Question

Group reagent for the precipitation of group II basic radicals for the qualitative analysis is:
A. $dil.\text{ }HCl\text{ }+\text{ }{{H}_{2}}S$
B. $N{{H}_{4}}C{{l}_{(s)}}\text{ }+\text{ }N{{H}_{4}}O{{H}_{(aq)}}\text{ }+\text{ }{{H}_{2}}S$
C. ${{\left( N{{H}_{4}} \right)}_{2}}C{{O}_{3}}$ solution
D. none of the above

Answer 1

Hint: So as to answer this question we need to have a basic idea about the group II basic radicals and their properties. Once we know it will be easier for us to know the group reagent for the precipitation of the group to basic radicals for the qualitative analysis.

Complete answer:
We should know that the main principle on which qualitative analysis is based on. It is focused on detecting ions in an aqua solution. The solution is treated with various reactions to test for reactions characteristics of certain ions, which may cause colour change, solid forming and other visible changes.
The group II basic radicals are $C{{u}^{2+}},C{{d}^{2+}},P{{b}^{2+}},H{{g}^{2+}}\left( II \right),B{{i}^{3+}},~A{{s}^{3+}},S{{b}^{3+}},S{{n}^{2+}}$.
Qualitative analysis is a part of chemistry that deals with figuring out the elements or ingredients of which a compound or mixture is made. We need to find a reaction where only the compounds formed by the group 2 cations precipitate while the compounds formed by other groups are soluble.
Chlorides of group 1 compounds are insoluble in water; all other chlorides of metals are usually soluble.
Sulphides of group 2 are insoluble in an acidic medium, thus we need to add compounds that would promote the formation of sulphides in an acidic medium.
To eliminate the possibility of any group 1 cations being present; add $HCl$ to the salt to form chlorides and separate the precipitate and the solution. The solution will have salts that do not have cations from group 1 since they have already precipitated out and separated.
Group reagent for the precipitation of group 2 basic radicals for the qualitative $dil.\text{ }HCl+{{H}_{2}}S$ in group II to sulphide, precipitated in acidic medium.

So, the correct answer is “Option A”.

Note: Know that the sulphides of group three group 3 cations are not soluble in basic medium, so add $N{{H}_{4}}OH$ to detect the presence of group 3 cations after separating the precipitate formed by the group 2 cations. Again, separate the precipitate by the group 3 cations and add ${{\left( N{{H}_{4}} \right)}_{2}}C{{O}_{3}}$ solution to form carbonates with group 4 cations that are insoluble in water. If you separate the solution from the precipitate again you will get a solution with group 5 cations.
Remember that qualitative analysis is a cyclic process that works on the process of elimination. Thus, we can separate the salts by precipitating them out one by one.