Question

A white powder A on heating gave a non-combustible gas and a white residue. The residue on heating turns yellow. The residue dissolves in dilute $HCl$ and the solution gives a white precipitate with ${{K}_{4}}[Fe{{(CN)}_{6}}]$. Compound A is:
(A) $CaC{{O}_{3}}$
(B) $ZnC{{O}_{3}}$
(C) $CaS{{O}_{3}}$
(D) $CuC{{O}_{3}}$

Answer 1

An attempt to this question can be made by taking one of the options and performing the above set of reactions and matching the observations. It is observed that zinc oxide on heating turns yellow and is white when cold. $Zn$ salts are known to precipitate as well. Based on this you can determine the carbonate salt which will have observations similar to the ones mentioned in the question.

Complete step by step answer:
Carbonate salts upon heating tend to release carbon dioxide in gaseous form and lead to the formation of metal oxide. So, the compound is a carbonate.
The next reaction is heating the metal oxide. It is observed that on heating the oxide turns yellow and returns back to white after cooling. This is a striking property of zinc oxide molecules.
We will now perform the next reaction given in question and verify if our assumption is correct.
${{K}_{4}}[Fe{{(CN)}_{6}}]$ reacts with Zinc(II) ions in aqueous solution. The reaction is mentioned below.
$Z{{n}^{2+}}\,+\,{{[Fe{{(CN)}_{6}}]}^{4-}}\,\to \,\,\,Z{{n}_{2}}{{[Fe{{(CN)}_{6}}]}_{ppt}}$
The product is a white precipitate. Thus, our assumption was right and the compound (A) is identified as $ZnC{{O}_{3}}$.

Therefore, the correct answer is option (B).

Note: It is important to know that zinc oxide turns yellow on heating as it loses oxygen reversibly from its crystal lattice.