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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.

Last updated date: 13th Jun 2024
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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.