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

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.