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Compound A reduced $HgC{{l}_{2}}$, to a white precipitate, turning to grey. Solution of A turns the yellow coloured solution of $FeC{{l}_{3}}$ to a green coloured solution. Solution of A gives white precipitate with NaOH which dissolves in excess of NaOH. Moreover, the solution of 'A' gives yellow precipitate, $FeC{{l}_{3}}$ when ${{H}_{2}}S$ gas is passed through it. This precipitate dissolved in yellow ammonium sulphide. Compound 'A' is found to give chromyl chloride test. Identify the compound 'A' and give the reaction involved.


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Last updated date: 13th Jun 2024
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Answer
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Hint: As it is given in the question that compound A reduces ferric chloride solution to a green solution, so compound A must be a good reducing agent.

Complete step-by step solution:
In order to answer our question, we need to notice each and every reaction properly. Now, let us start from the reaction of A and ${{H}_{2}}S$. It is given in the question that A reacts with ${{H}_{2}}S$ gas and a yellow precipitate is obtained. Moreover, this yellow precipitate is soluble in yellow ammonium sulphide. Let us try to write the reaction for it:
$A+{{H}_{2}}S\to yellow\,ppt$….(i)
Now, it is also said that A reduced the compound $HgC{{l}_{2}}$. The name of the test is called chromyl chloride test and it detects the presence of chlorine. Now, from (i), we can deduce that A is the cation of tin. A can be assumed as $S{{n}^{2+}}$. Also, as it passes the chromyl chloride test, so A must also contain chlorine. So, from all this information, we can say that the compound A is $SnC{{l}_{2}}$.Now, we have been also given that compound A turns $FeC{{l}_{3}}$ to a green coloured solution. As no more atoms are present, the green coloured solution has to be $FeC{{l}_{2}}$. So, the compound A acts as a reducing agent, in the reaction. So, we have confirmed A to be $SnC{{l}_{2}}$. Let us see the reactions that are involved throughout the question.
$\begin{align}
& A+HgC{{l}_{2}}\to white\,ppt \\
& SnC{{l}_{2}}+HgC{{l}_{2}}\to H{{g}_{2}}C{{l}_{2}} \\
\end{align}$
Here, the white precipitate is $H{{g}_{2}}C{{l}_{2}}$. Some amount of $SnC{{l}_{4}}$ is also formed.
$SnC{{l}_{2}}+FeC{{l}_{3}}\to FeC{{l}_{2}}$
Let us check the white precipitate that is formed by reacting with NaOH. So, we have
$SnC{{l}_{2}}+NaOH\to Sn{{(OH)}_{2}}$, which is a white ppt
$Sn{{(OH)}_{{}}}+NaOH\to N{{a}_{2}}Sn{{O}_{2}}$, this is a soluble compound
$SnC{{l}_{2}}+{{H}_{2}}S\to SnS$
It is a yellow ppt, HCl is also formed in the reaction. Finally, we have
$SnS+{{(N{{H}_{4}})}_{2}}S\to {{(N{{H}_{4}})}_{2}}Sn{{S}_{3}}$, which is again a soluble compound.

NOTE: In the final reaction, the compound ${{(N{{H}_{4}})}_{2}}S$, the name is ammonium sulphide and it is a yellow solid. $FeC{{l}_{3}}$ is a yellow coloured solution, whereas $FeC{{l}_{2}}$ is green coloured.