Question

Compound (A) is a greenish crystalline salt which gives the following results when tested:

(a) Addition of \[BaC{l_2}\] solution to a solution of (A) results in the formation of a white precipitate (B) which is insoluble in dilute \[HCl\] .
(b) On heating, water vapor and two gases (C) and (D) are liberated, leaving a red-brown residue (E).
(c) E dissolves in \[conc.HCl\;\] to give a yellow solution (F).
(d) With \[{H_2}S\] , the solution (F) yields a yellow-white ppt. (G) which when filtered, leaves a greenish filtrate (H).
(e) The solution (F) gives a green solution with \[SnC{l_2}\] ​, and when reacted with \[{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\;\] gives a blue ppt.
Thus, gases (C) and (D) are
A.$S{O_2} $, $S{O_3} $
B.$S{O_2} $, $C{O_2} $
C.$N{O_2}, MgO$
D.$ZnO,S{O_3}$

Answer 1

Hint:
For the compound B, we have to remember that sulphate compounds do not react with dilute \[HCl\] .
Compound C and D have a characteristic red color meaning it contains iron. Usually \[{K_4} \left [ {Fe {{\left({CN} \right)} _6}} \right] \; \] gives a blue precipitate when it comes in contact with ions of Iron.

Complete step by step answer:
As mentioned, the first equation leads to the formation of an insoluble white precipitate. Therefore, we have to consider that the compound has a sulphate group since the white precipitate formed will be $BaS{O_4} $ . Consider the component that is combined with the sulphate group to be $x$ an unknown variable. Which will mean that the reaction will look like this,
$xS{O_4} + BaC{l_2} \to xC{l_2} + BaS{O_4} $ . We can conclude that the variable $x$ will be $Fe$.
 In part b, we find out that on heating with water vapor we get a residue that is red brown. This occurs in compounds like,$F{e_2}{O_3}$ . The reaction will be as follows
\[FeS{O_4} \to F{e_2} {O_3} + S{O_2} + S{O_3} \]
 Here we find out that the residue is the compound E that is mentioned and the gases liberated are \[S{O_2} \] or C and \[S{O_3} \] that is D.
The next part that is part C reaction will be as follows,
\[F{e_2} {O_3} + 6HCl \to 2FeC{l_3} + 3{H_2} O\]
Here the yellow solution is \[FeC{l_3} \] .
Part d uses reducing agent \[{H_2} S\] giving the reaction
\[FeCl{_3} + {H_2}S \to FeC{l_2} + HCl + S\]
Here the compound G is sulphur and the greenish filtrate is \[FeC{l_2} \] .
Part E says that reaction with \[SnC{l_2} \] leads to formation of a green solution. This means that the compound reacting also contains chlorine ions. This reaction can be written as follows:
\[FeC{l_3} + SnC{l_2} \to FeC{l_2} + SnC{l_4} \]
Here, $FeC{l_2} $ is green in color.
that is part e as it is mentioning the test that produces a Prussian blue color. You may recall this test from the qualitative analysis part of organic chemistry where in \[{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\;\] reacted with a compound to form a Prussian blue color. The reaction will be as follows:
\[FeC{l_3} + {K_4} {\left [ {Fe (CN)} \right] _6} \to F{e_4} {[Fe{(CN)_6}] _3} + KCl\]
Here the complex \[F{e_4} {[Fe{(CN)_6}] _3} \] is blue.
Therefore, from all these reactions we can say that the answer is option A.

Note: It is important to remember that $BaS{O_4} $ is a precipitate that will not react with hydrochloric acid.
-The red brown residue is $F{e_2} {O_3} $.
-The yellow solution formed is \[FeC{l_3} \] due to the chlorine ions.
-The greenish filtrate is \[FeC{l_2} \] as ferrous ions have a green color.