since gas \( B \) turns lime water milky but the milkiness disappears with continued gas passage, \( B \) is likely \( \text{CO}_2 \). It is initially forming calcium carbonate \( \text{CaCO}_3 \), turning the lime water milky, and then forms calcium bicarbonate \( \text{Ca(HCO}_3)_2 \), which is soluble, clearing the milkiness.

Step 2: Identifying solid C

Solid \( C \) turns yellow when heated and white when cooled. This is characteristic of zinc oxide \( \text{ZnO} \), which is white at room temperature and yellow when heated. \( C \) dissolves in dilute acid to form a solution that gives a white precipitate with \( \text{K}_4[\text{Fe(CN)}_6] \), consistent with zinc ions.

Step 3: Inferring compound A

Since \( B \) is \( \text{CO}_2 \), compound \( A \) must contain \( \text{CO}_3^{2-} \). Given the decomposition of \( A \) to form \( \text{ZnO} \) and \( \text{CO}_2 \), \( A \) is likely zinc carbonate \( \text{ZnCO}_3 \). \( A \) dissolves in dilute \( \text{HCl} \) with the evolution of \( \text{CO}_2 \).

Step 4: Identifying precipitate D

When the solution of \( A \) is treated with \( \text{NH}_4\text{OH} \) and \( \text{H}_2\text{S} \), a white precipitate \( D \) forms. This indicates the formation of zinc sulfide \( \text{ZnS} \), which is known to form in such reactions.

Step 5: Identifying precipitate E

Upon adding \( \text{NaOH} \) to a solution derived from \( A \), a white precipitate \( E \) initially forms and then dissolves with excess base. This behavior is typical of zinc hydroxide \( \text{Zn(OH)}_2 \), which is soluble in excess \( \text{NaOH} \) forming \( \text{Na}_2\text{[Zn(OH)}_4] \).

Key Concepts

Zinc Compounds

Zinc compounds are fascinating and versatile in the field of inorganic chemistry. A common zinc compound you'll encounter is zinc carbonate \( \text{ZnCO}_3 \). When heated, it decomposes into zinc oxide \( \text{ZnO} \) and carbon dioxide \( \text{CO}_2 \). Zinc oxide itself is an interesting substance; it is white when cool but turns yellow upon heating, returning to white as it cools again.
This color change is due to a phenomenon called thermal expansion which affects the crystal lattice structure temporarily.

• Different zinc compounds have distinct chemical properties.
• Zinc ions \( \text{Zn}^{2+} \) can be identified through precipitation reactions where zinc ions react to form insoluble compounds.
• In the presence of ammonium hydroxide \( \text{NH}_4\text{OH} \), for instance, zinc ions will precipitate as zinc sulfide \( \text{ZnS} \).

These reactions illustrate the subtle but significant behaviors of zinc compounds when reacting with different chemicals.

Carbon Dioxide Reactions

Carbon dioxide \( \text{CO}_2 \) is a colorless, noncombustible gas released in various chemical reactions. One classic test involves its reaction with lime water \( \text{Ca(OH)}_2 \). Initially, carbon dioxide reacts with lime water to form calcium carbonate \( \text{CaCO}_3 \), a white precipitate, which turns the solution milky.
However, as more carbon dioxide is bubbled through the solution, the calcium carbonate reacts with water and additional \( \text{CO}_2 \) to form calcium bicarbonate \( \text{Ca(HCO}_3\text{)}_2 \), which is soluble and clears the solution.

• This two-step reaction helps in identifying carbon dioxide in laboratory settings.
• The disappearance of milkiness is a confirmation of the presence of carbon dioxide as it indicates the formation of calcium bicarbonate.

Such reactions illustrate the importance of \( \text{CO}_2 \) in chemical processes and its role in ecological phenomena like the carbon cycle.

Precipitation Reactions

Precipitation reactions are key in identifying and isolating compounds in chemistry, especially when dealing with metal ions. For zinc ions, these reactions are quite telling.
When you add sodium hydroxide \( \text{NaOH} \) to a zinc solution, it first forms zinc hydroxide \( \text{Zn(OH)}_2 \), a white precipitate.
But, add more hydroxide, and the zinc hydroxide dissolves, forming a soluble zincate complex \( \text{Na}_2[\text{Zn(OH}_4)] \). This behavior helps in the detection and analysis of zinc in various solutions.

• Precipitation allows chemists to check for the presence of certain metal ions.
• Operations like filtration can help separate precipitates for further study or usage.
• The reversible nature of some precipitation reactions, like that of zinc hydroxide, can be key to understanding complex chemical behaviors.

Understanding precipitation reactions is crucial in inorganic chemistry for both academic and industrial applications.