Choose the correct reason for the statement:
Sulphuric acid is not used for the preparation of the original solution in the analysis of basic radicals because?
A. It is a reducing agent
B. It forms insoluble sulfate with certain basic radicals
C. It forms a soluble complex
D. It is oxidizing in nature

Hint:A salt generally consists of two parts basic radicals and acidic radicals, during a qualitative analysis of salts we prepare a mixture of the salts with solvents to determine the acidic and basic counterparts of a given salt. The basic radical has the positive charge and the acidic radical has the negative charge.

Complete step-by-step solution:We have studied that Sulphuric acid (${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$) is a strong acid that liberates ${{\text{H}}^ + }$and ${\text{SO}}_4^{2 - }$ ions in the solution mixture.
But when we use sulphuric acid for the preparation of the original solution to analyze the basic radical of salt, it forms an insoluble sulfate with certain basic radicals. E.g., \[{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{, B}}{{\text{a}}^{{\text{2 + }}}}{\text{, C}}{{\text{a}}^{{\text{2 + }}}}\], etc.
Also, a dilute solution of sulphuric acid tends to oxidize certain basic radicals through a single displacement reaction, by forming a metal sulfate and ${{\text{H}}_2}$ gas. E.g., it oxidizes Iron into iron sulfate
${\text{Fe(s) + }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}(aq) \to {{\text{H}}_2}(g) + {\text{ FeS}}{{\text{O}}_4}(aq)$
Therefore, after reading these two properties of sulphuric acid we can conclude that two options are correct for the given statement.

Hence the correct options are (D) and (B) i.e., it forms insoluble sulfates with certain basic radicals and it is oxidizing in nature.

Note:Sulphuric acid as we know can act as an oxidizing agent but it is a weak oxidizing agent and can only oxidize certain metal ions that are above Cu in the reactivity series (the table of standard reduction potential). E.g., Fe, Al, Zn, Mg, and Ni.